Systems of Linear Equations

This chapter establishes the foundational principles for solving systems of linear equations, a cornerstone of linear algebra. Mastery of Gaussian Elimination and determinants is critical for subsequent advanced topics and frequently assessed in examinations for its direct application in problem-solving.

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Chapter Contents

| Topic |

|---|-------| | 1 | Gaussian Elimination | | 2 | Determinants |

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We begin with Gaussian Elimination.

Part 1: Gaussian Elimination

Gaussian elimination is a fundamental algorithm for solving systems of linear equations by systematically transforming the system's augmented matrix into row echelon form. This process allows for straightforward determination of solution existence and uniqueness.

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Core Concepts

1. Systems of Linear Equations and Augmented Matrices

We define a system of linear equations as a set of linear equations involving the same variables. An augmented matrix represents such a system by combining the coefficient matrix and the constant terms.

📖 System of Linear Equations

A system of $m$ linear equations in $n$ variables $x_1, \dots, x_n$ is given by:

\begin{aligned}a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n & = b_1 \\a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n & = b_2 \\ & \vdots \\a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n & = b_m\end{aligned}

📖 Augmented Matrix

The augmented matrix for a system of linear equations $A\mathbf{x} = \mathbf{b}$ is formed by appending the column vector $\mathbf{b}$ to the coefficient matrix $A$ .

[A | \mathbf{b}] = \left[\begin{array}{cccc|c}a_{11} & a_{12} & \cdots & a_{1n} & b_1 \\a_{21} & a_{22} & \cdots & a_{2n} & b_2 \\\vdots & \vdots & \ddots & \vdots & \vdots \\a_{m1} & a_{m2} & \cdots & a_{mn} & b_m\end{array}\right]

Worked Example:

Consider the system of equations:

\begin{aligned}2x - 3y + z & = 10 \\x + y - 2z & = -3 \\3x - 2y + 4z & = 15\end{aligned}

Step 1: Identify coefficients and constants.

> The coefficients are $a_{ij}$ and constants are $b_i$ .

Step 2: Construct the augmented matrix.

\left[\begin{array}{ccc|c}> 2 & -3 & 1 & 10 \\> 1 & 1 & -2 & -3 \\> 3 & -2 & 4 & 15> \end{array}\right]

Answer: The augmented matrix is $\left[\begin{array}{ccc|c} 2 & -3 & 1 & 10 \\ 1 & 1 & -2 & -3 \\ 3 & -2 & 4 & 15 \end{array}\right]$ .

:::question type="MCQ" question="Which of the following is the correct augmented matrix for the system of equations:

\begin{aligned}x_1 - 2x_2 & = 5 \\3x_1 + x_2 & = -1 \\\end{aligned}

" options=["

\left[\begin{array}{cc|c} 1 & -2 & 5 \\ 3 & 1 & -1 \end{array}\right]

","

\left[\begin{array}{cc|c} 1 & 3 & 5 \\ -2 & 1 & -1 \end{array}\right]

","

\begin{bmatrix} 1 & -2 \\ 3 & 1 \end{bmatrix}

","

\begin{bmatrix} 5 \\ -1 \end{bmatrix}

"] answer="

\left[\begin{array}{cc|c} 1 & -2 & 5 \\ 3 & 1 & -1 \end{array}\right]

" hint="The augmented matrix combines coefficients and constants, separated by a vertical line." solution="The system is $x_1 - 2x_2 = 5$ and $3x_1 + x_2 = -1$ .
The coefficients of $x_1$ are 1 and 3.
The coefficients of $x_2$ are -2 and 1.
The constants are 5 and -1.
Thus, the augmented matrix is formed by placing the coefficients on the left of the bar and constants on the right.

\left[\begin{array}{cc|c}> 1 & -2 & 5 \\> 3 & 1 & -1> \end{array}\right]

This matches the first option."
:::

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2. Elementary Row Operations

Elementary row operations (EROs) are fundamental transformations applied to the rows of an augmented matrix that do not change the solution set of the corresponding system of linear equations.

📖 Elementary Row Operations

The three types of elementary row operations are:

Row Swap: Interchanging two rows ( $R_i \leftrightarrow R_j$ ).

Row Scaling: Multiplying a row by a non-zero scalar ( $kR_i \to R_i$ , where $k \neq 0$ ).

Row Addition: Adding a scalar multiple of one row to another row ( $R_i + kR_j \to R_i$ ).

Worked Example:

Consider the matrix:

A = \begin{bmatrix}1 & 2 & 3 \\4 & 5 & 6 \\7 & 8 & 9\end{bmatrix}

Step 1: Apply the ERO $R_2 \leftrightarrow R_3$ .

\begin{bmatrix}> 1 & 2 & 3 \\> 7 & 8 & 9 \\> 4 & 5 & 6> \end{bmatrix}

Step 2: Apply the ERO $2R_1 \to R_1$ to the new matrix.

\begin{bmatrix}> 2 & 4 & 6 \\> 7 & 8 & 9 \\> 4 & 5 & 6> \end{bmatrix}

Step 3: Apply the ERO $R_3 - 4R_1 \to R_3$ to the new matrix. (Using original $R_1$ for context, or the current $R_1$ ). Let's use the matrix from Step 2.

R_3 \to R_3 - 2R_1 \text{ (using the current } R_1 \text{ from Step 2)}

\begin{bmatrix}> 2 & 4 & 6 \\> 7 & 8 & 9 \\> 4 - 2(2) & 5 - 2(4) & 6 - 2(6)> \end{bmatrix} = \begin{bmatrix}> 2 & 4 & 6 \\> 7 & 8 & 9 \\> 0 & -3 & -6> \end{bmatrix}

Answer: The resulting matrix after the operations is $\begin{bmatrix} 2 & 4 & 6 \\ 7 & 8 & 9 \\ 0 & -3 & -6 \end{bmatrix}$ .

:::question type="MCQ" question="Given the matrix $M = \begin{bmatrix} 1 & 0 & 3 \\ 2 & 4 & 1 \\ 0 & 1 & 5 \end{bmatrix}$ , which sequence of elementary row operations transforms $M$ into a matrix where the element in the $(2,1)$ position is 0?" options=[" $R_2 \to R_2 - 2R_1$ "," $R_1 \leftrightarrow R_2$ "," $R_2 \to 2R_2$ ","None of the above"] answer=" $R_2 \to R_2 - 2R_1$ " hint="The element in the $(2,1)$ position is 2. To make it 0, subtract a multiple of the first row from the second row." solution="The element in the $(2,1)$ position is 2.
We want to perform an operation to make this element 0.
Option A: $R_2 \to R_2 - 2R_1$ .
The new second row will be $(2 - 2(1), 4 - 2(0), 1 - 2(3)) = (0, 4, -5)$ .
The element in the $(2,1)$ position becomes 0.
Option B: $R_1 \leftrightarrow R_2$ . This swaps the rows, placing 2 in $(1,1)$ and 1 in $(2,1)$ , so the $(2,1)$ element is still 1, not 0.
Option C: $R_2 \to 2R_2$ . This scales the second row, making the $(2,1)$ element $2 \times 2 = 4$ , not 0.
Therefore, $R_2 \to R_2 - 2R_1$ is the correct operation.

\begin{bmatrix}> 1 & 0 & 3 \\> 2 & 4 & 1 \\> 0 & 1 & 5> \end{bmatrix} \xrightarrow{R_2 \to R_2 - 2R_1} \begin{bmatrix}> 1 & 0 & 3 \\> 0 & 4 & -5 \\> 0 & 1 & 5> \end{bmatrix}

"
:::

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3. Row Echelon Form (REF)

Gaussian elimination aims to transform an augmented matrix into Row Echelon Form (REF) using elementary row operations.

📖 Row Echelon Form (REF)

A matrix is in row echelon form if it satisfies the following conditions:

All non-zero rows are above any rows of all zeros.

The leading entry (pivot) of each non-zero row is to the right of the leading entry of the row above it.

All entries in a column below a leading entry are zero.

Worked Example:

Transform the following matrix into Row Echelon Form:

A = \begin{bmatrix}0 & 0 & -2 & 0 & 7 & 12 \\2 & 4 & -10 & 6 & 12 & 28 \\2 & 4 & -5 & 6 & -5 & -1\end{bmatrix}

Step 1: Swap $R_1$ and $R_2$ to get a non-zero leading entry in $R_1$ .

A \xrightarrow{R_1 \leftrightarrow R_2} \begin{bmatrix}> 2 & 4 & -10 & 6 & 12 & 28 \\> 0 & 0 & -2 & 0 & 7 & 12 \\> 2 & 4 & -5 & 6 & -5 & -1> \end{bmatrix}

Step 2: Make the leading entry of $R_1$ equal to 1 (optional, but good practice). $R_1 \to \frac{1}{2}R_1$ .

\xrightarrow{R_1 \to \frac{1}{2}R_1} \begin{bmatrix}> 1 & 2 & -5 & 3 & 6 & 14 \\> 0 & 0 & -2 & 0 & 7 & 12 \\> 2 & 4 & -5 & 6 & -5 & -1> \end{bmatrix}

Step 3: Eliminate the entry below the leading 1 in $R_1$ . $R_3 \to R_3 - 2R_1$ .

\xrightarrow{R_3 \to R_3 - 2R_1} \begin{bmatrix}> 1 & 2 & -5 & 3 & 6 & 14 \\> 0 & 0 & -2 & 0 & 7 & 12 \\> 0 & 0 & 5 & 0 & -17 & -29> \end{bmatrix}

Step 4: Make the leading entry of $R_2$ equal to 1. $R_2 \to -\frac{1}{2}R_2$ .

\xrightarrow{R_2 \to -\frac{1}{2}R_2} \begin{bmatrix}> 1 & 2 & -5 & 3 & 6 & 14 \\> 0 & 0 & 1 & 0 & -7/2 & -6 \\> 0 & 0 & 5 & 0 & -17 & -29> \end{bmatrix}

Step 5: Eliminate the entry below the leading 1 in $R_2$ . $R_3 \to R_3 - 5R_2$ .

\xrightarrow{R_3 \to R_3 - 5R_2} \begin{bmatrix}> 1 & 2 & -5 & 3 & 6 & 14 \\> 0 & 0 & 1 & 0 & -7/2 & -6 \\> 0 & 0 & 0 & 0 & 1/2 & 1> \end{bmatrix}

Step 6: Make the leading entry of $R_3$ equal to 1. $R_3 \to 2R_3$ .

\xrightarrow{R_3 \to 2R_3} \begin{bmatrix}> 1 & 2 & -5 & 3 & 6 & 14 \\> 0 & 0 & 1 & 0 & -7/2 & -6 \\> 0 & 0 & 0 & 0 & 1 & 2> \end{bmatrix}

Answer: The matrix in Row Echelon Form is $\begin{bmatrix} 1 & 2 & -5 & 3 & 6 & 14 \\ 0 & 0 & 1 & 0 & -7/2 & -6 \\ 0 & 0 & 0 & 0 & 1 & 2 \end{bmatrix}$ .

:::question type="MCQ" question="Which of the following matrices is in Row Echelon Form (REF)?" options=["

\begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}

","

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

","

\begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}

","

\begin{bmatrix} 0 & 1 & 2 \\ 1 & 0 & 3 \\ 0 & 0 & 0 \end{bmatrix}

"] answer="

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

" hint="Check the conditions for REF: non-zero rows above zero rows, leading entries move right, zeros below leading entries." solution="Let's check each option against the REF conditions:

\begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}

The leading entry of $R_3$ (1 in column 2) is to the left of the leading entry of $R_2$ (1 in column 3). This violates condition 2.

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

All conditions are met. Leading entries are 1s, they move to the right, and all entries below leading entries are zero. This is also in RREF.

\begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}

The row of all zeros ( $R_2$ ) is not at the bottom. This violates condition 1.

\begin{bmatrix} 0 & 1 & 2 \\ 1 & 0 & 3 \\ 0 & 0 & 0 \end{bmatrix}

The leading entry of $R_2$ (1 in column 1) is to the left of the leading entry of $R_1$ (1 in column 2). This violates condition 2.

Therefore, only the second option is in Row Echelon Form."
:::

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4. Reduced Row Echelon Form (RREF)

Gauss-Jordan elimination further transforms the matrix from REF into Reduced Row Echelon Form (RREF). This form simplifies back-substitution.

📖 Reduced Row Echelon Form (RREF)

A matrix is in reduced row echelon form if it satisfies the conditions for REF, plus:

Each leading entry is 1.

Each column containing a leading entry has zeros everywhere else.

Worked Example:

Transform the following matrix into Reduced Row Echelon Form:

A = \begin{bmatrix}1 & 2 & -5 & 3 & 6 & 14 \\0 & 0 & 1 & 0 & -7/2 & -6 \\0 & 0 & 0 & 0 & 1 & 2\end{bmatrix}

(This is the REF matrix from the previous example.)

Step 1: Use $R_3$ to create zeros above its leading 1 (in column 5).
$R_2 \to R_2 + \frac{7}{2}R_3$ and $R_1 \to R_1 - 6R_3$ .

\xrightarrow{R_2 \to R_2 + \frac{7}{2}R_3} \begin{bmatrix}> 1 & 2 & -5 & 3 & 6 & 14 \\> 0 & 0 & 1 & 0 & 0 & 1 \\> 0 & 0 & 0 & 0 & 1 & 2> \end{bmatrix}

\xrightarrow{R_1 \to R_1 - 6R_3} \begin{bmatrix}> 1 & 2 & -5 & 3 & 0 & 2 \\> 0 & 0 & 1 & 0 & 0 & 1 \\> 0 & 0 & 0 & 0 & 1 & 2> \end{bmatrix}

Step 2: Use $R_2$ to create zeros above its leading 1 (in column 3).
$R_1 \to R_1 + 5R_2$ .

\xrightarrow{R_1 \to R_1 + 5R_2} \begin{bmatrix}> 1 & 2 & 0 & 3 & 0 & 7 \\> 0 & 0 & 1 & 0 & 0 & 1 \\> 0 & 0 & 0 & 0 & 1 & 2> \end{bmatrix}

Answer: The matrix in Reduced Row Echelon Form is $\begin{bmatrix} 1 & 2 & 0 & 3 & 0 & 7 \\ 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 2 \end{bmatrix}$ .

:::question type="MCQ" question="Which of the following matrices is in Reduced Row Echelon Form (RREF)?" options=["

\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}

","

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}

","

\begin{bmatrix} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}

","

\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

"] answer="

\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}

" hint="Check all RREF conditions, especially that columns with leading entries have zeros elsewhere." solution="Let's check each option against the RREF conditions:

\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}

This matrix satisfies all REF conditions. The leading entries are 1s. Column 1 (pivot for $R_1$ ) has zeros below it. Column 2 (pivot for $R_2$ ) has zeros above and below it. This matrix is in RREF.

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}

The leading entry in $R_2$ is 1 in column 2. The entry above it in column 2 is 0. However, the element $(2,3)$ is 1, not 0. This is in REF, but not RREF because the column with a leading entry (column 2) does not have zeros everywhere else (it's fine, the 1 in (2,3) is not in a pivot column). But it's not RREF because the leading entry in $R_2$ is not the only non-zero entry in its column and the leading entry in $R_1$ (at (1,1)) is not the only non-zero entry in its column above it. The definition states 'Each column containing a leading entry has zeros everywhere else'. Column 2 contains a leading entry (1 at (2,2)), but the entry (1,2) is 0, which is correct. The matrix is in RREF. My initial thought process was incorrect for this option. Let's re-evaluate.
Okay, let's re-evaluate option 2 carefully.

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}

- Condition 1 (REF): All non-zero rows are above zero rows. (Met)
- Condition 2 (REF): Leading entry of each non-zero row is to the right of the leading entry of the row above it. (Met: (1,1) then (2,2))
- Condition 3 (REF): All entries in a column below a leading entry are zero. (Met)
- Condition 4 (RREF): Each leading entry is 1. (Met)
- Condition 5 (RREF): Each column containing a leading entry has zeros everywhere else.
- Column 1 (leading entry 1 at (1,1)): All other entries are 0. (Met)
- Column 2 (leading entry 1 at (2,2)): All other entries are 0. (Met)
So option 2 is in RREF. This implies there might be multiple correct answers or I misjudged option 1.

Let's re-check option 1:

\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}

- Condition 1 (REF): Met.
- Condition 2 (REF): Met.
- Condition 3 (REF): Met.
- Condition 4 (RREF): Met.
- Condition 5 (RREF):
- Column 1 (leading entry 1 at (1,1)): All other entries are 0. (Met)
- Column 2 (leading entry 1 at (2,2)): All other entries are 0. (Met)
So option 1 is also in RREF.

This means the question is flawed if it's MCQ asking for the RREF matrix. It should be MSQ. I need to make sure my questions have only one correct answer for MCQ. Let's adjust the options.

Let's make option 2 NOT RREF. Change $(2,3)$ to something that violates RREF.
Original Options:
1.

\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}

(RREF)
2.

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}

(RREF)
3.

\begin{bmatrix} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}

(REF, but not RREF because (1,2) is not 0, and column 2 is not a pivot column, so it's fine. But (1,2) is non-zero and not a pivot. This is RREF. Let's re-examine RREF rules. Ah, 'Each column containing a leading entry has zeros everywhere else.' Column 1 has a leading entry, (1,1), and is all zeros elsewhere. Column 3 has a leading entry (2,3), and is all zeros elsewhere. Column 2 does not have a leading entry. So this is RREF too!)

My understanding of RREF for question generation might be slightly off. Let's be very precise:
1. All non-zero rows are above any rows of all zeros.
2. The leading entry (pivot) of each non-zero row is to the right of the leading entry of the row above it.
3. All entries in a column below a leading entry are zero.
4. Each leading entry is 1.
5. Each column containing a leading entry has zeros everywhere else (above and below).

Let's re-check based on this strict definition:
1.

\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}

- Pivots: (1,1), (2,2). Both are 1.
- Column 1 (pivot): (1,1) is 1, (2,1)=0, (3,1)=0. OK.
- Column 2 (pivot): (2,2) is 1, (1,2)=0, (3,2)=0. OK.
- All other REF conditions met. This IS RREF.

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}

- Pivots: (1,1), (2,2). Both are 1.
- Column 1 (pivot): (1,1) is 1, (2,1)=0, (3,1)=0. OK.
- Column 2 (pivot): (2,2) is 1, (1,2)=0, (3,2)=0. OK.
- All other REF conditions met. This IS RREF.

\begin{bmatrix} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}

- Pivots: (1,1), (2,3). Both are 1.
- Column 1 (pivot): (1,1) is 1, (2,1)=0, (3,1)=0. OK.
- Column 3 (pivot): (2,3) is 1, (1,3)=0, (3,3)=0. OK.
- All other REF conditions met. This IS RREF.

This is problematic. I cannot have three RREF options for an MCQ. I need to make them clearly NOT RREF.

New options for RREF question:
A)

\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

(Not RREF: (1,2) should be 0)
B)

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

(RREF - Identity matrix)
C)

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}

(RREF - Pivots (1,1) and (2,3). Column 1 has zeros elsewhere. Column 3 has zeros elsewhere. This is RREF. Still two RREF options.)

Okay, I will ensure only one option is RREF.
Let's use a simpler example for the question.

Revised Question for RREF:
:::question type="MCQ" question="Which of the following matrices is in Reduced Row Echelon Form (RREF)?" options=["

\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}

","

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

","

\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}

","

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}

"] answer="

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

" hint="Check all RREF conditions, especially that columns with leading entries have zeros everywhere else." solution="Let's check each option against the RREF conditions:
1.

\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}

This matrix is in REF, but not RREF because the leading entry in $R_2$ (at (2,2)) has a non-zero entry above it (2 at (1,2)). Also, the leading entry in $R_3$ (at (3,3)) has non-zero entries above it (3 at (1,3) and 4 at (2,3)).
2.

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

This matrix satisfies all RREF conditions. Each leading entry is 1, and each column containing a leading entry has zeros everywhere else.
3.

\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}

This matrix is in RREF. My previous analysis was correct for this one. This means I need to make one of the other options clearly not RREF.
4.

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}

This matrix is in RREF.

This is still problematic for an MCQ. I must select one answer. I will stick to the simplest RREF for the answer, and make the others clearly not.

Let's make option 1 and 3 not RREF.
Option 1: Already not RREF.
Option 3: Make it not RREF.
Option 4: Make it not RREF.

Final attempt for RREF question options:
A)

\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

(Not RREF: (1,2) should be 0)
B)

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

(RREF - Identity matrix)
C)

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix}

(Not RREF: (2,3) is non-zero, but column 3 is not a pivot column. This is RREF. The condition is 'Each column containing a leading entry has zeros everywhere else'. Column 2 has a leading entry (2,2). It's fine for (2,3) to be non-zero. So this is RREF.)

Okay, I am overthinking the RREF question generation. I will pick one RREF matrix, and for the other options, I will introduce clear violations like non-zero entries above a pivot, or pivots not being 1, or rows of zeros not at the bottom. The identity matrix is always RREF and is a safe bet for the answer.

Let's use the current question and make sure ONLY one is RREF.
The current options are:
1.

\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}

(REF, NOT RREF) - Good.
2.

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

(RREF) - This will be the answer.
3.

\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}

(RREF) - Problematic. I need to change this. Let's make it REF but not RREF.
Change to:

\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}

(This is RREF too. Let's make it not REF. Swap rows 2 and 3:

\begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}

Not REF. Or make it so a pivot column has non-zero above.

\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}

This is REF, not RREF (1,2) is non-zero).
4.

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}

(RREF) - Problematic.

Okay, I will make the options extremely clear. Only one RREF.

Revised RREF question:
:::question type="MCQ" question="Which of the following matrices is in Reduced Row Echelon Form (RREF)?" options=["

\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

","

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

","

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}

","

\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}

"] answer="

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

" hint="A matrix in RREF must have leading 1s, zeros everywhere else in pivot columns, and pivots moving right with zero rows at the bottom." solution="Let's check each option against the RREF conditions:
1.

\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

This matrix is not in RREF because the leading entry in $R_2$ (at (2,2)) has a non-zero entry above it (2 at (1,2)). For RREF, all entries in a pivot column must be zero except for the pivot itself.
2.

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

This matrix satisfies all RREF conditions. Each leading entry is 1, and each column containing a leading entry has zeros everywhere else. This is the identity matrix, which is always in RREF.
3.

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}

This matrix is not in RREF because the row of all zeros is not at the bottom (it doesn't have a row of all zeros, but the leading entry of $R_3$ (1 at (3,2)) is to the left of the leading entry of $R_2$ (1 at (2,3)). This violates the pivot-right condition for REF.
4.

\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}

This matrix is not in RREF because the leading entry of $R_2$ (1 at (2,1)) is to the left of the leading entry of $R_1$ (1 at (1,2)). This violates the pivot-right condition for REF.

Therefore, only the second option is in Reduced Row Echelon Form."
:::

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5. Solving Systems using Gaussian Elimination (Unique Solution)

Once an augmented matrix is in REF or RREF, the system can be solved using back-substitution (from REF) or direct reading (from RREF). A unique solution exists if there is a pivot position in every column corresponding to a variable.

Worked Example:

Solve the system of equations using Gaussian Elimination:

\begin{aligned}x + y + z & = 6 \\-x + 2y + 3z & = 14 \\2x - y + z & = 3\end{aligned}

Step 1: Write the augmented matrix.

\left[\begin{array}{ccc|c}> 1 & 1 & 1 & 6 \\> -1 & 2 & 3 & 14 \\> 2 & -1 & 1 & 3> \end{array}\right]

Step 2: Apply EROs to get to REF.
$R_2 \to R_2 + R_1$
$R_3 \to R_3 - 2R_1$

\left[\begin{array}{ccc|c}> 1 & 1 & 1 & 6 \\> 0 & 3 & 4 & 20 \\> 0 & -3 & -1 & -9> \end{array}\right]

Step 3: Continue to REF.
$R_3 \to R_3 + R_2$

\left[\begin{array}{ccc|c}> 1 & 1 & 1 & 6 \\> 0 & 3 & 4 & 20 \\> 0 & 0 & 3 & 11> \end{array}\right]

Step 4: Perform back-substitution.
From $R_3$ : $3z = 11 \implies z = \frac{11}{3}$ .

From $R_2$ : $3y + 4z = 20 \implies 3y + 4(\frac{11}{3}) = 20 \implies 3y + \frac{44}{3} = 20 \implies 3y = 20 - \frac{44}{3} = \frac{60-44}{3} = \frac{16}{3} \implies y = \frac{16}{9}$ .

From $R_1$ : $x + y + z = 6 \implies x + \frac{16}{9} + \frac{11}{3} = 6 \implies x + \frac{16}{9} + \frac{33}{9} = 6 \implies x + \frac{49}{9} = 6 \implies x = 6 - \frac{49}{9} = \frac{54-49}{9} = \frac{5}{9}$ .

Answer: The unique solution is $x = \frac{5}{9}, y = \frac{16}{9}, z = \frac{11}{3}$ .

:::question type="NAT" question="Solve the following system of equations for $x$ :

\begin{aligned}x - 2y + z & = 0 \\2y - 8z & = 8 \\-4x + 5y + 9z & = -9\end{aligned}

" answer="1" hint="Convert to an augmented matrix and use Gaussian elimination to find $x, y, z$ . Then provide the value of $x$ ." solution="Step 1: Write the augmented matrix.

\left[\begin{array}{ccc|c}> 1 & -2 & 1 & 0 \\> 0 & 2 & -8 & 8 \\> -4 & 5 & 9 & -9> \end{array}\right]

Step 2: Apply EROs to get to REF.
$R_3 \to R_3 + 4R_1$

\left[\begin{array}{ccc|c}> 1 & -2 & 1 & 0 \\> 0 & 2 & -8 & 8 \\> 0 & -3 & 13 & -9> \end{array}\right]

Step 3: Continue to REF.
$R_2 \to \frac{1}{2}R_2$

\left[\begin{array}{ccc|c}> 1 & -2 & 1 & 0 \\> 0 & 1 & -4 & 4 \\> 0 & -3 & 13 & -9> \end{array}\right]

Step 4: Continue to REF.
$R_3 \to R_3 + 3R_2$

\left[\begin{array}{ccc|c}> 1 & -2 & 1 & 0 \\> 0 & 1 & -4 & 4 \\> 0 & 0 & 1 & 3> \end{array}\right]

Step 5: Perform back-substitution.
From $R_3$ : $z = 3$ .

From $R_2$ : $y - 4z = 4 \implies y - 4(3) = 4 \implies y - 12 = 4 \implies y = 16$ .

From $R_1$ : $x - 2y + z = 0 \implies x - 2(16) + 3 = 0 \implies x - 32 + 3 = 0 \implies x - 29 = 0 \implies x = 29$ .

Wait, there was a mistake in my manual calculation for $x$ .
$x - 2(16) + 3 = 0 \implies x - 32 + 3 = 0 \implies x - 29 = 0 \implies x = 29$ .
Let's re-check the question's answer. The question's answer is 1. My calculation is 29.
Let's trace again.
Original problem:
$x - 2y + z = 0$
$2y - 8z = 8$
$-4x + 5y + 9z = -9$

Augmented matrix:

\left[\begin{array}{ccc|c}1 & -2 & 1 & 0 \\0 & 2 & -8 & 8 \\-4 & 5 & 9 & -9\end{array}\right]

$R_3 \to R_3 + 4R_1$ :

\left[\begin{array}{ccc|c}1 & -2 & 1 & 0 \\0 & 2 & -8 & 8 \\0 & -3 & 13 & -9\end{array}\right]

$R_2 \to \frac{1}{2}R_2$ :

\left[\begin{array}{ccc|c}1 & -2 & 1 & 0 \\0 & 1 & -4 & 4 \\0 & -3 & 13 & -9\end{array}\right]

$R_3 \to R_3 + 3R_2$ :
$R_3 = (0, -3, 13, -9) + 3(0, 1, -4, 4) = (0, -3+3, 13-12, -9+12) = (0, 0, 1, 3)$ .
This is correct.

Matrix in REF:

\left[\begin{array}{ccc|c}1 & -2 & 1 & 0 \\0 & 1 & -4 & 4 \\0 & 0 & 1 & 3\end{array}\right]

Back substitution:
$z=3$ . (Correct)
$y - 4z = 4 \implies y - 4(3) = 4 \implies y - 12 = 4 \implies y = 16$ . (Correct)
$x - 2y + z = 0 \implies x - 2(16) + 3 = 0 \implies x - 32 + 3 = 0 \implies x - 29 = 0 \implies x = 29$ . (Correct)

The provided `answer="1"` for the question is incorrect based on my derivation. I will update the answer to `29`. It's crucial for the answer and solution to match.

Final check of the question and my solution for $x$ :
Question asks for $x$ . My solution gives $x=29$ . The answer should be $29$ .

Answer: 29"
:::

---

6. Inconsistent Systems (No Solution)

A system of linear equations has no solution if, during Gaussian elimination, a row in the augmented matrix transforms into a form where the coefficient part is all zeros, but the constant term is non-zero. This corresponds to an equation like $0 = k$ , where $k \neq 0$ .

Worked Example:

Determine if the following system has a solution:

\begin{aligned}x + y + 2z & = 9 \\2x + 4y - 3z & = 1 \\3x + 6y - 5z & = 0\end{aligned}

Step 1: Write the augmented matrix.

\left[\begin{array}{ccc|c}> 1 & 1 & 2 & 9 \\> 2 & 4 & -3 & 1 \\> 3 & 6 & -5 & 0> \end{array}\right]

Step 2: Apply EROs to get to REF.
$R_2 \to R_2 - 2R_1$
$R_3 \to R_3 - 3R_1$

\left[\begin{array}{ccc|c}> 1 & 1 & 2 & 9 \\> 0 & 2 & -7 & -17 \\> 0 & 3 & -11 & -27> \end{array}\right]

Step 3: Continue to REF.
$R_2 \to \frac{1}{2}R_2$

\left[\begin{array}{ccc|c}> 1 & 1 & 2 & 9 \\> 0 & 1 & -7/2 & -17/2 \\> 0 & 3 & -11 & -27> \end{array}\right]

Step 4: Continue to REF.
$R_3 \to R_3 - 3R_2$

\left[\begin{array}{ccc|c}> 1 & 1 & 2 & 9 \\> 0 & 1 & -7/2 & -17/2 \\> 0 & 0 & -1/2 & -3/2> \end{array}\right]

The last row corresponds to the equation $-\frac{1}{2}z = -\frac{3}{2}$ , which has a solution $z=3$ . This system is not inconsistent. My example needs to be truly inconsistent. Let me pick a different example or adjust this one.

Let's modify the example to be inconsistent:
Consider the system:

\begin{aligned}x + y + z & = 1 \\2x + 2y + 2z & = 3 \\3x + 3y + 3z & = 4\end{aligned}

Step 1: Write the augmented matrix.

\left[\begin{array}{ccc|c}> 1 & 1 & 1 & 1 \\> 2 & 2 & 2 & 3 \\> 3 & 3 & 3 & 4> \end{array}\right]

Step 2: Apply EROs to get to REF.
$R_2 \to R_2 - 2R_1$
$R_3 \to R_3 - 3R_1$

\left[\begin{array}{ccc|c}> 1 & 1 & 1 & 1 \\> 0 & 0 & 0 & 1 \\> 0 & 0 & 0 & 1> \end{array}\right]

Step 3: Observe the result.

> The second row corresponds to the equation $0x + 0y + 0z = 1$ , or $0 = 1$ . This is a contradiction.

Answer: The system is inconsistent and has no solution.

:::question type="MCQ" question="Which of the following augmented matrices in Row Echelon Form indicates an inconsistent system of linear equations?" options=["

\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 2 \end{array}\right]

","

\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right]

","

\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 1 \end{array}\right]

","

\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]

"] answer="

\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 1 \end{array}\right]

" hint="An inconsistent system will have a row representing $0 = k$ where $k \neq 0$ ." solution="Let's analyze each option:

\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 2 \end{array}\right]

This matrix represents a system with a unique solution ( $x+2y+3z=4, y+2z=3, z=2$ ).

\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right]

This matrix represents a system with infinitely many solutions (one free variable, $z$ ).

\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 1 \end{array}\right]

The last row of this matrix represents the equation $0x + 0y + 0z = 1$ , or $0 = 1$ . This is a contradiction, indicating that the system has no solution.

\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]

This matrix represents a system with infinitely many solutions (two free variables, $y$ and $z$ ).

Therefore, the third option indicates an inconsistent system."
:::

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7. Consistent Dependent Systems (Infinite Solutions)

A system has infinitely many solutions if, after reducing the augmented matrix to REF, there are fewer pivot positions (leading 1s) than variables, and no row represents a contradiction ( $0=k, k \neq 0$ ). The variables corresponding to columns without pivots are called free variables.

Worked Example:

Solve the system of equations:

\begin{aligned}x + 2y - z & = 2 \\2x + 4y + z & = 4 \\-x - 2y + 2z & = -2\end{aligned}

Step 1: Write the augmented matrix.

\left[\begin{array}{ccc|c}> 1 & 2 & -1 & 2 \\> 2 & 4 & 1 & 4 \\> -1 & -2 & 2 & -2> \end{array}\right]

Step 2: Apply EROs to get to REF.
$R_2 \to R_2 - 2R_1$
$R_3 \to R_3 + R_1$

\left[\begin{array}{ccc|c}> 1 & 2 & -1 & 2 \\> 0 & 0 & 3 & 0 \\> 0 & 0 & 1 & 0> \end{array}\right]

Step 3: Continue to REF.
$R_2 \to \frac{1}{3}R_2$

\left[\begin{array}{ccc|c}> 1 & 2 & -1 & 2 \\> 0 & 0 & 1 & 0 \\> 0 & 0 & 1 & 0> \end{array}\right]

Step 4: Continue to REF.
$R_3 \to R_3 - R_2$

\left[\begin{array}{ccc|c}> 1 & 2 & -1 & 2 \\> 0 & 0 & 1 & 0 \\> 0 & 0 & 0 & 0> \end{array}\right]

Step 5: Convert to RREF (optional, but makes expressing solutions easier).
$R_1 \to R_1 + R_2$

\left[\begin{array}{ccc|c}> 1 & 2 & 0 & 2 \\> 0 & 0 & 1 & 0 \\> 0 & 0 & 0 & 0> \end{array}\right]

Step 6: Express the solution.
From $R_2$ : $z = 0$ .
From $R_1$ : $x + 2y = 2 \implies x = 2 - 2y$ .
Here, $y$ is a free variable. Let $y = t$ , where $t \in \mathbb{R}$ .

Answer: The solution set is $x = 2 - 2t, y = t, z = 0$ for any real number $t$ .

:::question type="MCQ" question="Consider the system of equations represented by the following augmented matrix in RREF:

\left[\begin{array}{cccc|c}1 & -1 & 0 & 2 & 5 \\0 & 0 & 1 & 3 & 2 \\0 & 0 & 0 & 0 & 0\end{array}\right]

Which of the following describes the solution set?" options=["Unique solution","No solution","Infinitely many solutions with one free variable","Infinitely many solutions with two free variables"] answer="Infinitely many solutions with two free variables" hint="Count the number of variables and the number of pivot positions." solution="The augmented matrix is in RREF:

\left[\begin{array}{cccc|c}1 & -1 & 0 & 2 & 5 \\0 & 0 & 1 & 3 & 2 \\0 & 0 & 0 & 0 & 0\end{array}\right]

There are 4 variables (

x_1, x_2, x_3, x_4

) and 2 pivot positions (in columns 1 and 3, corresponding to

x_1

and

x_3

).
The number of free variables is the number of variables minus the number of pivot positions:

4 - 2 = 2

.
Since there are free variables and no inconsistent rows (

0=k, k \neq 0

), the system has infinitely many solutions.
The free variables are

x_2

and

x_4

.
From

R_2

x_3 + 3x_4 = 2 \implies x_3 = 2 - 3x_4

.
From

R_1

x_1 - x_2 + 2x_4 = 5 \implies x_1 = 5 + x_2 - 2x_4

.
Let

x_2 = s

and

x_4 = t

.
Then

x_1 = 5 + s - 2t

x_2 = s

x_3 = 2 - 3t

x_4 = t

.
This confirms infinitely many solutions with two free variables.

Thus, the system has infinitely many solutions with two free variables."
:::

---

8. Homogeneous Systems

A homogeneous system of linear equations is one where all the constant terms are zero. Such systems always have at least one solution, the trivial solution where all variables are zero. Non-trivial solutions exist if there are free variables.

📖 Homogeneous System

A system of linear equations $A\mathbf{x} = \mathbf{0}$ is homogeneous if the constant vector $\mathbf{b}$ is the zero vector.

\begin{aligned}a_{11}x_1 + \cdots + a_{1n}x_n & = 0 \\ & \vdots \\a_{m1}x_1 + \cdots + a_{mn}x_n & = 0\end{aligned}

Worked Example:

Solve the homogeneous system:

\begin{aligned}x_1 - 2x_2 + 3x_3 & = 0 \\-2x_1 + 4x_2 - 6x_3 & = 0 \\3x_1 - 6x_2 + 9x_3 & = 0\end{aligned}

Step 1: Write the augmented matrix.

\left[\begin{array}{ccc|c}> 1 & -2 & 3 & 0 \\> -2 & 4 & -6 & 0 \\> 3 & -6 & 9 & 0> \end{array}\right]

Step 2: Apply EROs to get to REF.
$R_2 \to R_2 + 2R_1$
$R_3 \to R_3 - 3R_1$

\left[\begin{array}{ccc|c}> 1 & -2 & 3 & 0 \\> 0 & 0 & 0 & 0 \\> 0 & 0 & 0 & 0> \end{array}\right]

Step 3: Express the solution.
The matrix is in RREF. There is one pivot variable ( $x_1$ ) and two free variables ( $x_2, x_3$ ).
From $R_1$ : $x_1 - 2x_2 + 3x_3 = 0 \implies x_1 = 2x_2 - 3x_3$ .
Let $x_2 = s$ and $x_3 = t$ , where $s, t \in \mathbb{R}$ .

Answer: The solution set is $x_1 = 2s - 3t, x_2 = s, x_3 = t$ for any real numbers $s, t$ . This system has infinitely many non-trivial solutions in addition to the trivial solution $(0,0,0)$ .

:::question type="MSQ" question="Which of the following statements are true about a homogeneous system of linear equations $A\mathbf{x} = \mathbf{0}$ ?" options=["It always has at least one solution.","It can have no solution.","It has infinitely many solutions if there are free variables.","If $\operatorname{det}(A) \neq 0$ , it has only the trivial solution."] answer="It always has at least one solution.,It has infinitely many solutions if there are free variables.,If $\operatorname{det}(A) \neq 0$ , it has only the trivial solution." hint="Recall the definition of a homogeneous system and how the number of solutions is determined by pivots and rank." solution="Let's analyze each statement:

It always has at least one solution. True. The trivial solution

\mathbf{x} = \mathbf{0}

(where all variables are zero) always satisfies

A\mathbf{0} = \mathbf{0}

It can have no solution. False. As established in statement 1, the trivial solution always exists. Homogeneous systems are always consistent.

It has infinitely many solutions if there are free variables. True. If, after row reduction, there are columns without pivot positions (free variables), then these variables can take on any real value, leading to infinitely many solutions. This includes the trivial solution as a specific case.

If $\operatorname{det}(A) \neq 0$ , it has only the trivial solution. True. If

\operatorname{det}(A) \neq 0

, then the matrix

A

is invertible. Multiplying

A\mathbf{x} = \mathbf{0}

A^{-1}

gives

A^{-1}A\mathbf{x} = A^{-1}\mathbf{0} \implies I\mathbf{x} = \mathbf{0} \implies \mathbf{x} = \mathbf{0}

. This means the only solution is the trivial one. This also implies that the RREF of

A

is the identity matrix, meaning every variable is a pivot variable, so there are no free variables.

Therefore, statements 1, 3, and 4 are true."
:::

---

Advanced Applications

Gaussian elimination can be used to analyze systems with parameters, determining conditions for unique, no, or infinite solutions.

Worked Example:

For what values of $k$ does the following system have (a) a unique solution, (b) no solution, (c) infinitely many solutions?

\begin{aligned}x + y + z & = 1 \\x + 2y + 4z & = k \\x + 4y + 10z & = k^2\end{aligned}

Step 1: Write the augmented matrix.

\left[\begin{array}{ccc|c}> 1 & 1 & 1 & 1 \\> 1 & 2 & 4 & k \\> 1 & 4 & 10 & k^2> \end{array}\right]

Step 2: Apply EROs to get to REF.
$R_2 \to R_2 - R_1$
$R_3 \to R_3 - R_1$

\left[\begin{array}{ccc|c}> 1 & 1 & 1 & 1 \\> 0 & 1 & 3 & k-1 \\> 0 & 3 & 9 & k^2-1> \end{array}\right]

Step 3: Continue to REF.
$R_3 \to R_3 - 3R_2$

\left[\begin{array}{ccc|c}> 1 & 1 & 1 & 1 \\> 0 & 1 & 3 & k-1 \\> 0 & 0 & 0 & k^2 - 1 - 3(k-1)> \end{array}\right]

Step 4: Simplify the last entry in $R_3$ .
$k^2 - 1 - 3k + 3 = k^2 - 3k + 2 = (k-1)(k-2)$ .

The augmented matrix in REF is:
>

\left[\begin{array}{ccc|c}> 1 & 1 & 1 & 1 \\> 0 & 1 & 3 & k-1 \\> 0 & 0 & 0 & (k-1)(k-2)> \end{array}\right]

Step 5: Analyze for different solution types based on the last row.
The last row corresponds to the equation $0x + 0y + 0z = (k-1)(k-2)$ .

(a) Unique solution: A unique solution exists if there are no free variables and no inconsistent rows. Here, the third column does not have a pivot, so $z$ is a free variable. This system can never have a unique solution.

(b) No solution: This occurs if $0 = (k-1)(k-2)$ is false, i.e., $(k-1)(k-2) \neq 0$ .
This means $k \neq 1$ and $k \neq 2$ .
In this case, the last row becomes $0 = (\text{non-zero value})$ , which is a contradiction.

(c) Infinitely many solutions: This occurs if $0 = (k-1)(k-2)$ is true, i.e., $(k-1)(k-2) = 0$ .
This means $k = 1$ or $k = 2$ .
In these cases, the last row becomes $0 = 0$ , which is consistent. Since there is a free variable ( $z$ ), there will be infinitely many solutions.

Answer:
(a) Unique solution: Never.
(b) No solution: $k \neq 1$ and $k \neq 2$ .
(c) Infinitely many solutions: $k = 1$ or $k = 2$ .

:::question type="MCQ" question="For what value of $a$ does the following system have no solution?

\begin{aligned}x + y - z & = 1 \\2x + 3y + az & = 3 \\x + 2y + (a+1)z & = 2\end{aligned}

" options=[" $a=0$ "," $a=1$ "," $a=-1$ "," $a=2$ "] answer=" $a=0$ " hint="Form the augmented matrix and reduce it to REF. Look for a condition where $0 = \text{non-zero}$ ." solution="Step 1: Write the augmented matrix.

\left[\begin{array}{ccc|c}> 1 & 1 & -1 & 1 \\> 2 & 3 & a & 3 \\> 1 & 2 & a+1 & 2> \end{array}\right]

Step 2: Apply EROs to get to REF.
$R_2 \to R_2 - 2R_1$
$R_3 \to R_3 - R_1$

\left[\begin{array}{ccc|c}> 1 & 1 & -1 & 1 \\> 0 & 1 & a+2 & 1 \\> 0 & 1 & a+2 & 1> \end{array}\right]

Step 3: Continue to REF.
$R_3 \to R_3 - R_2$

\left[\begin{array}{ccc|c}> 1 & 1 & -1 & 1 \\> 0 & 1 & a+2 & 1 \\> 0 & 0 & 0 & 0> \end{array}\right]

Step 4: Analyze for no solution.
The last row is $0 = 0$ , which means the system is always consistent. There are no values of $a$ for which this system has no solution.

My example for the question needs to yield an inconsistent case for a specific value of $a$ .
Let me modify the question or the solution.

Let's modify the system to make it inconsistent for some $a$ .
Consider:
$x + y - z = 1$
$2x + 3y + az = 3$
$x + 2y + (a+1)z = 5$ (changed constant in 3rd equation)

Step 1: Augmented matrix.

\left[\begin{array}{ccc|c}> 1 & 1 & -1 & 1 \\> 2 & 3 & a & 3 \\> 1 & 2 & a+1 & 5> \end{array}\right]

Step 2: EROs.
$R_2 \to R_2 - 2R_1$
$R_3 \to R_3 - R_1$

\left[\begin{array}{ccc|c}> 1 & 1 & -1 & 1 \\> 0 & 1 & a+2 & 1 \\> 0 & 1 & a+2 & 4> \end{array}\right]

Step 3: Continue to REF.
$R_3 \to R_3 - R_2$

\left[\begin{array}{ccc|c}> 1 & 1 & -1 & 1 \\> 0 & 1 & a+2 & 1 \\> 0 & 0 & 0 & 3> \end{array}\right]

Step 4: Analyze for no solution.
The last row corresponds to $0x + 0y + 0z = 3$ , or $0 = 3$ . This is a contradiction.
This means the system is always inconsistent, regardless of $a$ . This is not what the question is asking for.

I need a system where $0 = \text{something involving } a$ on the right.

New example for the question:

\begin{aligned}x + y + z & = 1 \\2x + 2y + az & = 3 \\3x + 3y + 3z & = 4\end{aligned}

This is similar to my worked example for inconsistent systems.
Let's make sure the options are relevant.

Consider this system:

\begin{aligned}x - y + 2z & = 1 \\2x - 2y + (a+4)z & = 3 \\-x + y - 2z & = -1\end{aligned}

Step 1: Augmented matrix.

\left[\begin{array}{ccc|c}> 1 & -1 & 2 & 1 \\> 2 & -2 & a+4 & 3 \\> -1 & 1 & -2 & -1> \end{array}\right]

Step 2: EROs.
$R_2 \to R_2 - 2R_1$
$R_3 \to R_3 + R_1$

\left[\begin{array}{ccc|c}> 1 & -1 & 2 & 1 \\> 0 & 0 & a & 1 \\> 0 & 0 & 0 & 0> \end{array}\right]

Step 3: Analyze for no solution.
The second row corresponds to $az = 1$ .
The third row is $0=0$ .
For the system to have no solution, the equation $az=1$ must be a contradiction. This happens if $a=0$ , because then it becomes $0z=1$ , or $0=1$ , which is false.
If $a \neq 0$ , then $z = 1/a$ , and we would have infinitely many solutions (with $y$ as a free variable).

So, the system has no solution when $a=0$ .
This matches the provided options and desired logic.

Answer: $a=0$ "
:::

---

Problem-Solving Strategies

💡 CMI Strategy

When solving systems with parameters, always reduce the augmented matrix to Row Echelon Form. The conditions for unique, no, or infinite solutions will emerge from the last non-zero row(s) and the presence of free variables. Pay close attention to expressions that can become zero or non-zero based on the parameter.

---

Common Mistakes

⚠️ Watch Out

❌ Mistake: Forgetting to apply an ERO to the constant column in an augmented matrix.
✅ Correct approach: Every elementary row operation must be applied across the entire row, including the augmented part.

❌ Mistake: Confusing REF and RREF conditions, especially regarding zeros above leading entries.
✅ Correct approach: In RREF, every column containing a leading entry must have zeros everywhere else (above and below the leading entry). In REF, only zeros below the leading entry are required.

❌ Mistake: Incorrectly interpreting a row of zeros in REF/RREF.
✅ Correct approach: A row like $\left[\begin{array}{ccc|c} 0 & 0 & 0 & k \end{array}\right]$ implies inconsistency if $k \neq 0$ and infinite solutions if $k=0$ (provided other rows don't yield a unique solution). A row like $\left[\begin{array}{ccc|c} 0 & 0 & 0 & 0 \end{array}\right]$ always indicates a consistent system, potentially with free variables.

---

Practice Questions

:::question type="MCQ" question="Which of the following operations is NOT an elementary row operation?" options=["Multiplying a row by a non-zero scalar.","Adding a multiple of one row to another row.","Swapping two rows.","Multiplying a row by zero."] answer="Multiplying a row by zero." hint="Elementary row operations must be reversible to preserve the solution set of the system." solution="The three elementary row operations are:

Swapping two rows.

Multiplying a row by a non-zero scalar.

Adding a scalar multiple of one row to another row.

Multiplying a row by zero would eliminate that row, potentially changing the rank of the matrix and altering the solution set in a non-reversible way. Thus, it is not an elementary row operation."
:::

:::question type="NAT" question="After applying Gaussian elimination, an augmented matrix for a system of 3 equations in 3 variables ( $x, y, z$ ) is in Row Echelon Form as:

\left[\begin{array}{ccc|c}1 & -1 & 2 & 5 \\0 & 1 & -3 & -2 \\0 & 0 & 1 & 1\end{array}\right]

What is the value of

x

?" answer="4" hint="Use back-substitution starting from the last row." solution="The system of equations represented by the matrix is:

x - y + 2z = 5

y - 3z = -2

z = 1

Step 1: From equation (3), we have $z = 1$ .

Step 2: Substitute $z=1$ into equation (2):
$y - 3(1) = -2$
$y - 3 = -2$
$y = 1$

Step 3: Substitute $y=1$ and $z=1$ into equation (1):
$x - 1 + 2(1) = 5$
$x - 1 + 2 = 5$
$x + 1 = 5$
$x = 4$

The value of $x$ is 4."
:::

:::question type="MSQ" question="Consider the augmented matrix of a system of linear equations that has been reduced to Row Echelon Form:

\left[\begin{array}{cccc|c}1 & 3 & 0 & -2 & 4 \\0 & 0 & 1 & 5 & -1 \\0 & 0 & 0 & 0 & 0\end{array}\right]

Which of the following statements are true about the solution to this system?" options=["The system has a unique solution.","The system is consistent.","There are two free variables.","The variable

x_3

is a free variable."] answer="The system is consistent.,There are two free variables." hint="Identify pivot variables and free variables. Check for consistency." solution="Let the variables be

x_1, x_2, x_3, x_4

.
The matrix is:

\left[\begin{array}{cccc|c}1 & 3 & 0 & -2 & 4 \\0 & 0 & 1 & 5 & -1 \\0 & 0 & 0 & 0 & 0\end{array}\right]

The system has a unique solution. False. There are 4 variables and only 2 pivot positions (leading 1s in columns 1 and 3). This means there are

4 - 2 = 2

free variables, leading to infinitely many solutions, not a unique one.

The system is consistent. True. There is no row of the form

[0 \ 0 \ 0 \ 0 | k]

where

k \neq 0

. The last row is

[0 \ 0 \ 0 \ 0 | 0]

, which indicates consistency.

There are two free variables. True. The pivot variables are

x_1

and

x_3

. The non-pivot variables are

x_2

and

x_4

. These are the free variables.

The variable $x_3$ is a free variable. False.

x_3

corresponds to a pivot position (the leading 1 in the second row is in the

x_3

column). Thus,

x_3

is a pivot variable, not a free variable.

Therefore, statements 2 and 3 are true."
:::

:::question type="NAT" question="Determine the value of $k$ for which the following system has infinitely many solutions:

\begin{aligned}x + y + z & = 3 \\x + 2y + kz & = 4 \\2x + 3y + 4z & = 7\end{aligned}

" answer="2" hint="Reduce the augmented matrix to REF. For infinitely many solutions, you need a row of zeros on both sides of the bar in the coefficient matrix, and no inconsistent rows." solution="Step 1: Write the augmented matrix.

\left[\begin{array}{ccc|c}> 1 & 1 & 1 & 3 \\> 1 & 2 & k & 4 \\> 2 & 3 & 4 & 7> \end{array}\right]

Step 2: Apply EROs.
$R_2 \to R_2 - R_1$
$R_3 \to R_3 - 2R_1$

\left[\begin{array}{ccc|c}> 1 & 1 & 1 & 3 \\> 0 & 1 & k-1 & 1 \\> 0 & 1 & 2 & 1> \end{array}\right]

Step 3: Continue to REF.
$R_3 \to R_3 - R_2$

\left[\begin{array}{ccc|c}> 1 & 1 & 1 & 3 \\> 0 & 1 & k-1 & 1 \\> 0 & 0 & 2-(k-1) & 1-1> \end{array}\right]

\left[\begin{array}{ccc|c}> 1 & 1 & 1 & 3 \\> 0 & 1 & k-1 & 1 \\> 0 & 0 & 3-k & 0> \end{array}\right]

Step 4: Analyze for infinitely many solutions.
For infinitely many solutions, the last row must represent $0=0$ and there must be free variables. This requires the coefficient $3-k$ to be zero.
$3 - k = 0 \implies k = 3$ .

Wait, let's re-check my example.
If $k=3$ , the matrix is:

\left[\begin{array}{ccc|c}1 & 1 & 1 & 3 \\0 & 1 & 2 & 1 \\0 & 0 & 0 & 0\end{array}\right]

Here,

z

is a free variable, so there are infinitely many solutions.
So

k=3

is the answer.

My solution and answer do not match. The question asks for $k$ such that the system has infinitely many solutions. My derivation yields $k=3$ . The provided answer is $2$ .
Let's check for $k=2$ .
If $k=2$ , the matrix is:

\left[\begin{array}{ccc|c}1 & 1 & 1 & 3 \\0 & 1 & 1 & 1 \\0 & 0 & 1 & 0\end{array}\right]

This system has a unique solution:

z=0

y=1

x=2

.
So

k=2

results in a unique solution, not infinitely many.

The question or the answer needs to be corrected. I will correct the answer to $3$ .

Answer: 3"
:::

:::question type="MCQ" question="A system of linear equations is represented by an augmented matrix $A$ . If, after Gaussian elimination, $A$ is reduced to an REF where the last row is $\left[\begin{array}{cccc|c} 0 & 0 & 0 & 0 & 7 \end{array}\right]$ , what can be concluded about the system?" options=["It has a unique solution.","It has infinitely many solutions.","It is inconsistent.","It is a homogeneous system."] answer="It is inconsistent." hint="A row of zeros on the left side of the bar and a non-zero on the right side indicates a contradiction." solution="The row $\left[\begin{array}{cccc|c} 0 & 0 & 0 & 0 & 7 \end{array}\right]$ corresponds to the equation $0x_1 + 0x_2 + 0x_3 + 0x_4 = 7$ , or simply $0 = 7$ . This is a false statement (a contradiction). When Gaussian elimination leads to a contradiction, the system of linear equations has no solution. Such a system is called inconsistent.

A unique solution would have a pivot in every variable column and no inconsistent rows.

Infinitely many solutions would have free variables and no inconsistent rows.

A homogeneous system has all constant terms equal to zero."

:::

---

Summary

❗ Key Formulas & Takeaways

| Concept | Expression |

|---|----------------|------------| | 1 | Augmented Matrix |

[A | \mathbf{b}]

| | 2 | Elementary Row Operations |

R_i \leftrightarrow R_j

kR_i \to R_i

R_i + kR_j \to R_i

[0 \ \cdots \ 0 | k]

with

k \neq 0

A\mathbf{x}=\mathbf{0}

) | Always consistent (trivial solution

\mathbf{x}=\mathbf{0}

). Non-trivial solutions if free variables exist. |

---

What's Next?

💡 Continue Learning

This topic connects to:

Matrix Inverses: Gaussian elimination (specifically Gauss-Jordan) can be used to find the inverse of a matrix by augmenting it with the identity matrix $[A | I]$ .

Determinants: The value of the determinant can indicate whether a system has a unique solution (non-zero determinant) or not (zero determinant).

Vector Spaces and Subspaces: The solution set of a homogeneous system forms a null space, which is a vector subspace.

Linear Transformations: Understanding how systems of equations relate to mapping vectors in one space to another.

---

💡 Next Up

Proceeding to Determinants.

---

Part 2: Determinants

Determinants are scalar values associated with square matrices, providing crucial information about matrix invertibility, linear independence of column/row vectors, and the scaling factor of linear transformations. We utilize determinants to solve systems of linear equations and characterize matrix properties.

---

Core Concepts

1. Definition for Small Matrices

For a square matrix $A$ , its determinant, denoted $\det(A)$ or $|A|$ , is a scalar value calculated from its elements. We begin with $2 \times 2$ and $3 \times 3$ matrices.

📐 Determinant of a 2x2 Matrix

For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ ,

\det(A) = ad - bc

When to use: Calculating determinants for

2 \times 2

matrices.

Worked Example: Calculate the determinant of $A = \begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix}$ .

Step 1: Apply the $2 \times 2$ determinant formula.

\det(A) = (3)(2) - (1)(4)

Step 2: Simplify the expression.

> $
\det(A) = 6 - 4

>

\det(A) = 2

2

\det(B) = 1((1)(0) - (4)(6)) - 2((0)(0) - (4)(5)) + 3((0)(6) - (1)(5))

* * S t e p 2 : * * S i m p l i f y e a c h t e r m . >

\det(B) = 1(0 - 24) - 2(0 - 20) + 3(0 - 5)

>

\det(B) = 1(-24) - 2(-20) + 3(-5)

>

\det(B) = -24 + 40 - 15

* * S t e p 3 : * * S u m t h e t e r m s . >

\det(B) = 16 - 15

>

\det(B) = 1

1

\det(M) = 2((1)(-1) - (2)(0)) - (-1)((3)(-1) - (2)(1)) + 0((3)(0) - (1)(1))

* * S t e p 2 : * * S i m p l i f y t h e t e r m s . >

\det(M) = 2(-1 - 0) + 1(-3 - 2) + 0(0 - 1)

>

\det(M) = 2(-1) + 1(-5) + 0

>

\det(M) = -2 - 5

>

\det(M) = -7

n \times n

a_{11} = 1, a_{12} = 2, a_{13} = 0

* * S t e p 2 : * * C a l c u l a t e t h e c o f a c t o r s . >

C_{11} = (-1)^{1+1} \det\begin{bmatrix} 1 & 4 \\ 5 & 2 \end{bmatrix} = 1((1)(2) - (4)(5)) = 2 - 20 = -18

>

C_{12} = (-1)^{1+2} \det\begin{bmatrix} 3 & 4 \\ 0 & 2 \end{bmatrix} = -1((3)(2) - (4)(0)) = -1(6 - 0) = -6

>

C_{13} = (-1)^{1+3} \det\begin{bmatrix} 3 & 1 \\ 0 & 5 \end{bmatrix} = 1((3)(5) - (1)(0)) = 15 - 0 = 15

* * S t e p 3 : * * A p p l y t h e c o f a c t o r e x p a n s i o n f o r m u l a . >

\det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}

>

\det(A) = 1(-18) + 2(-6) + 0(15)

>

\det(A) = -18 - 12 + 0

>

\det(A) = -30

-30

\det(B) = a_{31}C_{31} + a_{32}C_{32} + a_{33}C_{33} + a_{34}C_{34}

>

\det(B) = 0 \cdot C_{31} + 0 \cdot C_{32} + 2 \cdot C_{33} + 0 \cdot C_{34}

>

\det(B) = 2 \cdot C_{33}

C_{33}

C_{33} = (-1)^{3+3} M_{33} = 1 \cdot \det\begin{bmatrix} 4 & 0 & 0 \\ 2 & 1 & 3 \\ 1 & 0 & 1 \end{bmatrix}

&#x27; in math mode at position 46: …rminant of the̲3 \times 3min…" style="color:#cc0000">Step 3:** Calculate the determinant of the3 \times 3$ minor matrix. Expand along the second column of this minor.>

\det\begin{bmatrix} 4 & 0 & 0 \\ 2 & 1 & 3 \\ 1 & 0 & 1 \end{bmatrix} = 0 \cdot C'_{12} + 1 \cdot C'_{22} + 0 \cdot C'_{32}

>

C'_{22} = (-1)^{2+2} \det\begin{bmatrix} 4 & 0 \\ 1 & 1 \end{bmatrix} = 1((4)(1) - (0)(1)) = 4

\det\begin{bmatrix} 4 & amp; 0 & amp; 0 \\ 2 & amp; 1 & amp; 3 \\ 1 & amp; 0 & amp; 1 \end{bmatrix} = 1 \cdot 4 = 4

\det(B) = 2 \cdot 4

>

\det(B) = 8

&#x27; in math mode at position 35: …evaluating the̲3 \times 3min…" style="color:#cc0000">Correction:** Re-evaluating the3 \times 3$ minor determinant:>

\det\begin{bmatrix} 4 & 0 & 0 \\ 2 & 1 & 3 \\ 1 & 0 & 1 \end{bmatrix}

4 \det\begin{bmatrix} 1 & amp; 3 \\ 0 & amp; 1 \end{bmatrix} - 0 \det(\dots) + 0 \det(\dots)

= 4((1)(1) - (3)(0)) = 4(1) = 4

C_{33}

C_{33} = (-1)^{3+3} \det\begin{bmatrix} 4 & 0 & 0 \\ 2 & 1 & 3 \\ 1 & 0 & 1 \end{bmatrix}

&#x27; in math mode at position 15: Expanding the̲3 \times 3mat…" style="color:#cc0000">Expanding the3 \times 3$ matrix along the first row:>

4 \cdot \det\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} - 0 \cdot \det\begin{bmatrix} 2 & 3 \\ 1 & 1 \end{bmatrix} + 0 \cdot \det\begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix}

>

= 4(1 \cdot 1 - 3 \cdot 0) - 0 + 0

>

= 4(1) = 4

C_{33} = 4

\det(B) = a_{31}C_{31} + a_{32}C_{32} + a_{33}C_{33} + a_{34}C_{34}

>

\det(B) = 0 \cdot C_{31} + 0 \cdot C_{32} + 2 \cdot C_{33} + 0 \cdot C_{34}

>

\det(B) = 2 \cdot C_{33}

C_{33}

C_{33} = (-1)^{3+3} M_{33} = 1 \cdot \det\begin{bmatrix} 4 & 0 & 0 \\ 2 & 1 & 3 \\ 1 & 0 & 1 \end{bmatrix}

minor matrix

\det\begin{bmatrix} 4 & 0 & 0 \\ 2 & 1 & 3 \\ 1 & 0 & 1 \end{bmatrix} = 4 \cdot \det\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} - 0 \cdot \det\begin{bmatrix} 2 & 3 \\ 1 & 1 \end{bmatrix} + 0 \cdot \det\begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix}

>

= 4((1)(1) - (3)(0))

>

= 4(1)

>

= 4

back into the expression for

\det(B) = 2 \cdot 4

>

\det(B) = 8

a

\begin{aligned}
x + y + z &= 1 \\
x + 2y + 3z &= 4 \\
2x + 3y + (a+1)z &= 6
\end{aligned}

with

\begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 1 & 2 & 3 & | & 4 \\ 2 & 3 & a+1 & | & 6 \end{pmatrix}

R_2 \to R_2 - R_1

\begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 2 & | & 3 \\ 0 & 1 & a-1 & | & 4 \end{pmatrix}

R_3 \to R_3 - R_2

\begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 2 & | & 3 \\ 0 & 0 & a-3 & | & 1 \end{pmatrix}

0x + 0y + 0z = 1

\det(A) = (1)(4)(6) = 24

\boxed{24}

\begin{aligned}
\det(2A^T B^{-1}) &= \det(2A^T) \det(B^{-1}) \\
&= (2^3 \det(A^T)) \left(\frac{1}{\det(B)}\right) \\
&= (8 \det(A)) \left(\frac{1}{\det(B)}\right) \\
&= (8 \times 2) \left(\frac{1}{-3}\right) \\
&= 16 \times \left(-\frac{1}{3}\right) \\
&= -\frac{16}{3}
\end{aligned}

k

\begin{aligned}
(k-1)x + 2y &= 0 \\
3x + ky &= 0
\end{aligned}

has non-trivial solutions if and only if

A = \begin{pmatrix} k-1 & 2 \\ 3 & k \end{pmatrix}

&#x27; in math mode at position 30: \dotsdeterminant of ̲A :" style="color:#cc0000">Calculate the determinant of A$:

\det(A) = (k-1)(k) - (2)(3) = k^2 - k - 6

to find the values of

k^2 - k - 6 = 0

Factor the quadratic equation:

(k-3)(k+2) = 0

are

3 + (-2) = 1
$$
The final answer is

\boxed{1}

."
:::

---

What's Next?

💡 Continue Your CMI Journey

Building on the foundational concepts of solving linear systems and understanding matrix properties, your next steps in Linear Algebra will delve into more abstract structures. You'll explore Vector Spaces, where the solution sets of homogeneous systems form subspaces, and develop a deeper understanding of Linear Independence, Span, and Basis. These concepts are crucial for understanding the geometry of linear transformations and will eventually lead to Eigenvalues and Eigenvectors, where determinants play a central role in finding characteristic polynomials.

Systems of Linear Equations

Systems of Linear Equations

Chapter Contents

| Topic |

Part 1: Gaussian Elimination

Core Concepts

1. Systems of Linear Equations and Augmented Matrices

2. Elementary Row Operations

3. Row Echelon Form (REF)

4. Reduced Row Echelon Form (RREF)

5. Solving Systems using Gaussian Elimination (Unique Solution)

6. Inconsistent Systems (No Solution)

7. Consistent Dependent Systems (Infinite Solutions)

8. Homogeneous Systems

Advanced Applications

Problem-Solving Strategies

Common Mistakes

Practice Questions

Summary

| Concept | Expression |

What's Next?

Part 2: Determinants

Core Concepts

1. Definition for Small Matrices

2. Cofactor Expansion (Laplace Expansion)

Chapter Summary

Chapter Review Questions

What's Next?

🎯 Key Points to Remember

Related Topics in Linear Algebra

Eigenvalues and Eigenvectors

Matrix Operations

Orthogonality

Linear Transformations

More Resources

Study Notes

Short Notes

Test Series

Mock Tests

Previous Year Papers

Chapter-wise PYQs

Chapter Practice

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