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Mathematical Methods Chapter 3 Ex 3.3 Solved Notes & Quiz

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Mathematical Methods Chapter 3 Ex 3.3 Solved Notes & Quiz

Official Academic Study Notes • Published: October 10, 2026 • Free Printable Resource

What are the Core Formulas and Definitions for this Exercise?

In linear algebra, understanding matrix transformations and vector space properties relies heavily on row operations and matrix rank. Below are the foundational concepts required to solve Exercise 3.3 of Mathematical Methods by S.M. Yusuf.

Elementary Row Operations (EROs): Operations on matrices that preserve row equivalence:
1. Interchanging two rows ($R_i \leftrightarrow R_j$).
2. Multiplying a row by a non-zero scalar ($k R_i$).
3. Adding a scalar multiple of one row to another ($R_i + k R_j$).
Row Echelon Form (REF): A matrix is in row echelon form if:
1. All non-zero rows are above any rows of all zeros.
2. The leading entry (first non-zero number from the left) of a non-zero row is strictly to the right of the leading entry of the row above it.
Reduced Row Echelon Form (RREF): A matrix is in RREF if it satisfies all conditions of REF, and additionally:
1. The leading entry in each non-zero row is $1$.
2. Each column containing a leading $1$ has zeros in all other positions.
Rank of a Matrix ($\text{rank}(A)$): The number of non-zero rows in the row echelon form of matrix $A$. It equals the maximum number of linearly independent rows or columns.
Rouché–Capelli Theorem (Consistency Criterion):
- System $A \mathbf{x} = \mathbf{b}$ is consistent if $\text{rank}(A) = \text{rank}([A|\mathbf{b}])$.
- Unique solution if $\text{rank}(A) = n$ (number of variables).
- Infinitely many solutions if $\text{rank}(A) < n$.
- Inconsistent (no solution) if $\text{rank}(A) < \text{rank}([A|\mathbf{b}])$.

Concept / Quantity	Mathematical Notation / Criterion	Key Significance
Augmented Matrix	$[A\|\mathbf{b}]$	Combines coefficients and constant vector for system analysis.
Rank Condition for Unique Solution	$\text{rank}(A) = \text{rank}([A\|\mathbf{b}]) = n$	Guarantees a single unique point solution in $\mathbb{R}^n$.
Rank Condition for Infinite Solutions	$\text{rank}(A) = \text{rank}([A\|\mathbf{b}]) < n$	Degree of freedom equals $n - \text{rank}(A)$.
Linear Independence	$\sum c_i \mathbf{v}_i = \mathbf{0} \implies c_i = 0 \; \forall i$	Determines if vectors form a minimal spanning set.

How to Solve All Exercise Questions Step-by-Step?

Question 1: Reduce the given matrix $A$ to row echelon form and find its rank:

$$A = \begin{bmatrix} 1 & 2 & -1 & 3 \\ 2 & 4 & -1 & 7 \\ -1 & -2 & 3 & -1 \end{bmatrix}$$

Solution:
We apply Elementary Row Operations (EROs) to convert $A$ into Row Echelon Form.
Step 1: Create zeros in column 1 below row 1 using $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 + R_1$: $$A \sim \begin{bmatrix} 1 & 2 & -1 & 3 \\ 2 - 2(1) & 4 - 2(2) & -1 - 2(-1) & 7 - 2(3) \\ -1 + 1 & -2 + 2 & 3 + (-1) & -1 + 3 \end{bmatrix}$$ $$A \sim \begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 2 & 2 \end{bmatrix}$$ Step 2: Create zeros in column 3 below row 2 using $R_3 \to R_3 - 2R_2$: $$A \sim \begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 2 - 2(1) & 2 - 2(1) \end{bmatrix}$$ $$A \sim \begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ The matrix is now in Row Echelon Form.
Number of non-zero rows = $2$.
Answer: $$\text{Echelon Form} = \begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}, \quad \text{rank}(A) = 2$$

Question 2: Reduce the matrix $B$ to reduced row echelon form (RREF) and find its rank:

$$B = \begin{bmatrix} 1 & -1 & 2 & 3 \\ 2 & 2 & 0 & 2 \\ 4 & 0 & 4 & 8 \end{bmatrix}$$

Solution:
Step 1: Perform $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 4R_1$: $$B \sim \begin{bmatrix} 1 & -1 & 2 & 3 \\ 0 & 4 & -4 & -4 \\ 0 & 4 & -4 & -4 \end{bmatrix}$$ Step 2: Divide Row 2 by $4$ ($R_2 \to \frac{1}{4} R_2$): $$B \sim \begin{bmatrix} 1 & -1 & 2 & 3 \\ 0 & 1 & -1 & -1 \\ 0 & 4 & -4 & -4 \end{bmatrix}$$ Step 3: Perform $R_3 \to R_3 - 4R_2$: $$B \sim \begin{bmatrix} 1 & -1 & 2 & 3 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ Step 4: To get RREF, create zero above the leading $1$ in row 2 using $R_1 \to R_1 + R_2$: $$B \sim \begin{bmatrix} 1 & 0 & 1 & 2 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ The matrix is in Reduced Row Echelon Form with 2 non-zero rows.
Answer: $$\text{RREF} = \begin{bmatrix} 1 & 0 & 1 & 2 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 \end{bmatrix}, \quad \text{rank}(B) = 2$$

Question 3: Find the rank of the $4 \times 4$ matrix $C$:

$$C = \begin{bmatrix} 1 & 3 & -2 & 1 \\ 2 & 1 & 3 & 2 \\ 3 & 4 & 1 & 3 \\ 5 & 5 & 4 & 5 \end{bmatrix}$$

Solution:
Apply EROs to clear column 1:

$R_2 \to R_2 - 2R_1 = [0, -5, 7, 0]$
$R_3 \to R_3 - 3R_1 = [0, -5, 7, 0]$
$R_4 \to R_4 - 5R_1 = [0, -10, 14, 0]$

Matrix becomes: $$C \sim \begin{bmatrix} 1 & 3 & -2 & 1 \\ 0 & -5 & 7 & 0 \\ 0 & -5 & 7 & 0 \\ 0 & -10 & 14 & 0 \end{bmatrix}$$ Now clear rows 3 and 4 using row 2:

$R_3 \to R_3 - R_2 = [0, 0, 0, 0]$
$R_4 \to R_4 - 2R_2 = [0, 0, 0, 0]$

$$C \sim \begin{bmatrix} 1 & 3 & -2 & 1 \\ 0 & -5 & 7 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ There are exactly 2 non-zero rows remaining.
Answer: $$\text{rank}(C) = 2$$

Question 4: Find the value of $\lambda$ for which the rank of matrix $D$ is less than $3$:

$$D = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & \lambda \end{bmatrix}$$

Solution:
For a $3 \times 3$ matrix to have rank less than 3, its determinant must be zero, or its row echelon form must contain at least one row of zeros.
Using row operations:

$R_2 \to R_2 - R_1 = [0, 1, 3]$
$R_3 \to R_3 - R_1 = [0, 3, \lambda - 1]$

$$D \sim \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & 3 & \lambda - 1 \end{bmatrix}$$ Next, perform $R_3 \to R_3 - 3R_2$: $$D \sim \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & (\lambda - 1) - 3(3) \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & \lambda - 10 \end{bmatrix}$$ For $\text{rank}(D) < 3$, the third row must be completely zero: $$\lambda - 10 = 0 \implies \lambda = 10$$ Answer: $$\lambda = 10$$

Question 5: Solve the following systems of linear equations using Gaussian elimination / Row reduction:

(Part i) Non-homogeneous System:

$$\begin{aligned} x + y + z &= 6 \\ x + 2y + 3z &= 14 \\ x + 4y + 9z &= 36 \end{aligned}$$

Solution (Part i):
Form the augmented matrix $[A|\mathbf{b}]$: $$[A|\mathbf{b}] = \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 1 & 2 & 3 & | & 14 \\ 1 & 4 & 9 & | & 36 \end{bmatrix}$$ Apply $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$: $$[A|\mathbf{b}] \sim \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 8 \\ 0 & 3 & 8 & | & 30 \end{bmatrix}$$ Apply $R_3 \to R_3 - 3R_2$: $$[A|\mathbf{b}] \sim \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 8 \\ 0 & 0 & 2 & | & 6 \end{bmatrix}$$ Divide $R_3$ by $2$ ($R_3 \to \frac{1}{2} R_3$): $$[A|\mathbf{b}] \sim \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 8 \\ 0 & 0 & 1 & | & 3 \end{bmatrix}$$ By back-substitution:

From Row 3: $z = 3$
From Row 2: $y + 2z = 8 \implies y + 2(3) = 8 \implies y = 2$
From Row 1: $x + y + z = 6 \implies x + 2 + 3 = 6 \implies x = 1$

Answer (Part i): $$(x, y, z) = (1, 2, 3)$$

(Part ii) Homogeneous System:

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

Solution (Part ii):
Write the coefficient matrix $A$: $$A = \begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & -3 \\ 4 & -3 & -1 \end{bmatrix}$$ Perform $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 4R_1$: $$A \sim \begin{bmatrix} 1 & -2 & 1 \\ 0 & 5 & -5 \\ 0 & 5 & -5 \end{bmatrix}$$ Perform $R_2 \to \frac{1}{5} R_2$ and then $R_3 \to R_3 - R_2$: $$A \sim \begin{bmatrix} 1 & -2 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}$$ Here, $\text{rank}(A) = 2 < 3$ (number of variables). Hence, non-trivial infinite solutions exist.
Let $x_3 = t$ where $t \in \mathbb{R}$.
From Row 2: $x_2 - x_3 = 0 \implies x_2 = t$.
From Row 1: $x_1 - 2x_2 + x_3 = 0 \implies x_1 - 2t + t = 0 \implies x_1 = t$.
Answer (Part ii): $$(x_1, x_2, x_3) = (t, t, t), \quad \text{for any } t \in \mathbb{R}$$

Question 6: Determine whether the vectors $\mathbf{v}_1 = (1, 2, 3)$, $\mathbf{v}_2 = (2, 5, 7)$, and $\mathbf{v}_3 = (1, 3, 5)$ in $\mathbb{R}^3$ are linearly independent.

Solution:
Construct the vector equation $c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 = \mathbf{0}$.
This forms a matrix equation $A \mathbf{c} = \mathbf{0}$ where the columns of $A$ are vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$: $$A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 5 & 3 \\ 3 & 7 & 5 \end{bmatrix}$$ Compute the determinant $\det(A)$: $$\det(A) = 1 \cdot \begin{vmatrix} 5 & 3 \\ 7 & 5 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 3 \\ 3 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 5 \\ 3 & 7 \end{vmatrix}$$ $$\det(A) = 1(25 - 21) - 2(10 - 9) + 1(14 - 15)$$ $$\det(A) = 1(4) - 2(1) + 1(-1) = 4 - 2 - 1 = 1$$ Since $\det(A) = 1 \neq 0$, matrix $A$ has full rank ($3$).
The only solution to $A \mathbf{c} = \mathbf{0}$ is the trivial solution $c_1 = c_2 = c_3 = 0$.
Therefore, the vectors are linearly independent.
Answer: $$\text{The vectors are Linearly Independent because } \det(A) = 1 \neq 0.$$

Interactive Practice Quiz: Test Your Understanding (Clickable MCQs)

Q1: What is the rank of a $3 \times 4$ zero matrix?

Explanation: A zero matrix has no non-zero rows, so its rank is strictly $0$.

Q2: What is the maximum possible rank of a matrix of size $4 \times 3$?