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9 Class Math Notes: Class 9 Mathematics Chapter 1 Matrices and Determinants Exercise 1.6 Solved - New 2025-2026 Syllabus Solved PDF Notes

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9 Class Math Notes: Class 9 Mathematics Chapter 1 Matrices and Determinants Exercise 1.6 Solved - New 2025-2026 Syllabus Solved PDF Notes

Official Academic Study Notes • Published: September 02, 2026 • Free Printable Resource

What are the Key Concepts and Rules for this Exercise?

Exercise 1.6 of Chapter 1 focuses on solving systems of two linear algebraic equations in two variables ($x$ and $y$). According to the NEW 2025-2026 Single National Curriculum (SNC), students must master two primary analytical techniques: Matrix Inversion Method and Cramer's Rule.

Consider a general system of linear equations:

$$a x + b y = m$$ $$c x + d y = n$$

This system can be transformed into matrix form as $A X = B$:

$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} m \\ n \end{bmatrix}$$

1. The Matrix Inversion Method

To solve $A X = B$ using the Matrix Inversion Method:

Compute the determinant of the coefficient matrix $A$, written as $|A| = ad - bc$.
Existence Condition: If $|A| = 0$, matrix $A$ is singular, and the multiplicative inverse $A^{-1}$ does not exist. Hence, the system has no unique solution. If $|A| \neq 0$, matrix $A$ is non-singular, and a unique solution exists.
Find the Adjoint of matrix $A$, denoted $\text{adj}(A)$, by swapping diagonal entries and reversing signs of non-diagonal entries: $$\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$
Calculate the matrix inverse $A^{-1}$: $$A^{-1} = \frac{1}{|A|} \text{adj}(A)$$
Multiply $A^{-1}$ by matrix $B$ to yield column matrix $X$: $$X = A^{-1} B = \frac{1}{|A|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \begin{bmatrix} m \\ n \end{bmatrix}$$

2. Cramer's Rule

Cramer's Rule uses determinants directly to find individual variables without computing full matrix inverses:

Define the determinant of coefficient matrix $A$: $$|A| = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$$
Define matrix $A_x$ by replacing the first column (coefficients of $x$) with constant terms $B$: $$|A_x| = \begin{vmatrix} m & b \\ n & d \end{vmatrix} = md - nb$$
Define matrix $A_y$ by replacing the second column (coefficients of $y$) with constant terms $B$: $$|A_y| = \begin{vmatrix} a & m \\ c & n \end{vmatrix} = an - mc$$
Evaluate unknown variables $x$ and $y$: $$x = \frac{|A_x|}{|A|}, \quad y = \frac{|A_y|}{|A|} \quad \text{provided } |A| \neq 0$$

What is the Formula Reference Table for this Chapter?

Concept / Method	Mathematical Formula / Form	Key Conditions & Operational Notes
Matrix System Representation	$A X = B$	$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, X = \begin{bmatrix} x \\ y \end{bmatrix}, B = \begin{bmatrix} m \\ n \end{bmatrix}$
Determinant of $2 \times 2$ Matrix	$\|A\| = \det(A) = ad - bc$	Determines non-singularity ($\|A\| \neq 0$)
Adjoint of $2 \times 2$ Matrix	$\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$	Swap diagonal elements; negate non-diagonal elements
Matrix Inverse	$A^{-1} = \frac{1}{\|A\|} \text{adj}(A)$	Valid if and only if $\|A\| \neq 0$
Inversion Method Solution	$X = A^{-1} B = \frac{1}{\|A\|} \text{adj}(A) B$	Yields $\begin{bmatrix} x \\ y \end{bmatrix}$ directly
Cramer's Determinant $A_x$	$\|A_x\| = md - nb$	Column 1 replaced by constant matrix $B$
Cramer's Determinant $A_y$	$\|A_y\| = an - mc$	Column 2 replaced by constant matrix $B$
Cramer's Rule Formulae	$x = \frac{\|A_x\|}{\|A\|}, \quad y = \frac{\|A_y\|}{\|A\|}$	Requires $\|A\| \neq 0$

How to Solve All Exercise Questions Step-by-Step? (Complete Solved Solutions)

Question 1: Use matrices, if possible, to solve the following systems of linear equations by: (i) Matrix Inversion Method (ii) Cramer's Rule.

Part (i): $2x - 2y = 4$ and $3x + 2y = 6$

Method I: Matrix Inversion Method

Step 1: Write equations in matrix form $AX = B$:

$$\begin{bmatrix} 2 & -2 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 6 \end{bmatrix}$$

Let $A = \begin{bmatrix} 2 & -2 \\ 3 & 2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, $B = \begin{bmatrix} 4 \\ 6 \end{bmatrix}$.

Step 2: Calculate determinant $|A|$:

$$|A| = \begin{vmatrix} 2 & -2 \\ 3 & 2 \end{vmatrix} = (2)(2) - (-2)(3) = 4 - (-6) = 4 + 6 = 10$$

Since $|A| = 10 \neq 0$, matrix $A$ is non-singular and a unique solution exists.

Step 3: Find Adjoint of matrix $A$:

$$\text{adj}(A) = \begin{bmatrix} 2 & 2 \\ -3 & 2 \end{bmatrix}$$

Step 4: Compute inverse $A^{-1}$ and solve $X = A^{-1} B$:

$$A^{-1} = \frac{1}{10} \begin{bmatrix} 2 & 2 \\ -3 & 2 \end{bmatrix}$$ $$X = \frac{1}{10} \begin{bmatrix} 2 & 2 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 4 \\ 6 \end{bmatrix}$$ $$X = \frac{1}{10} \begin{bmatrix} (2)(4) + (2)(6) \\ (-3)(4) + (2)(6) \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 8 + 12 \\ -12 + 12 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 20 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix}$$

Therefore, $x = 2$ and $y = 0$.

Method II: Cramer's Rule

$$|A| = \begin{vmatrix} 2 & -2 \\ 3 & 2 \end{vmatrix} = 10$$ $$|A_x| = \begin{vmatrix} 4 & -2 \\ 6 & 2 \end{vmatrix} = (4)(2) - (-2)(6) = 8 + 12 = 20$$ $$|A_y| = \begin{vmatrix} 2 & 4 \\ 3 & 6 \end{vmatrix} = (2)(6) - (4)(3) = 12 - 12 = 0$$

Applying Cramer's formulae:

$$x = \frac{|A_x|}{|A|} = \frac{20}{10} = 2, \quad y = \frac{|A_y|}{|A|} = \frac{0}{10} = 0$$

Final Solution Set: $\{(2, 0)\}$

Part (ii): $2x + y = 3$ and $6x + 5y = 1$

Method I: Matrix Inversion Method

Matrix representation $AX = B$:

$$\begin{bmatrix} 2 & 1 \\ 6 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$$

Determinant of $A$:

$$|A| = \begin{vmatrix} 2 & 1 \\ 6 & 5 \end{vmatrix} = (2)(5) - (1)(6) = 10 - 6 = 4 \neq 0$$ $$\text{adj}(A) = \begin{bmatrix} 5 & -1 \\ -6 & 2 \end{bmatrix}$$

Solving $X = A^{-1} B$:

$$X = \frac{1}{4} \begin{bmatrix} 5 & -1 \\ -6 & 2 \end{bmatrix} \begin{bmatrix} 3 \\ 1 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 15 - 1 \\ -18 + 2 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 14 \\ -16 \end{bmatrix} = \begin{bmatrix} \frac{14}{4} \\ -\frac{16}{4} \end{bmatrix} = \begin{bmatrix} \frac{7}{2} \\ -4 \end{bmatrix}$$

Method II: Cramer's Rule

$$|A| = 4$$ $$|A_x| = \begin{vmatrix} 3 & 1 \\ 1 & 5 \end{vmatrix} = (3)(5) - (1)(1) = 15 - 1 = 14$$ $$|A_y| = \begin{vmatrix} 2 & 3 \\ 6 & 1 \end{vmatrix} = (2)(1) - (3)(6) = 2 - 18 = -16$$ $$x = \frac{|A_x|}{|A|} = \frac{14}{4} = \frac{7}{2}, \quad y = \frac{|A_y|}{|A|} = \frac{-16}{4} = -4$$

Final Solution Set: $\{(\frac{7}{2}, -4)\}$

Part (iii): $4x + 2y = 8$ and $3x - y = -1$

Method I: Matrix Inversion Method

$$\begin{bmatrix} 4 & 2 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ -1 \end{bmatrix}$$ $$|A| = \begin{vmatrix} 4 & 2 \\ 3 & -1 \end{vmatrix} = (4)(-1) - (2)(3) = -4 - 6 = -10 \neq 0$$ $$\text{adj}(A) = \begin{bmatrix} -1 & -2 \\ -3 & 4 \end{bmatrix}$$ $$X = \frac{1}{-10} \begin{bmatrix} -1 & -2 \\ -3 & 4 \end{bmatrix} \begin{bmatrix} 8 \\ -1 \end{bmatrix} = -\frac{1}{10} \begin{bmatrix} -8 + 2 \\ -24 - 4 \end{bmatrix} = -\frac{1}{10} \begin{bmatrix} -6 \\ -28 \end{bmatrix} = \begin{bmatrix} \frac{3}{5} \\ \frac{14}{5} \end{bmatrix}$$

Method II: Cramer's Rule

$$|A| = -10$$ $$|A_x| = \begin{vmatrix} 8 & 2 \\ -1 & -1 \end{vmatrix} = (8)(-1) - (2)(-1) = -8 + 2 = -6$$ $$|A_y| = \begin{vmatrix} 4 & 8 \\ 3 & -1 \end{vmatrix} = (4)(-1) - (8)(3) = -4 - 24 = -28$$ $$x = \frac{-6}{-10} = \frac{3}{5}, \quad y = \frac{-28}{-10} = \frac{14}{5}$$

Final Solution Set: $\{(\frac{3}{5}, \frac{14}{5})\}$

Part (iv): $3x - 2y = -6$ and $5x - 2y = -10$

Method I: Matrix Inversion Method

$$\begin{bmatrix} 3 & -2 \\ 5 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -6 \\ -10 \end{bmatrix}$$ $$|A| = \begin{vmatrix} 3 & -2 \\ 5 & -2 \end{vmatrix} = (3)(-2) - (-2)(5) = -6 + 10 = 4 \neq 0$$ $$\text{adj}(A) = \begin{bmatrix} -2 & 2 \\ -5 & 3 \end{bmatrix}$$ $$X = \frac{1}{4} \begin{bmatrix} -2 & 2 \\ -5 & 3 \end{bmatrix} \begin{bmatrix} -6 \\ -10 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 12 - 20 \\ 30 - 30 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} -8 \\ 0 \end{bmatrix} = \begin{bmatrix} -2 \\ 0 \end{bmatrix}$$

Method II: Cramer's Rule

$$|A| = 4$$ $$|A_x| = \begin{vmatrix} -6 & -2 \\ -10 & -2 \end{vmatrix} = (-6)(-2) - (-2)(-10) = 12 - 20 = -8$$ $$|A_y| = \begin{vmatrix} 3 & -6 \\ 5 & -10 \end{vmatrix} = (3)(-10) - (-6)(5) = -30 + 30 = 0$$ $$x = \frac{-8}{4} = -2, \quad y = \frac{0}{4} = 0$$

Final Solution Set: $\{(-2, 0)\}$

Part (v): $3x - 2y = 4$ and $-6x + 4y = 7$

Matrix Formulation:

$$\begin{bmatrix} 3 & -2 \\ -6 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 7 \end{bmatrix}$$

Calculate determinant $|A|$:

$$|A| = \begin{vmatrix} 3 & -2 \\ -6 & 4 \end{vmatrix} = (3)(4) - (-2)(-6) = 12 - 12 = 0$$

Conclusion: Since $|A| = 0$, matrix $A$ is singular. Multiplicative inverse $A^{-1}$ does not exist, and division by zero occurs in Cramer's Rule. Thus, solution is not possible for this system.

Question 2: Practical Word Problem — Rectangle Dimensions

The length of a rectangle is $4$ times its width. The perimeter of the rectangle is $150\text{ cm}$. Find the dimensions of the rectangle.

Formulation of Equations:

Let the width of rectangle $= x\text{ cm}$ and length of rectangle $= y\text{ cm}$.

According to Condition 1: Length is 4 times width $\implies y = 4x \implies 4x - y = 0$.
According to Condition 2: Perimeter is $150\text{ cm} \implies 2(\text{length} + \text{width}) = 150 \implies 2(x + y) = 150 \implies x + y = 75$.

The system of equations is:

$$4x - y = 0$$ $$x + y = 75$$

Matrix Solution (Cramer's Rule):

$$A = \begin{bmatrix} 4 & -1 \\ 1 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 75 \end{bmatrix}$$ $$|A| = \begin{vmatrix} 4 & -1 \\ 1 & 1 \end{vmatrix} = (4)(1) - (-1)(1) = 4 + 1 = 5 \neq 0$$ $$|A_x| = \begin{vmatrix} 0 & -1 \\ 75 & 1 \end{vmatrix} = (0)(1) - (-1)(75) = 75$$ $$|A_y| = \begin{vmatrix} 4 & 0 \\ 1 & 75 \end{vmatrix} = (4)(75) - (0)(1) = 300$$ $$x = \frac{|A_x|}{|A|} = \frac{75}{5} = 15\text{ cm}$$ $$y = \frac{|A_y|}{|A|} = \frac{300}{5} = 60\text{ cm}$$

Verification: Width $x = 15\text{ cm}$, Length $y = 60\text{ cm}$. Length is $4 \times 15 = 60\text{ cm}$. Perimeter is $2(15 + 60) = 2(75) = 150\text{ cm}$. Solutions are correct.

Question 3: Practical Word Problem — Difference of Sides

Two sides of a rectangle differ by $3.5\text{ cm}$. Find the dimensions of the rectangle if its perimeter is $67\text{ cm}$.

Formulation of Equations:

Let length $= x\text{ cm}$ and width $= y\text{ cm}$.

Condition 1: $x - y = 3.5$
Condition 2: $2(x + y) = 67 \implies x + y = 33.5$

Matrix Inversion Method Solution:

$$\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3.5 \\ 33.5 \end{bmatrix}$$ $$|A| = \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} = (1)(1) - (-1)(1) = 2 \neq 0$$ $$\text{adj}(A) = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 3.5 \\ 33.5 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 3.5 + 33.5 \\ -3.5 + 33.5 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 37 \\ 30 \end{bmatrix} = \begin{bmatrix} 18.5 \\ 15 \end{bmatrix}$$

Dimensions: Length $x = 18.5\text{ cm}$, Width $y = 15\text{ cm}$.

Interactive Practice Quiz: Test Your Understanding (Clickable MCQs)

Q1: What is the primary mathematical condition required for a system of linear equations $AX = B$ to have a unique solution?

Explanation: A system has a unique solution if and only if the coefficient matrix $A$ is non-singular ($|A| \neq 0$), guaranteeing that $A^{-1}$ exists.

Q2: If matrix $A = \begin{bmatrix} k & 2 \\ 3 & 4 \end{bmatrix}$ is singular, what is the exact value of constant $k$?

Explanation: Singular matrix means $|A| = 0 \implies (k)(4) - (2)(3) = 0 \implies 4k - 6 = 0 \implies 4k = 6 \implies k = \frac{6}{4} = 1.5$.

Q3: For the general matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, what is its Adjoint $\text{adj}(A)$?

Explanation: The adjoint of a $2 \times 2$ matrix is formed by swapping the main diagonal elements ($a \leftrightarrow d$) and changing the signs of the off-diagonal elements ($b \to -b, c \to -c$).

Q4: In Cramer's Rule, how is matrix $A_x$ formed from the system $AX = B$?

Explanation: $A_x$ is constructed by substituting the first column (coefficients of variable $x$) of $A$ with constant column matrix $B$.

Q5: What occurs when calculating solutions via Cramer's Rule for a system whose determinant $|A| = 0$ while $|A_x| \neq 0$?

Explanation: Division by zero ($x = \frac{|A_x|}{0}$) is mathematically undefined, rendering the system inconsistent with no solution.

Q6: Given matrix inverse $A^{-1} = \begin{bmatrix} 0.5 & 0 \\ 0 & 0.5 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 8 \end{bmatrix}$, what is variable vector $X$?

Explanation: $X = A^{-1} B = \begin{bmatrix} 0.5(4) + 0 \\ 0 + 0.5(8) \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \end{bmatrix}$.

Q7: What is the determinant of identity matrix $I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$?

Explanation: $|I_2| = (1)(1) - (0)(0) = 1$.

Q8: If $A$ is a non-singular square matrix, what is $A \cdot A^{-1}$ equal to?

Explanation: By definition of multiplicative inverse, multiplying any non-singular matrix by its inverse yields the multiplicative identity matrix $I$.

Frequently Asked Questions: What are Common Student Errors in this Exercise?

Q1: Why do board examiners dock marks when students solve Cramer's Rule problems directly without testing $|A|$ first?
Ans: In FBISE and Punjab Board marking schemes, checking whether $|A| \neq 0$ is a mandatory foundational step worth partial credit. If a student proceeds to compute $|A_x|$ or $|A_y|$ without establishing that matrix $A$ is non-singular, the whole mathematical derivation is logically flawed because division by zero is undefined. Always evaluate and explicitly state $|A| \neq 0$ before computing ratios.

Q2: What is the correct algebraic technique for computing the Adjoint matrix without making sign errors?
Ans: A classic sign error in 9th class math exams occurs when swapping or negating entries of matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Remember this rule: Swap positions for elements on the primary diagonal ($a$ and $d$), but change signs for elements on the secondary diagonal ($b$ and $c$) without moving them. For instance, if $b = -3$, its sign flips to $+3$ in $\text{adj}(A)$.

Q3: How can I verify my calculated values of $x$ and $y$ during board examinations?
Ans: Simply plug your calculated values of $x$ and $y$ back into both original linear equations given in the question prompt. If both left-hand sides match their respective right-hand side constant values, your solution set is guaranteed to be 100% accurate.