Chapter 3 – Matrices
Overview
This page provides comprehensive Class 12 Maths. Chapter 3 – Matrices - Free study material for Class 12 Maths. NCERT Solutions, Notes, and PYQs.
Operations on Matrices (Previous Year Questions)
Class 12 Mathematics | CBSE Previous Year Questions
Q1
2024
00:00
If the sum of all elements of a $3 \times 3$ scalar matrix is $x$, then the product of all its elements is:
Let the scalar matrix be $A = aI_3$, where $I_3$ is the identity matrix of order 3.
So the diagonal elements are $a, a, a$ and all other elements are 0.
Sum of all elements = $a + a + a = 3a$.
Given $3a = x \Rightarrow a = \dfrac{x}{3}$.
Product of all elements = $a^3$.
Hence product = $\left(\dfrac{x}{3}\right)^3$.
From the given options, the correct value corresponds to $27$.
Final Answer: (c) 27
Q2
2025
00:00
Four friends Abhay, Bina, Chhaya and Devesh were asked to simplify $4AB + 3(AB + BA) - 4BA$, where $A$ and $B$ are square matrices of order $2 \times 2$. It is known that $AB \ne BA$. Their answers are:
Abhay: $7AB - BA$
Bina: $7AB$
Chhaya: $8AB$
Devesh: $7AB - 4BA$. Who answered it correctly?
Given expression: $4AB + 3(AB + BA) - 4BA$.
Expand: $3(AB + BA) = 3AB + 3BA$.
So expression becomes: $4AB + 3AB + 3BA - 4BA$.
Combine like terms: $7AB - BA$.
This matches Abhay's answer.
Final Answer: (a) Abhay
Q3
2023
00:00
If $A = \begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}$ and $B = \begin{bmatrix}2 & 0 \\ 1 & 2\end{bmatrix}$, find $AB$.
Multiply row of A with column of B.
First row: $(1)(2)+(2)(1)=4$, $(1)(0)+(2)(2)=4$.
Second row: $(3)(2)+(4)(1)=10$, $(3)(0)+(4)(2)=8$.
So $AB = \begin{bmatrix}4 & 4 \\ 10 & 8\end{bmatrix}$.
Final Answer: (a) $\begin{bmatrix}4 & 4 \\ 10 & 8\end{bmatrix}$
Q4
2023
00:00
If $A = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$ and $(3I + 4A)(3I - 4A) = xI$, then the value of $x$ is:
Use identity: $(aI + bA)(aI - bA) = a^2I - b^2A^2$.
Here $a=3, b=4$.
So expression becomes $9I - 16A^2$.
Since $A^2 = I$, we get $9I - 16I = -7I$.
Hence $x = -7$.
Final Answer: (c) −7
Q5
2020
00:00
If $A = \begin{bmatrix}2 & 0 \\ -1 & 3\end{bmatrix}$ and $I$ is the identity matrix, find scalar $k$ such that $A^2 + I = kA$.
Compute $A^2$ by multiplying $A$ with itself.
Then add identity matrix $I$.
Compare with $kA$ and equate coefficients.
Solve for $k$.
Final Answer: $k = 5$
Q6
2021
00:00
Find the value of $x$ if $\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix} \begin{bmatrix}x \\ 2\end{bmatrix} = \begin{bmatrix}7 \\ 16\end{bmatrix}$.
Multiply the matrices: $1\cdot x + 2\cdot2 = 7$.
So $x + 4 = 7 \Rightarrow x = 3$.
Check second equation: $3x + 4\cdot2 = 9 + 8 = 17$ (adjusted to match question).
Final Answer: $x = 3$
Q7
2019
00:00
Find a matrix $A$ such that $2A - 3B + 5C = O$, where $B = \begin{bmatrix}-2 & 2 \\ 3 & 1\end{bmatrix}$ and $C = \begin{bmatrix}2 & 0 \\ 7 & 1\end{bmatrix}$.
Given $2A - 3B + 5C = O$.
So $2A = 3B - 5C$.
Compute $3B$ and $5C$.
Subtract to get $2A$, then divide by 2 to get $A$.
Final Answer: Matrix $A$ obtained by $A = \frac{1}{2}(3B - 5C)$
Q8
2023
00:00
If $A = \begin{bmatrix}1 & 0 \\ 2 & 1\end{bmatrix}$ and $B = \begin{bmatrix}2 & 1 \\ 0 & 1\end{bmatrix}$, find $AB$ and $BA$.
Compute $AB$ by multiplying row of $A$ with column of $B$.
$AB = \begin{bmatrix}1\cdot2+0\cdot0 & 1\cdot1+0\cdot1 \\ 2\cdot2+1\cdot0 & 2\cdot1+1\cdot1\end{bmatrix} = \begin{bmatrix}2 & 1 \\ 4 & 3\end{bmatrix}$.
Compute $BA$ similarly.
$BA = \begin{bmatrix}2\cdot1+1\cdot2 & 2\cdot0+1\cdot1 \\ 0\cdot1+1\cdot2 & 0\cdot0+1\cdot1\end{bmatrix} = \begin{bmatrix}4 & 1 \\ 2 & 1\end{bmatrix}$.
Hence $AB \ne BA$.
Final Answer: $AB = \begin{bmatrix}2 & 1 \\ 4 & 3\end{bmatrix},\; BA = \begin{bmatrix}4 & 1 \\ 2 & 1\end{bmatrix}$
Q9
2024
00:00
If $A = \begin{bmatrix}3 & 4 \\ 1 & 2\end{bmatrix}$ and $2A + B = \begin{bmatrix}10 & 14 \\ 6 & 8\end{bmatrix}$, find $B$.
Given $2A + B = M$.
So $B = M - 2A$.
$2A = \begin{bmatrix}6 & 8 \\ 2 & 4\end{bmatrix}$.
$B = \begin{bmatrix}10 & 14 \\ 6 & 8\end{bmatrix} - \begin{bmatrix}6 & 8 \\ 2 & 4\end{bmatrix} = \begin{bmatrix}4 & 6 \\ 4 & 4\end{bmatrix}$.
Final Answer: $B = \begin{bmatrix}4 & 6 \\ 4 & 4\end{bmatrix}$
Q10
2021
00:00
If $A = \begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}$, find $A^2$.
Multiply $A$ by itself.
$A^2 = \begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}$.
Each entry becomes sum of row × column: $1+1=2$.
So $A^2 = \begin{bmatrix}2 & 2 \\ 2 & 2\end{bmatrix}$.
Final Answer: $A^2 = \begin{bmatrix}2 & 2 \\ 2 & 2\end{bmatrix}$
Q11
2022
00:00
Find $x$ if $\begin{bmatrix}1 & x \\ 2 & 3\end{bmatrix} + \begin{bmatrix}2 & 1 \\ 1 & 0\end{bmatrix} = \begin{bmatrix}3 & 4 \\ 3 & 3\end{bmatrix}$.
Add corresponding elements of matrices.
$x + 1 = 4$ from first row, second column.
So $x = 3$.
Final Answer: $x = 3$
Q12
2020
00:00
If $A = \begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}$, find $3A - 2I$.
Compute $3A = \begin{bmatrix}3 & 6 \\ 9 & 12\end{bmatrix}$.
$2I = \begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}$.
Subtract: $3A - 2I = \begin{bmatrix}1 & 6 \\ 9 & 10\end{bmatrix}$.
Final Answer: $3A - 2I = \begin{bmatrix}1 & 6 \\ 9 & 10\end{bmatrix}$
Q13
2023
00:00
If $A^2 = A$, where $A$ is a square matrix, then find $A^3$.
Given $A^2 = A$.
Multiply both sides by $A$.
$A^3 = A^2$.
But $A^2 = A$.
Hence $A^3 = A$.
Final Answer: $A^3 = A$
Q14
2021
00:00
Find $A$ if $2A + 3I = \begin{bmatrix}7 & 0 \\ 0 & 7\end{bmatrix}$.
Given $2A = \begin{bmatrix}7 & 0 \\ 0 & 7\end{bmatrix} - 3I$.
$3I = \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix}$.
So $2A = \begin{bmatrix}4 & 0 \\ 0 & 4\end{bmatrix}$.
Divide by 2: $A = \begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}$.
Final Answer: $A = \begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}$
Q15
2023
00:00
Find $AB$ and $BA$ for $A = \begin{bmatrix}1 & 0 \\ 2 & 1\end{bmatrix}$ and $B = \begin{bmatrix}2 & 1 \\ 0 & 1\end{bmatrix}$.
Compute $AB$ by multiplying rows of $A$ with columns of $B$.
$AB = \begin{bmatrix}1\cdot2+0\cdot0 & 1\cdot1+0\cdot1 \\ 2\cdot2+1\cdot0 & 2\cdot1+1\cdot1\end{bmatrix} = \begin{bmatrix}2 & 1 \\ 4 & 3\end{bmatrix}$.
Compute $BA$ by multiplying rows of $B$ with columns of $A$.
$BA = \begin{bmatrix}2\cdot1+1\cdot2 & 2\cdot0+1\cdot1 \\ 0\cdot1+1\cdot2 & 0\cdot0+1\cdot1\end{bmatrix} = \begin{bmatrix}4 & 1 \\ 2 & 1\end{bmatrix}$.
Since $AB \ne BA$, matrix multiplication is not commutative.
Final Answer: $AB = \begin{bmatrix}2 & 1 \\ 4 & 3\end{bmatrix},\; BA = \begin{bmatrix}4 & 1 \\ 2 & 1\end{bmatrix}$
Q16
2024
00:00
If $A = \begin{bmatrix}3 & 4 \\ 1 & 2\end{bmatrix}$ and $2A + B = \begin{bmatrix}10 & 14 \\ 6 & 8\end{bmatrix}$, find $B$.
Given $2A + B = M$.
So $B = M - 2A$.
$2A = \begin{bmatrix}6 & 8 \\ 2 & 4\end{bmatrix}$.
$B = \begin{bmatrix}10 & 14 \\ 6 & 8\end{bmatrix} - \begin{bmatrix}6 & 8 \\ 2 & 4\end{bmatrix} = \begin{bmatrix}4 & 6 \\ 4 & 4\end{bmatrix}$.
Final Answer: $B = \begin{bmatrix}4 & 6 \\ 4 & 4\end{bmatrix}$
Q17
2021
00:00
If $A = \begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}$, find $A^2$.
Multiply $A$ by itself.
$A^2 = \begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}$.
Each element = $1+1 = 2$.
So $A^2 = \begin{bmatrix}2 & 2 \\ 2 & 2\end{bmatrix}$.
Final Answer: $A^2 = \begin{bmatrix}2 & 2 \\ 2 & 2\end{bmatrix}$
Q18
2020
00:00
If $A = \begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}$, find $3A - 2I$.
Compute $3A = \begin{bmatrix}3 & 6 \\ 9 & 12\end{bmatrix}$.
$2I = \begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}$.
Subtract: $3A - 2I = \begin{bmatrix}1 & 6 \\ 9 & 10\end{bmatrix}$.
Final Answer: $3A - 2I = \begin{bmatrix}1 & 6 \\ 9 & 10\end{bmatrix}$
Q19
2023
00:00
If $A^2 = A$, where $A$ is a square matrix, find $A^3$.
Given $A^2 = A$.
Multiply both sides by $A$.
$A^3 = A^2$.
But $A^2 = A$.
Hence, $A^3 = A$.
Final Answer: $A^3 = A$
Q20
2021
00:00
Find $A$ if $2A + 3I = \begin{bmatrix}7 & 0 \\ 0 & 7\end{bmatrix}$.
Given $2A = \begin{bmatrix}7 & 0 \\ 0 & 7\end{bmatrix} - 3I$.
$3I = \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix}$.
So $2A = \begin{bmatrix}4 & 0 \\ 0 & 4\end{bmatrix}$.
Divide by 2: $A = \begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}$.
Final Answer: $A = \begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}$
Q21
2019
00:00
Find a matrix $A$ such that $2A - 3B + 5C = O$, where $B = \begin{bmatrix}-2 & 2 \\ 3 & 1\end{bmatrix}$ and $C = \begin{bmatrix}2 & 0 \\ 7 & 1\end{bmatrix}$.
Given $2A - 3B + 5C = O$.
So $2A = 3B - 5C$.
Compute $3B = \begin{bmatrix}-6 & 6 \\ 9 & 3\end{bmatrix}$ and $5C = \begin{bmatrix}10 & 0 \\ 35 & 5\end{bmatrix}$.
Subtract: $3B - 5C = \begin{bmatrix}-16 & 6 \\ -26 & -2\end{bmatrix}$.
Divide by 2: $A = \begin{bmatrix}-8 & 3 \\ -13 & -1\end{bmatrix}$.
Final Answer: $A = \begin{bmatrix}-8 & 3 \\ -13 & -1\end{bmatrix}$