NCERT Solutions for Class 12 Maths Matrices Exercise 3.2

Fundamental Operations on Matrices

To solve the problems in Exercise 3.2 of the NCERT textbook, you must fully master four fundamental matrix operations:

1. Matrix Addition and Subtraction: Two matrices $A$ and $B$ can be added or subtracted if and only if they have the same order (i.e., same number of rows and columns). The resulting matrix is formed by adding or subtracting the corresponding elements: $(A \pm B)_{ij} = a_{ij} \pm b_{ij}$.

1. Scalar Multiplication: When a matrix $A$ is multiplied by a real number (scalar) $k$, every single element inside the matrix is multiplied by $k$. Thus, $(kA)_{ij} = k \cdot a_{ij}$.

1. Matrix Multiplication: This is the most critical and complex operation. Two matrices $A$ and $B$ can be multiplied to form $AB$ if and only if the number of columns in $A$ equals the number of rows in $B$. If $A$ is of order $m \times n$ and $B$ is of order $n \times p$, then the product matrix $C = AB$ will have the order $m \times p$.

Step-by-Step Matrix Multiplication Method

1. Check Compatibility — Before multiplying, verify the orders. If Matrix A is $m \times n$ and Matrix B is $r \times p$, multiplication is possible only if $n = r$. The resulting matrix will have order $m \times p$.
2. Identify Row and Column Vectors — To find the element at position $(i, j)$ in the product matrix, you must multiply the $i$-th row of Matrix A with the $j$-th column of Matrix B.
3. Multiply and Accumulate — Multiply corresponding elements of the selected row and column, then sum them up: $c_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + \dots + a_{in}b_{nj}$.
4. Repeat for All Cells — Perform this operation systematically for every row of Matrix A combined with every column of Matrix B to fill the product matrix.

Worked Examples from Exercise Concepts

• Example 1: Let $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$. Find $2A + B$. Solution Step-by-Step: Step 1: Find $2A$ by multiplying each element of $A$ by 2. $2A = \begin{bmatrix} 2(2) & 2(4) \\ 2(3) & 2(2) \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 6 & 4 \end{bmatrix}$. Step 2: Add $2A$ and $B$ by summing their corresponding elements. $2A + B = \begin{bmatrix} 4 + 1 & 8 + 3 \\ 6 + (-2) & 4 + 5 \end{bmatrix} = \begin{bmatrix} 5 & 11 \\ 4 & 9 \end{bmatrix}$.
• Example 2: Find $AB$ if $A = \begin{bmatrix} 6 & 9 \\ 2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 6 & 0 \\ 7 & 9 & 8 \end{bmatrix}$. Solution Step-by-Step: Step 1: Check compatibility. $A$ is $2 \times 2$ and $B$ is $2 \times 3$. Since the columns of $A$ (2) equal the rows of $B$ (2), multiplication is defined. The resulting matrix will be of order $2 \times 3$. Step 2: Compute $c_{11} = 6(2) + 9(7) = 12 + 63 = 75$. Step 3: Compute $c_{12} = 6(6) + 9(9) = 36 + 81 = 117$. Step 4: Compute $c_{13} = 6(0) + 9(8) = 0 + 72 = 72$. Step 5: Compute $c_{21} = 2(2) + 3(7) = 4 + 21 = 25$. Step 6: Compute $c_{22} = 2(6) + 3(9) = 12 + 27 = 39$. Step 7: Compute $c_{23} = 2(0) + 3(8) = 0 + 24 = 24$. Thus, $AB = \begin{bmatrix} 75 & 117 & 72 \\ 25 & 39 & 24 \end{bmatrix}$.

Crucial Exam Traps & Properties to Remember

• Matrix multiplication is NOT commutative: In general, $AB \neq BA$. Even if both products are defined, they are rarely equal. Always preserve the multiplication order!
• Zero Product Rule Property: In real algebra, if $ab = 0$, then either $a = 0$ or $b = 0$. However, in matrix algebra, the product of two non-zero matrices can be a zero matrix. Keep this exception in mind during proof-based questions.
• Distributive Law: Multiplication is distributive over addition: $A(B + C) = AB + AC$.
• Associative Law: $(AB)C = A(BC)$, provided the multiplications are defined.

Practice Questions with Solutions

• Q: If $X + Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}$ and $X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$, solve for $X$ and $Y$. A: Step 1: Add the two given matrix equations. $(X + Y) + (X - Y) = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$ $2X = \begin{bmatrix} 7+3 & 0+0 \\ 2+0 & 5+3 \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix}$ Step 2: Solve for $X$ by multiplying the scalar $\frac{1}{2}$. $X = \frac{1}{2} \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}$ Step 3: Substitute $X$ back into the first equation to find $Y$. $Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} - X = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} - \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}$ $Y = \begin{bmatrix} 7-5 & 0-0 \\ 2-1 & 5-4 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$ Final answer: $X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}$ and $Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$.
• Q: Simplify the expression: $\cos\theta \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta \begin{bmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{bmatrix}$. A: Step 1: Distribute $\cos\theta$ and $\sin\theta$ into their respective matrices. First term: $\begin{bmatrix} \cos^2\theta & \cos\theta\sin\theta \\ -\sin\theta\cos\theta & \cos^2\theta \end{bmatrix}$ Second term: $\begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix}$ Step 2: Add the corresponding terms of the two resulting matrices. Result $= \begin{bmatrix} \cos^2\theta + \sin^2\theta & \cos\theta\sin\theta - \sin\theta\cos\theta \\ -\sin\theta\cos\theta + \sin\theta\cos\theta & \cos^2\theta + \sin^2\theta \end{bmatrix}$ Step 3: Apply the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$. Result $= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$ (Identity Matrix) Final answer: $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
• Q: Given $A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$. Compute the matrix $A^2 - 5A + 6I$, where $I$ is the identity matrix of order 3. A: Step 1: Calculate $A^2 = A \cdot A$. $A^2 = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 2(2)+0(2)+1(1) & 2(0)+0(1)+1(-1) & 2(1)+0(3)+1(0) \\ 2(2)+1(2)+3(1) & 2(0)+1(1)+3(-1) & 2(1)+1(3)+3(0) \\ 1(2)+(-1)(2)+0(1) & 1(0)+(-1)(1)+0(-1) & 1(1)+(-1)(3)+0(0) \end{bmatrix}$ $A^2 = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}$ Step 2: Compute $5A$ and $6I$. $5A = \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix}$ and $6I = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$ Step 3: Put them together: $A^2 - 5A + 6I$. $= \begin{bmatrix} 5-10+6 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+6 & 5-15+0 \\ 0-5+0 & -1-(-5)+0 & -2-0+6 \end{bmatrix} = \begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}$ Final answer: $\begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}$
• Q: Find the value of $x$ and $y$ from the matrix equation: $2 \begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$. A: Step 1: Multiply the scalar 2 with the first matrix. $\begin{bmatrix} 2x & 10 \\ 14 & 2y-6 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$ Step 2: Add the LHS matrices. $\begin{bmatrix} 2x+3 & 6 \\ 15 & 2y-4 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$ Step 3: Compare corresponding entries to set up equations for variables. $2x + 3 = 7 \implies 2x = 4 \implies x = 2$ $2y - 4 = 14 \implies 2y = 18 \implies y = 9$ Final answer: $x = 2$ and $y = 9$.

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