Question

If \( A = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \), \( B = \begin{bmatrix} -1 & 2 \\ 0 & 4 \end{bmatrix} \), then match the following and choose the correct answer.

\[ \begin{array}{ll} \text{List - I} & \text{List - II} \\ \\ (a)\; A+B & (i)\; \begin{bmatrix} -1 & 10 \\ 1 & 10 \end{bmatrix} \\ (b)\; A-B & (ii)\; \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} \\ (c)\; AB & (iii)\; \begin{bmatrix} 1 & 4 \\ 0 & 10 \end{bmatrix} \\ (d)\; 2A + B' & (iv)\; \begin{bmatrix} 2 & 0 \\ -1 & -1 \end{bmatrix} \end{array} \]

• A. a - ii, b - iv, c - i, d - iii
• B. a - iii, b - iv, c - i, d - ii
• C. a - iii, b - i, c - v, d - ii
• D. a - ii, b - i, c - iv, d - iii

Hint:

In matching-type matrix questions, compute the easiest expressions first. Matrix addition and subtraction often eliminate most incorrect options before matrix multiplication is even required.

Answer 1

The Correct Option is A

Concept:

Matrix addition and subtraction are performed element-wise. Matrix multiplication follows the row-by-column rule. The transpose of a matrix is obtained by interchanging rows and columns.

Given:

\[ A= \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}, \qquad B= \begin{bmatrix} -1 & 2 \\ 0 & 4 \end{bmatrix} \]

We evaluate each expression step by step.

Step 1: Find \(A + B\) \[ A+B= \begin{bmatrix} 1+(-1) & 2+2 \\ -1+0 & 3+4 \end{bmatrix} = \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} \]

This matches item (ii). Hence, \(a \rightarrow ii\).

Step 2: Find \(A - B\) \[ A-B= \begin{bmatrix} 1-(-1) & 2-2 \\ -1-0 & 3-4 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ -1 & -1 \end{bmatrix} \]

This matches item (iv). Hence, \(b \rightarrow iv\).

Step 3: Find \(AB\) \[ AB= \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} -1 & 2 \\ 0 & 4 \end{bmatrix} \]

Now compute each entry:

\[ AB= \begin{bmatrix} (1)(-1)+(2)(0) & (1)(2)+(2)(4) \\ (-1)(-1)+(3)(0) & (-1)(2)+(3)(4) \end{bmatrix} \] \[ = \begin{bmatrix} -1 & 10 \\ 1 & 10 \end{bmatrix} \]

This matches item (i). Hence, \(c \rightarrow i\).

Step 4: Find \(2A + B'\) \[ 2A= \begin{bmatrix} 2 & 4 \\ -2 & 6 \end{bmatrix}, \qquad B'= \begin{bmatrix} -1 & 0 \\ 2 & 4 \end{bmatrix} \] \[ 2A+B'= \begin{bmatrix} 2+(-1) & 4+0 \\ -2+2 & 6+4 \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 0 & 10 \end{bmatrix} \]

This matches item (iii). Hence, \(d \rightarrow iii\).

Step 5: Final Matching \[ a \rightarrow ii,\quad b \rightarrow iv,\quad c \rightarrow i,\quad d \rightarrow iii \] \[ \boxed{\text{a - ii,\; b - iv,\; c - i,\; d - iii}} \]

Final Answer: \(\boxed{(A)}\)