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

Step 1: Find $A+B$. Add corresponding elements: \[ A+B = \begin{bmatrix} 1+(-1) & 2+2 -1+0 & 3+4 \end{bmatrix} \] \[ = \begin{bmatrix} 0 & 4 -1 & 7 \end{bmatrix} \] This matches item (ii). Therefore, \[ a \rightarrow ii \]

Step 2: Find $A-B$. Subtract corresponding elements: \[ A-B = \begin{bmatrix} 1-(-1) & 2-2 -1-0 & 3-4 \end{bmatrix} \] \[ = \begin{bmatrix} 2 & 0 -1 & -1 \end{bmatrix} \] This matches item (iv). Therefore, \[ b \rightarrow iv \]

Step 3: Find $AB$. Using row-by-column multiplication: \[ AB= \begin{bmatrix} 1 & 2 -1 & 3 \end{bmatrix} \begin{bmatrix} -1 & 2 0 & 4 \end{bmatrix} \] First row, first column: \[ (A)(-1)+(B)(0)=-1 \] First row, second column: \[ (A)(B)+(B)(D)=2+8=10 \] Second row, first column: \[ (-1)(-1)+(C)(0)=1 \] Second row, second column: \[ (-1)(B)+(C)(D)=-2+12=10 \] Therefore, \[ AB= \begin{bmatrix} -1 & 10 1 & 10 \end{bmatrix} \] This matches item (i). Hence, \[ c \rightarrow i \]

Step 4: Find $2A+B'$. First calculate $2A$: \[ 2A= \begin{bmatrix} 2 & 4 -2 & 6 \end{bmatrix} \] Next calculate the transpose of $B$: \[ B'= \begin{bmatrix} -1 & 0 2 & 4 \end{bmatrix} \] Now add them: \[ 2A+B' = \begin{bmatrix} 2+(-1) & 4+0 -2+2 & 6+4 \end{bmatrix} \] \[ = \begin{bmatrix} 1 & 4 0 & 10 \end{bmatrix} \] This matches item (iii). Hence, \[ d \rightarrow iii \]

Step 5: Compile all matches. \[ a \rightarrow ii \] \[ b \rightarrow iv \] \[ c \rightarrow i \] \[ d \rightarrow iii \] Therefore, \[ {\text{a - ii,\; b - iv,\; c - i,\; d - iii}} \] which corresponds to \[ {(A)} \]