Q.Compute the following:

Showing the 12 most recent of 13 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $A = \begin{bmatrix}1 & 2 & 3\\ 2 & 3 & 1\end{bmatrix}$ and $B = \begin{bmatrix}3 & -1 & 3\\ -1 & 0 & 2\end{bmatrix}$, then the $(2A - B)$ will be

   (a) $\begin{bmatrix}-1 & 5 & 3\\ 5 & 6 & 0\end{bmatrix}$

   (b) $\begin{bmatrix}5 & -1 & 3\\ 5 & 6 & 0\end{bmatrix}$

   (c) $\begin{bmatrix}2 & 4 & 6\\ 4 & 6 & 2\end{bmatrix}$

   (d) $\begin{bmatrix}-3 & 1 & -3\\ 5 & 6 & 2\end{bmatrix}$

   ›Reveal solutionSolution

   Scale $A$ by 2 and subtract $B$ entry-by-entry.

   $2A = \begin{bmatrix}2&4&6\\4&6&2\end{bmatrix}$.

   $2A-B = \begin{bmatrix}2-3 & 4-(-1) & 6-3\\ 4-(-1) & 6-0 & 2-2\end{bmatrix} = \begin{bmatrix}-1&5&3\\5&6&0\end{bmatrix}$.

   ✓Final answer

   The correct option is (a) $\begin{bmatrix}-1&5&3\\5&6&0\end{bmatrix}$.

2. CBSE 2026Set ANNUAL1 mark
   Q.If A = \begin{pmatrix}1 & 2 & 3\ 2 & 3 & 1\end{pmatrix} and B = \begin{pmatrix}3 & -1 & 3\ -1 & 0 & 2\end{pmatrix}, then find the value of 2A - B.

   ›Reveal solutionSolution

   Scale $A$ by $2$ (multiply every entry by 2), then subtract $B$ entry-by-entry.

   Working:

   $A = \begin{pmatrix}1 & 2 & 3\\ 2 & 3 & 1\end{pmatrix},\quad 2A = \begin{pmatrix}2 & 4 & 6\\ 4 & 6 & 2\end{pmatrix}$

   $2A - B = \begin{pmatrix}2 & 4 & 6\\ 4 & 6 & 2\end{pmatrix} - \begin{pmatrix}3 & -1 & 3\\ -1 & 0 & 2\end{pmatrix} = \begin{pmatrix}2-3 & 4-(-1) & 6-3\\ 4-(-1) & 6-0 & 2-2\end{pmatrix}$

   $= \begin{pmatrix}-1 & 5 & 3\\ 5 & 6 & 0\end{pmatrix}$

   ✓Final answer

   $2A-B = \begin{pmatrix}-1 & 5 & 3\\ 5 & 6 & 0\end{pmatrix}$.

3. CBSE 2024Set ANNUAL1 markMCQ
   Q.If $A = \begin{bmatrix}1 & 2 & 3\\ 2 & 3 & 1\end{bmatrix}$ and $B = \begin{bmatrix}3 & -1 & 3\\ -1 & 0 & 2\end{bmatrix}$, then $(2A-B)$ will be:

   (a) $\begin{bmatrix}1 & -5 & 2\\ 5 & 6 & 0\end{bmatrix}$

   (b) $\begin{bmatrix}5 & 6 & 0\\ 1 & -5 & 3\end{bmatrix}$

   (c) $\begin{bmatrix}-1 & 5 & 3\\ 5 & 6 & 0\end{bmatrix}$

   (d) $\begin{bmatrix}-1 & 3 & 5\\ 5 & 6 & 0\end{bmatrix}$

   ›Reveal solutionSolution

   Scale $A$ by 2 entrywise, then subtract $B$ entrywise.

   $A = \begin{bmatrix}1 & 2 & 3\\ 2 & 3 & 1\end{bmatrix}$, so $2A = \begin{bmatrix}2 & 4 & 6\\ 4 & 6 & 2\end{bmatrix}$.

   $B = \begin{bmatrix}3 & -1 & 3\\ -1 & 0 & 2\end{bmatrix}$.

   $2A - B = \begin{bmatrix}2-3 & 4-(-1) & 6-3\\ 4-(-1) & 6-0 & 2-2\end{bmatrix} = \begin{bmatrix}-1 & 5 & 3\\ 5 & 6 & 0\end{bmatrix}$

   ✓Final answer

   (c) $\begin{bmatrix}-1 & 5 & 3\\ 5 & 6 & 0\end{bmatrix}$

4. CBSE 2024Set ANNUAL1 mark
   Q.If A = [[2, 4], [3, 2]], B = [[-2, 5], [3, 4]] then find 3A - B.

   ›Reveal solutionSolution

   Scale $A$ by 3, then subtract $B$ entrywise.

   Given $A=\begin{pmatrix}2&4\\3&2\end{pmatrix}$ and $B=\begin{pmatrix}-2&5\\3&4\end{pmatrix}$.

   $3A=\begin{pmatrix}6&12\\9&6\end{pmatrix}$

   $3A-B=\begin{pmatrix}6-(-2)&12-5\\9-3&6-4\end{pmatrix}=\begin{pmatrix}8&7\\6&2\end{pmatrix}$

   ✓Final answer

   $3A - B = \begin{pmatrix}8 & 7\\ 6 & 2\end{pmatrix}$

5. CBSE 2024Set ANNUAL1 markMCQ
   Q.Let [ILLEGIBLE — page defect obscures the matrix's own name/label] = [[2, 4], [3, 2]], C = [[−2, 5], [3, 4]], the value of [ILLEGIBLE — page defect obscures the operator between the two matrices] C is :

   (a) [[0, 9], [6, 6]]

   (b) [[0, 6], [?, 6]] — one cell illegible (page defect)

   (c) [[?, 7], [?, 2]] — two cells illegible (page defect)

   (d) None of these

   ›Reveal solutionSolution

   Although the matrix's own name/label and the operator symbol between the two matrices are obscured by a scan defect, the arithmetic can be recovered: treating the given matrix and C as being added, the sum matches option (a) exactly.

   Note on the source scan: the label of the first matrix (e.g. 'A' or 'B') and the operator between the two matrices are lost to a page-defect blot. Given matrix (unnamed) = [[2, 4], [3, 2]] and C = [[−2, 5], [3, 4]].

   Testing addition: [[2 + (−2), 4 + 5], [3 + 3, 2 + 4]] = [[0, 9], [6, 6]]. This matches option (a) exactly — so the intended operation was addition (the '+' between the two matrices, degraded by the same blot).

   ✓Final answer

   [[0, 9], [6, 6]] — option (a) (recovered by testing which operation reproduces a printed option).

6. CBSE 2024Set ANNUAL1 markMCQ
   Q.If $\begin{bmatrix}3 & -4\\ 1 & 2\end{bmatrix} + X = \begin{bmatrix}3 & 1\\ 5 & 0\end{bmatrix}$, then the matrix X is ............... .

   (a) $\begin{bmatrix}6 & 5\\ 4 & 4\end{bmatrix}$

   (b) $\begin{bmatrix}0 & 5\\ 4 & 2\end{bmatrix}$

   (c) $\begin{bmatrix}0 & 5\\ 4 & -2\end{bmatrix}$

   (d) $\begin{bmatrix}0 & 5\\ -4 & -2\end{bmatrix}$

   ›Reveal solutionSolution

   Isolate X by subtracting the given matrix from the right-hand side matrix.

   Given $\begin{bmatrix}3 & -4\\ 1 & 2\end{bmatrix} + X = \begin{bmatrix}3 & 1\\ 5 & 0\end{bmatrix}$

   $X = \begin{bmatrix}3 & 1\\ 5 & 0\end{bmatrix} - \begin{bmatrix}3 & -4\\ 1 & 2\end{bmatrix} = \begin{bmatrix}3-3 & 1-(-4)\\ 5-1 & 0-2\end{bmatrix} = \begin{bmatrix}0 & 5\\ 4 & -2\end{bmatrix}$

   ✓Final answer

   $X = \begin{bmatrix}0 & 5\\ 4 & -2\end{bmatrix}$ (option c).

7. CBSE 2023Set ANNUAL1 mark
   Q.If A = [[1, 3], [-2, 5]] and B = [[2, 4], [3, 2]] then find A + B.

   ›Reveal solutionSolution

   Add corresponding entries of the two matrices.

   Matrix addition is defined entry-by-entry: if $A=[a_{ij}]$ and $B=[b_{ij}]$ are matrices of the same order, then $A+B=[a_{ij}+b_{ij}]$.

   Here $A = [[1, 3], [-2, 5]]$ and $B = [[2, 4], [3, 2]]$.

   Adding corresponding entries:

   • Row 1: $1+2=3$, $3+4=7$

   • Row 2: $-2+3=1$, $5+2=7$

   ✓Final answer

   $A+B = [[3, 7], [1, 7]]$

8. CBSE 2022Set ANNUAL1 markMCQ
   Q.If $A=\begin{bmatrix}1&2\\0&1\end{bmatrix}$ and $B=\begin{bmatrix}2&1\\-1&0\end{bmatrix}$, then find $2A+3B$.

   (a) $\begin{bmatrix}6&5\\0&2\end{bmatrix}$

   (b) $\begin{bmatrix}8&7\\-3&2\end{bmatrix}$

   (c) $\begin{bmatrix}7&7\\-3&2\end{bmatrix}$

   (d) $\begin{bmatrix}8&7\\-2&3\end{bmatrix}$

   ›Reveal solutionSolution

   Scale each matrix entrywise, then add corresponding entries.

   $A=\begin{bmatrix}1&2\\0&1\end{bmatrix}$, $B=\begin{bmatrix}2&1\\-1&0\end{bmatrix}$.

   $2A=\begin{bmatrix}2&4\\0&2\end{bmatrix}$, $3B=\begin{bmatrix}6&3\\-3&0\end{bmatrix}$.

   $2A+3B=\begin{bmatrix}2+6&4+3\\0-3&2+0\end{bmatrix}=\begin{bmatrix}8&7\\-3&2\end{bmatrix}$.

   ✓Final answer

   (b) $\begin{bmatrix}8&7\\-3&2\end{bmatrix}$.

9. CBSE 2022Set ANNUAL1 mark
   Q.If $A = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$, then $A - B =$ ______.

   ›Reveal solutionSolution

   Subtract matrices entry-by-entry.

   $A=\begin{bmatrix}1 & 3\\ -2 & 5\end{bmatrix}$, $B=\begin{bmatrix}2 & 4\\ 3 & 2\end{bmatrix}$.

   $A-B = \begin{bmatrix}1-2 & 3-4\\ -2-3 & 5-2\end{bmatrix} = \begin{bmatrix}-1 & -1\\ -5 & 3\end{bmatrix}$.

   ✓Final answer

   $A-B = \begin{bmatrix}-1 & -1\\ -5 & 3\end{bmatrix}$.

10. CBSE 2022Set ANNUAL1 mark
    Q.If matrix $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$, then find the value of $3A - B$.

    ›Reveal solutionSolution

    Scale $A$ by 3 entrywise, then subtract $B$ entrywise.

    Given $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$, $B = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$.

    Step 1: Compute $3A$ by multiplying every entry by 3:

    $3A = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix}$

    Step 2: Subtract $B$ entrywise (matrices must have the same order to add/subtract, which they do here, both $2\times2$):

    $3A - B = \begin{bmatrix} 6-(-2) & 12-5 \\ 9-3 & 6-4 \end{bmatrix} = \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}$

    ✓Final answer

    $3A - B = \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}$

11. CBSE 2021Set I1 markMCQ
    Q.$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} =$

    (a) $\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

    (b) $\begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}$

    (c) $\begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$

    (d) $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$

    ›Reveal solutionSolution

    Matrix addition is component-wise.

    Add corresponding entries:

    $\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}+\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}=\begin{bmatrix}1+0 & 0+1\\ 0+1 & 1+0\end{bmatrix}=\begin{bmatrix}1 & 1\\ 1 & 1\end{bmatrix}$.

    ✓Final answer

    The correct option is (d) $\begin{bmatrix}1 & 1\\ 1 & 1\end{bmatrix}$.

12. CBSE 2020Set ANNUAL1 mark
    Q.If A = [[1, 3], [−2, 5]] and B = [[−2, 5], [3, 4]], find the value of A − B.

    ›Reveal solutionSolution

    Subtract corresponding entries of the two matrices.

    $A-B = \begin{bmatrix}1&3\\-2&5\end{bmatrix} - \begin{bmatrix}-2&5\\3&4\end{bmatrix} = \begin{bmatrix}1-(-2) & 3-5\\ -2-3 & 5-4\end{bmatrix} = \begin{bmatrix}3 & -2\\ -5 & 1\end{bmatrix}$

    ✓Final answer

    $A-B=\begin{bmatrix}3&-2\\-5&1\end{bmatrix}$.