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Q.If A=[10−21]A = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} and B=[−510−10−5]B = \begin{bmatrix} -5 & 10 \\ -10 & -5 \end{bmatrix}, then ABAB is :

(a) [−5100−5]\begin{bmatrix} -5 & 10 \\ 0 & -5 \end{bmatrix}
(b) [0−52510]\begin{bmatrix} 0 & -5 \\ 25 & 10 \end{bmatrix}
(c) [10−25−50]\begin{bmatrix} 10 & -25 \\ -5 & 0 \end{bmatrix}
(d) [−5100−25]\begin{bmatrix} -5 & 10 \\ 0 & -25 \end{bmatrix}
CBSECBSE Class XII Board 2023MCQ· 1mImportance★★★★★
✓ Free question

Standard row-into-column matrix product gives AB=[−5100−25]AB=\begin{bmatrix}-5&10\\0&-25\end{bmatrix}.

For 2×22\times2 matrices, (AB)ij=∑kAikBkj(AB)_{ij}=\sum_k A_{ik}B_{kj} — each entry is a row of AA dotted with a column of BB.

  1. A=[10−21]A=\begin{bmatrix}1&0\\-2&1\end{bmatrix}, B=[−510−10−5]B=\begin{bmatrix}-5&10\\-10&-5\end{bmatrix}.
  2. (AB)11=1(−5)+0(−10)=−5(AB)_{11}=1(-5)+0(-10)=-5; (AB)12=1(10)+0(−5)=10(AB)_{12}=1(10)+0(-5)=10.
  3. (AB)21=(−2)(−5)+1(−10)=10−10=0(AB)_{21}=(-2)(-5)+1(-10)=10-10=0; (AB)22=(−2)(10)+1(−5)=−20−5=−25(AB)_{22}=(-2)(10)+1(-5)=-20-5=-25.
  4. Thus AB=[−5100−25]AB=\begin{bmatrix}-5&10\\0&-25\end{bmatrix}, matching option (d).
✓Final answer

(d) [−5100−25]\begin{bmatrix}-5 & 10 \\ 0 & -25\end{bmatrix}

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