Q44 / 160 · TS EAMCET 2023 (13 May Shift 1) · Mathematics

If $A=\left[\begin{array}{ccc}1 & 2 & -1 \\ -1 & 0 & 2 \\ 1 & 2 & 0\end{array}\right]$ and $B=\left[\begin{array}{ccc}-3 & -2 & 4 \\ 2 & 2 & -1 \\ -2 & 0 & 3\end{array}\right]$, then $A^2=$

  • A
    $A-B$
  • B
    $B-A$
  • Correct answer
    C
    $A+B$
  • D
    $\mathrm{B}^2$

Solution

$A=\left[\begin{array}{ccc}1 & 2 & -1 \\ -1 & 0 & 2 \\ 1 & 2 & 0\end{array}\right], B=\left[\begin{array}{ccc}-3 & -2 & 4 \\ 2 & 2 & -1 \\ -2 & 0 & 3\end{array}\right]$ $\begin{aligned} & A^2=\left[\begin{array}{ccc}-2 & 0 & 3 \\ 1 & 2 & 1 \\ -1 & 2 & 3\end{array}\right] \text { and } A+B=\left[\begin{array}{ccc}-2 & 0 & 3 \\ 1 & 2 & 1 \\ -1 & 2 & 3\end{array}\right] \\ & \therefore A^2=A+B .\end{aligned}$

Hence, the correct answer is option: 'C'.

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