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If A=[10001023−1]A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 3 & -1 \end{bmatrix}, then A2A^2 is equal to

Solution

✅ Correct Option: 3

The first two rows of AA are those of the identity, so they are unchanged in A2A^2. The third row of A2A^2 is 2R1+3R2−R32R_1+3R_2-R_3 of AA, which is (2,0,0)+(0,3,0)+(−2,−3,1)=(0,0,1)(2,0,0)+(0,3,0)+(-2,-3,1)=(0,0,1). Hence A2=IA^2=I, a unit matrix of order 3.

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