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If A=[2−3−5−1451−3−4]A = \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix}, then A2A^2 is equal to :

Solution

✅ Correct Option: 1

Multiplying AA by itself, the first row gives 4+3−5=24+3-5 = 2, −6−12+15=−3-6-12+15 = -3, −10−15+20=−5-10-15+20 = -5, and the remaining rows reproduce themselves similarly. So A2=AA^2 = A, i.e. AA is idempotent.

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