Aspire Faculty ID #15162 · Topic: JAMIA MCA 2017 · Just now
JAMIA MCA 2017

If $A = \left[\begin{array}{cc} x & 2 \\ 2 & x \end{array}\right]$ and $|A^2| = 0$, then $x$ is equal to

Solution

Since $|A^2| = |A|^2 = 0$, we have $|A| = 0$. $\therefore |A| = x^2 - 4 = 0 \Rightarrow x = \pm 2.$

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