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If A = \(\begin{bmatrix}ab & b^2 \\[0.3em]-a^2 & -ab \\[0.3em]\end{bmatrix}\), show that A2 = 0 |
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Answer» Given, A = \(\begin{bmatrix} ab & b^2 \\[0.3em] -a^2 & -ab \\[0.3em] \end{bmatrix}\) A2 = \(\begin{bmatrix} ab & b^2 \\[0.3em] -a^2 & -ab \\[0.3em] \end{bmatrix}\)\(\begin{bmatrix} ab & b^2 \\[0.3em] -a^2 & -ab \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix} a^2b^2-a^2b^2 & ab^3-ab^3 \\[0.3em] -a^3b+a^3b & -a^2b^2+a^2b^2 \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix} 0 &0 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}\) = 0 Hence, A2 = 0 |
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