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Question

# If A and B are invertible matrices, then which one of the following is not correct? (a) adj A = $\left|A\right|$ A-1 (b) det (A-1) = [det(A)]-1 (c) (AB)-1 = B-1A-1 (d) (A+B)-1 = B-1 + A-1

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Solution

## (a) adj A = $\left|A\right|$ A−1 As we know, ${A}^{-1}=\frac{\mathrm{adj}A}{\left|A\right|}\phantom{\rule{0ex}{0ex}}⇒{A}^{-1}\left|A\right|=\mathrm{adj}A$ Thus, adj A = $\left|A\right|$ A−1 is correct. (b) det(A−1) = [det(A)]−1 As we know, $\left|{A}^{-1}\right|=\frac{1}{\left|A\right|}\phantom{\rule{0ex}{0ex}}⇒\left|{A}^{-1}\right|={\left|A\right|}^{-1}\phantom{\rule{0ex}{0ex}}$ Thus, det(A−1) = [det(A)]−1 is correct. (c) (AB)−1 = B−1A−1 As we know, By reversal law of inverse (AB)−1 = B−1A−1 Thus, (AB)−1 = B−1A−1 is correct. (d) (A + B)−1 = B−1 + A−1 ${\left(A+B\right)}^{-1}=\frac{1}{\left|A+B\right|}\mathrm{adj}\left(A+B\right)\phantom{\rule{0ex}{0ex}}\ne \frac{1}{\left|A\right|}\mathrm{adj}\left(A\right)+\frac{1}{\left|B\right|}\mathrm{adj}\left(B\right)\phantom{\rule{0ex}{0ex}}\ne {\mathrm{A}}^{-1}+{\mathrm{B}}^{-1}$ Thus, (A + B)−1 = B−1 + A−1 is incorrect. Hence, the correct option is (d).

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