Q91 / 180 · KCET 2025 · Mathematics

If $A$ is a square matrix satisfying the equation $A^2-5 A+7 I=0$, where $I$ is the $I$ dentity matrix and 0 is null matrix of same order, then $\mathrm{A}^{-1}=$

  • Correct answer
    A
    $\frac{1}{7}(5 \mathrm{I}-\mathrm{A})$
  • B
    $\frac{1}{7}(\mathrm{~A}-5 \mathrm{I})$
  • C
    $7(5 I-A)$
  • D
    $\frac{1}{5}(7 \mathrm{I}-\mathrm{A})$

Solution

Given $\begin{aligned} & \mathrm{A}^2-5 \mathrm{~A}+7 \mathrm{I}=0 \quad\left(\because \text { Multiply by } \mathrm{A}^{-1} \text { both sides }|\mathrm{A}| \neq 0\right) \\ & \mathrm{A}-5 \mathrm{I}+7 \mathrm{~A}^{-1}=0 \\ & \mathrm{~A}^{-1}=\frac{(5 \mathrm{I}-\mathrm{A})}{7} \end{aligned}$

Hence, the correct answer is option: 'A'.

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