ExamBro
ExamBro
KCET · Maths · Matrices

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 \(\mathrm{A}^{-1}=\)

  1. A \(\frac{1}{7}(5 \mathrm{I}-\mathrm{A})\)
  2. B \(\frac{1}{7}(\mathrm{~A}-5 \mathrm{I})\)
  3. C \(7(5 I-A)\)
  4. D \(\frac{1}{5}(7 \mathrm{I}-\mathrm{A})\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{7}(5 \mathrm{I}-\mathrm{A})\)

Step-by-step Solution

Detailed explanation

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