If a matrix $A$ satisfies the equation $A^3-6A^2+11A-6I=0$,then $A^{-1}$ can be expressed in terms of $A$ as:
- A$\frac{1}{6}(A^2-6A+11I)$
- B$\frac{1}{6}(A^2+6A-11I)$
- C$\frac{1}{6}(-A^2+6A-11I)$
- D$\frac{1}{6}(A^2-6A-11I)$
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Similar Questions
If $A = \begin{bmatrix} 0 & 1+2i & i-2 \\ -1-2i & 0 & K \\ 2-i & -7 & 0 \end{bmatrix}$ and $A^{-1}$ does not exist,then $K = $ (where $i = \sqrt{-1}$)
If $A$ is a square matrix of order $3$,then consider the following statements.
$I$. If $|A|=0$,then $|\operatorname{Adj} A|=0$
$II$. If $|A| \neq 0$,then $|A^{-1}|=|A|^{-1}$
Which of the above statements is/are true?
$I$. If $|A|=0$,then $|\operatorname{Adj} A|=0$
$II$. If $|A| \neq 0$,then $|A^{-1}|=|A|^{-1}$
Which of the above statements is/are true?
Find $adj$ $A$ for $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$.
Easy
View SolutionIf $A=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right]$ and $A^{-1}=x A+y I$,where $I$ is the unit matrix of order $2$,then the values of $x$ and $y$ are respectively:
If $A=\left[\begin{array}{cc}2 & 3 \\ 1 & -4\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & -2 \\ -1 & 3\end{array}\right],$ then verify that $(AB)^{-1}=B^{-1} A^{-1}$.
Medium
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