If $A = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$,then $(A+B)^{-1} = $ . . . . . . .
- A$\frac{1}{25} I_3$
- B$\frac{1}{5} I_3$
- C$-\frac{1}{5} I_3$
- D$-\frac{1}{25} I_3$
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Similar Questions
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?
If $A = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$,then $\operatorname{adj} A = $
If $A$ is a non-singular matrix and $A^2 - A + I = 0$, then $A^{-1} = \dots$
DifficultMHT CET 2026
View SolutionIf $A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$ and $B^{-1} = \begin{bmatrix} 1/5 & 2/5 \\ 2/5 & -1/5 \end{bmatrix}$, then $(AB)^{-1} = $
DifficultMHT CET 2026
View SolutionGiven $A = \begin{bmatrix} x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z \end{bmatrix}$,$xyz = 60$ and $8x + 4y + 3z = 20$,then $A \cdot (\text{adj } A)$ is equal to
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