Question:
If A is a matrix of order 2 and I is the identity matrix of order 2 such that $A^{2}-4A+3I=0$ then $(A+3I)^{-1}=$
If A is a matrix of order 2 and I is the identity matrix of order 2 such that $A^{2}-4A+3I=0$ then $(A+3I)^{-1}=$
Show Hint
To find $(A+kI)^{-1}$, substitute $A = B-kI$ into the original equation and solve for $B^{-1}$.
Updated On: Jun 19, 2026
- $\frac{A}{24}-\frac{7}{24}I$
- $\frac{A}{21}-\frac{7}{21}I$
- $\frac{7I}{24}-\frac{1}{24}A$
- $A-3I$
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Concept
Use the characteristic equation to find the inverse by manipulating the given matrix equation.
Step 2: Analysis
Let $B = A + 3I$. Then $A = B - 3I$. Substitute this into $A^2 - 4A + 3I = 0$: $(B-3I)^2 - 4(B-3I) + 3I = 0$
Step 3: Calculation
$B^2 - 6B + 9I - 4B + 12I + 3I = 0$
$B^2 - 10B + 24I = 0$
Divide by $B$: $B - 10I + 24B^{-1} = 0$
$24B^{-1} = 10I - B = 10I - (A + 3I) = 7I - A$
$B^{-1} = \frac{7I - A}{24}$
Step 4: Conclusion
Hence, $(A+3I)^{-1} = \frac{7}{24}I - \frac{1}{24}A$. Final Answer: (C)
Use the characteristic equation to find the inverse by manipulating the given matrix equation.
Step 2: Analysis
Let $B = A + 3I$. Then $A = B - 3I$. Substitute this into $A^2 - 4A + 3I = 0$: $(B-3I)^2 - 4(B-3I) + 3I = 0$
Step 3: Calculation
$B^2 - 6B + 9I - 4B + 12I + 3I = 0$
$B^2 - 10B + 24I = 0$
Divide by $B$: $B - 10I + 24B^{-1} = 0$
$24B^{-1} = 10I - B = 10I - (A + 3I) = 7I - A$
$B^{-1} = \frac{7I - A}{24}$
Step 4: Conclusion
Hence, $(A+3I)^{-1} = \frac{7}{24}I - \frac{1}{24}A$. Final Answer: (C)
Was this answer helpful?
0
0
Top MHT CET Mathematics Questions
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- If $\int\frac{f\left(x\right)}{log \left(sin\,x\right)}dx = log\left[log\,sin\,x\right]+c$ then $f\left(x\right)=$
- If $\int\limits^{K}_0 \frac{dx}{2 + 18 x^2} = \frac{\pi}{24}$, then the value of K is
- If $\int^{\pi/2}_{0} \log\cos x dx =\frac{\pi}{2} \log\left(\frac{1}{2}\right)$ then $ \int^{\pi/2}_{0} \log\sec x dx = $
- The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0 $ is
View More Questions
Top MHT CET Invertible Matrices Questions
- Consider two matrices \(A=\begin{bmatrix}3 & -4 1 & -1\end{bmatrix}\) and \(B=\begin{bmatrix}6 & -13 5 & -10\end{bmatrix}\). If the following matrix equation holds true: \[ \left((A^{-1})^2 + B\right)\begin{bmatrix}x y\end{bmatrix} = \begin{bmatrix}0 0\end{bmatrix} \] Find the values of \(x\) and \(y\).
- If $A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}, B = \begin{bmatrix} 1 & 1 \\ 4 & -1 \end{bmatrix}$, then $(A+B)^{-1}$ is
- If $A=\begin{bmatrix}2 & -1\\ -1 & 3\end{bmatrix}$, then the inverse of $(2A^{2}+5A)$ is
- If $A = \begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$ then $A(I + \text{adj } A) =$
- If $A = \begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$ then $A(I + \text{adj } A) =$
View More Questions
Top MHT CET Questions
- During r- DNA technology, which one of the following enzymes is used for cleaving DNA molecule ?
- In non uniform circular motion, the ratio of tangential to radial acceleration is (r = radius of circle, $v =$ speed of the particle, $\alpha =$ angular acceleration)
- The temperature of $32^{\circ}C$ is equivalent to
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- The heat of formation of water is $ 260\, kJ $ . How much $ H_2O $ is decomposed by $ 130\, kJ $ of heat ?
View More Questions