Question:
Matrix A is non-singular matrix and \((A-3 I)(A-5 I)=0\), then \(\frac{15}{8} A^{-1}=\ldots \ldots\)
Matrix A is non-singular matrix and \((A-3 I)(A-5 I)=0\), then \(\frac{15}{8} A^{-1}=\ldots \ldots\)
Show Hint
Expand the given product and multiply the resulting equation by the inverse of A.
Updated On: Nov 13, 2026
\(\mathrm{I}-8\text{ }\mathrm{A}\)
\(2\mathrm{I}-\frac{1}{15}\text{ }\mathrm{A}\)
\(\mathrm{I}-\frac{1}{8}\text{ }\mathrm{A}\)
\(8\mathrm{I}-15\text{ }\mathrm{A}\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
The relation \((A-3I)(A-5I)=0\) is a quadratic matrix equation in A. Multiplying by \(A^{-1}\) lets us express \(A^{-1}\) in terms of \(A\) and \(I\).
Step 2: Expand
\(A^{2}-5A-3A+15I=0\), so \(A^{2}-8A+15I=0\).
Step 3: Multiply by the inverse
Multiply by \(A^{-1}\) (A is non-singular): \(A-8I+15A^{-1}=0\), so \(15A^{-1}=8I-A\).
Step 4: Calculation
Divide by 8: \(\dfrac{15}{8}A^{-1}=I-\dfrac{1}{8}A\).
Final Answer:
The expression equals \(I-\frac{1}{8}A\). \[ \boxed{\text{(C) }\mathrm{I}-\dfrac{1}{8}\mathrm{A}} \]
The relation \((A-3I)(A-5I)=0\) is a quadratic matrix equation in A. Multiplying by \(A^{-1}\) lets us express \(A^{-1}\) in terms of \(A\) and \(I\).
Step 2: Expand
\(A^{2}-5A-3A+15I=0\), so \(A^{2}-8A+15I=0\).
Step 3: Multiply by the inverse
Multiply by \(A^{-1}\) (A is non-singular): \(A-8I+15A^{-1}=0\), so \(15A^{-1}=8I-A\).
Step 4: Calculation
Divide by 8: \(\dfrac{15}{8}A^{-1}=I-\dfrac{1}{8}A\).
Final Answer:
The expression equals \(I-\frac{1}{8}A\). \[ \boxed{\text{(C) }\mathrm{I}-\dfrac{1}{8}\mathrm{A}} \]
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