Find the inverse,by elementary row operations (if possible),of the following matrix: $\left[\begin{array}{cc}1 & 3 \\ -5 & 7\end{array}\right]$
(D) Suppose $A = \left[\begin{array}{cc}1 & 3 \\ -5 & 7\end{array}\right]$.
To find the inverse using elementary row operations,we use the relation $A = IA$.
$\left[\begin{array}{cc}1 & 3 \\ -5 & 7\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] A$
Apply $R_2 \rightarrow R_2 + 5R_1$:
$\left[\begin{array}{cc}1 & 3 \\ 0 & 22\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ 5 & 1\end{array}\right] A$
Apply $R_2 \rightarrow \frac{1}{22} R_2$:
$\left[\begin{array}{cc}1 & 3 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ \frac{5}{22} & \frac{1}{22}\end{array}\right] A$
Apply $R_1 \rightarrow R_1 - 3R_2$:
$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}1 - 3(\frac{5}{22}) & 0 - 3(\frac{1}{22}) \\ \frac{5}{22} & \frac{1}{22}\end{array}\right] A$
$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}\frac{7}{22} & -\frac{3}{22} \\ \frac{5}{22} & \frac{1}{22}\end{array}\right] A$
Thus,$A^{-1} = \frac{1}{22} \left[\begin{array}{cc}7 & -3 \\ 5 & 1\end{array}\right]$.
To find the inverse using elementary row operations,we use the relation $A = IA$.
$\left[\begin{array}{cc}1 & 3 \\ -5 & 7\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] A$
Apply $R_2 \rightarrow R_2 + 5R_1$:
$\left[\begin{array}{cc}1 & 3 \\ 0 & 22\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ 5 & 1\end{array}\right] A$
Apply $R_2 \rightarrow \frac{1}{22} R_2$:
$\left[\begin{array}{cc}1 & 3 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}1 & 0 \\ \frac{5}{22} & \frac{1}{22}\end{array}\right] A$
Apply $R_1 \rightarrow R_1 - 3R_2$:
$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}1 - 3(\frac{5}{22}) & 0 - 3(\frac{1}{22}) \\ \frac{5}{22} & \frac{1}{22}\end{array}\right] A$
$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}\frac{7}{22} & -\frac{3}{22} \\ \frac{5}{22} & \frac{1}{22}\end{array}\right] A$
Thus,$A^{-1} = \frac{1}{22} \left[\begin{array}{cc}7 & -3 \\ 5 & 1\end{array}\right]$.
Explore More
Similar Questions
If $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$, then which of the following is true?
DifficultMHT CET 2026
View SolutionIf $A$ and $B$ are square matrices of order $3$ such that $|A|=2$ and $|B|=4$,then $|A(\operatorname{adj} B)| = \dots$
If $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$ and $A \cdot \text{adj}(A) = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$,then $k$ is equal to
Medium
View SolutionIf possible,using elementary row transformations,find the inverse of the following matrix:
$\left[\begin{array}{ccc}2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{array}\right]$
$\left[\begin{array}{ccc}2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{array}\right]$
Medium
View SolutionLet $A = \begin{bmatrix} 1 & 1 & 2 \\ -2 & 0 & 1 \\ 1 & 3 & 5 \end{bmatrix}$. Then the sum of all elements of the matrix $\text{adj}(\text{adj}(2(\text{adj} A)^{-1}))$ is equal to:
DifficultJEE MAIN 2026
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo