By using elementary operations,find the inver | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To find the inverse using elementary row operations,we write $A=IA$.$\left[\begin{array}{rr}1 & 2 \ 2 & -1\end{array}ight]=\left[\begin{array}{

Vedclass · Accepted Answer

$A^{-1}=\left[\begin{array}{ll}\frac{1}{5} & \frac{2}{5} \ \frac{2}{5} & \frac{-1}{5}\end{array}ight]$

Vedclass · Answer

(A) To find the inverse using elementary row operations,we write $A=IA$.$\left[\begin{array}{rr}1 & 2 \ 2 & -1\end{array}ight]=\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}ight] A$Applying $R_{2} ightarrow R_{2}-2 R_{1}$:$\left[\begin{array}{rr}1 & 2 \ 0 & -5\end{array}ight]=\left[\begin{array}{rr}1 & 0 \ -2 & 1\end{array}ight] A$Applying $R_{2} ightarrow -\frac{1}{5} R_{2}$:$\left[\begin{array}{ll}1 & 2 \ 0 & 1\end{array}ight]=\left[\begin{array}{cc}1 & 0 \ \frac{2}{5} & \frac{-1}{5}\end{array}ight] A$Applying $R_{1} ightarrow R_{1}-2 R_{2}$:$\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}ight]=\left[\begin{array}{ll}\frac{1}{5} & \frac{2}{5} \ \frac{2}{5} & \frac{-1}{5}\end{array}ight] A$Thus,$A^{-1}=\left[\begin{array}{ll}\frac{1}{5} & \frac{2}{5} \ \frac{2}{5} & \frac{-1}{5}\end{array}ight]$.

By using elementary operations,find the inverse of the matrix $A=\left[\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right]$.

Similar Questions

If $A^2-A+I=0$,then the inverse of the matrix $A$ is

Inverse of the matrix $\begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}$ is

Suppose $A$ is any $3 \times 3$ non-singular matrix and $(A-3 I)(A-5 I)=O$,where $I=I_3$ and $O=O_3$. Here $O_3$ represents the zero matrix of order $3$ and $I_3$ is the identity matrix of order $3$. If $\alpha A+\beta A^{-1}=4 I$,then $\alpha+\beta$ is equal to:

If $Q$ is the inverse of $A$,where $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$ and $10Q = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & x \\ 1 & -2 & 3 \end{bmatrix}$,find $x$.

If $A = \begin{bmatrix} 1 & \tan(\theta/2) \\ -\tan(\theta/2) & 1 \end{bmatrix}$ and $AB = I$,then $B = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module