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Find inverse of matrix A=\begin{bmatrix} \;\;1 &3 \\ -3 & 4 \end{bmatrix}

Option: 1

A^{-1}=\begin{bmatrix} \frac{7}{13} & \frac{2}{13}\\ \frac{3}{13} & \frac{1}{13} \end{bmatrix}


Option: 2

A^{-1}=\begin{bmatrix} \frac{4}{13} & \frac{3}{13}\\ \frac{3}{13} & \frac{1}{13} \end{bmatrix}


Option: 3

A^{-1}=\begin{bmatrix} \frac{7}{13} & -\frac{2}{13}\\ \frac{3}{13} & \frac{1}{13} \end{bmatrix}


Option: 4

A^{-1}=\begin{bmatrix} \frac{4}{13} & -\frac{3}{13}\\ \frac{3}{13} & \frac{1}{13} \end{bmatrix}


Answers (1)

 

 

Using elementary operations to compute the inverse of matrix of order 2 -

Steps for finding the inverse of a matrix of order 2 by elementary row operation

 

\begin{array}{l}{\text { Step } 1 : \text { Write } A=I_{n} A} \\ {\text { Step II : Perform a sequence of elementary row operations }} \\ {\text { successively on A on the LHS and the pre factor } I_{n} \text { on the RHS }} \\ {\text { till we obtain the result } I_{n}=B A} \\ {\text { Step III: Write } A^{-1}=B}\end{array}

 

 

-

Use AA-1=I

\\\begin{bmatrix} \;\;1 &3 \\ -3 & 4 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}A\\\\R_2\rightarrow 3R_1+R_2 \\\\\begin{bmatrix} 1 & 3\\ 0 & 13 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 3 & 1 \end{bmatrix}A\\\\R_2\rightarrow \frac{1}{13}R_2\\\\ \begin{bmatrix} 1 & 3\\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ \frac{3}{13} & \frac{1}{13} \end{bmatrix}A\\\\R_1\rightarrow R_1-3R_2\\\\ \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}=\begin{bmatrix} \frac{4}{13} & -\frac{3}{13}\\ \frac{3}{13} & \frac{1}{13} \end{bmatrix}A

hence, 

A^{-1}=\begin{bmatrix} \frac{4}{13} & -\frac{3}{13}\\ \frac{3}{13} & \frac{1}{13} \end{bmatrix}

 

Posted by

Sumit Saini

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