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Find the inverse of matrix A=\begin{bmatrix} \;\;2 & 1\\ -2 & 3 \end{bmatrix}

Option: 1

A^{-1}=\begin{bmatrix} \frac{3}{8} & \frac{1}{8}\\ \frac{1}{4} & \frac{1}{4} \end{bmatrix}


Option: 2

A^{-1}=\begin{bmatrix} -\frac{3}{8} & -\frac{1}{8}\\ \frac{1}{4} & \frac{1}{4} \end{bmatrix}


Option: 3

A^{-1}=\begin{bmatrix} \frac{1}{8} & -\frac{1}{8}\\ \frac{1}{4} & \frac{1}{4} \end{bmatrix}


Option: 4

A^{-1}=\begin{bmatrix} \frac{3}{8} & -\frac{1}{8}\\ \frac{1}{4} & \frac{1}{4} \end{bmatrix}


Answers (1)

best_answer

 

 

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}

 

-

 

 AA^{-1}=I

Now 

\\\begin{bmatrix} \;\;2 & 1\\ -2 & 3 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}A\\R_1\rightarrow \frac{1}{2}R_1\\\\\begin{bmatrix} \;\;1 & \frac{1}{2}\\ -2 & 3 \end{bmatrix}=\begin{bmatrix} \frac{1}{2} & 0\\ 0 & 1 \end{bmatrix}A\\\\R_2\rightarrow 2R_1+R_2 \\\\\begin{bmatrix} 1 & \frac{1}{2}\\ 0 & 4 \end{bmatrix}=\begin{bmatrix} \frac{1}{2} & 0\\ 1 & 1 \end{bmatrix}A\\\\R_2\rightarrow \frac{1}{4}R_2\\\\ \begin{bmatrix} 1 & \frac{1}{2}\\ 0 & 1 \end{bmatrix}=\begin{bmatrix} \frac{1}{2} & 0\\ \frac{1}{4} & \frac{1}{4} \end{bmatrix}A

\\R_1\rightarrow R_1-\frac{1}{2}R_2\\\\ \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}=\begin{bmatrix} \frac{3}{8} & -\frac{1}{8}\\ \frac{1}{4} & \frac{1}{4} \end{bmatrix}A

hence, A^{-1}=\begin{bmatrix} \frac{3}{8} & -\frac{1}{8}\\ \frac{1}{4} & \frac{1}{4} \end{bmatrix}

 

 

Posted by

rishi.raj

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