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# Using elementary transformations, find the inverse of the matrix if it exists - $\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$

Toolbox:
• There are six operations (transformations) on a matrix,three of which are due to rows and three due to columns which are known as elementary operations or transformations.
• Row/Column Switching: Interchange of any two rows or two columns, i.e, $R_i\leftrightarrow R_j$ or $\;C_i\leftrightarrow C_j$
• Row/Column Multiplication: The multiplication of the elements of any row or column by a non zero number: i.e, i.e $R_i\rightarrow kR_i$ where $k\neq 0$ or $\;C_j\rightarrow kC_j$ where $k\neq 0$
• Row/Column Addition:The addition to the element of any row or column ,the corresponding elements of any other row or column multiplied by any non zero number: i.e $R_i\rightarrow R_i+kR_j$ or $\;C_i\rightarrow C_i+kC_j$, where $i \neq j$.
• If A is a matrix such that A$^{-1}$ exists, then to find A$^{-1}$ using elementary row operations, write A = IA and apply a sequence of row operation on A = IA till we get, I = BA. The matrix B will be the inverse of A. Similarly, if we wish to find A$^{-1}$ using column operations, then, write A = AI and apply a sequence of column operations on A = AI till we get, I = AB.
Given:$A=\begin{bmatrix}3 & 1\\5 &2\end{bmatrix}$
Step 1:In order to use row elementary transformation we write as A=IA.
$\begin{bmatrix}3 & 1\\5& 2\end{bmatrix}=\begin{bmatrix}1 & 0\\0&1\end{bmatrix}A$
Step 2: Applying $R_1\rightarrow 2R_1-R_2$
$\begin{bmatrix}2(3)-5 & 2(1)-2\\5 &2\end{bmatrix}=\begin{bmatrix}2(1)-0 & 2(0)-1\\0&1\end{bmatrix}A$
$\begin{bmatrix}1 & 0\\5 &2\end{bmatrix}=\begin{bmatrix}2 & -1\\0&1\end{bmatrix}A$
Step 3:Applying $R_2\rightarrow R_2-5R_1$
$\begin{bmatrix}1 & 0\\5-5(1) & 2-5(0)\end{bmatrix}=\begin{bmatrix}2 & -1\\0-5(2)&1-5(-1)\end{bmatrix}A$
$\begin{bmatrix}1 & 0\\0 &2\end{bmatrix}=\begin{bmatrix}2 & -1\\-10&6\end{bmatrix}A$
Step 4:Applying $R_2\rightarrow \frac{1}{2}R_2$
$\begin{bmatrix}1 &0\\0 &\frac{2}{2}\end{bmatrix}=\begin{bmatrix}2 & -1\\\frac{-10}{2}&\frac{6}{2}\end{bmatrix}A$
$\begin{bmatrix}1 & 0\\0 &1\end{bmatrix}=\begin{bmatrix}2 & -1\\-5&3\end{bmatrix}A$
$A^{-1}=\begin{bmatrix}2 &-1\\-5 & 3\end{bmatrix}$
answered Feb 16, 2013
edited Mar 18, 2013