Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Matrices

# Using elementary transformations, find the inverse of the matrix if it exists - $\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}$

Can you answer this question?

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:$A=\begin{bmatrix}2 & 1\\7 &4\end{bmatrix}$
Step 1: In order to use the elementary row transformations, we may write A=IA
$\begin{bmatrix}2 & 1\\7 &4\end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}A$
Step 2: Applying $R_1\leftrightarrow R_2$
$\Rightarrow \begin{bmatrix}2 & 1\\7 &4\end{bmatrix}=\begin{bmatrix}1& 0\\0 &1\end{bmatrix}A$
$\Rightarrow \begin{bmatrix}7 & 4\\2 &1\end{bmatrix}=\begin{bmatrix}0& 1\\1 &0\end{bmatrix}A$
Step 3:Applying $R_1\rightarrow R_1-3R_2$
$\begin{bmatrix}7-3(2)& 4-3(1)\\2 &1\end{bmatrix}=\begin{bmatrix}0-3(1)& 1-3(0)\\1 &0\end{bmatrix}$
$\begin{bmatrix}1& 1\\2 &1\end{bmatrix}=\begin{bmatrix}-3& 1\\1 &0\end{bmatrix}A$
Step 4: Applying $R_2= R_2-2R_1$
$\begin{bmatrix}1& 1\\2-2(1) &1-2(1)\end{bmatrix}=\begin{bmatrix}-3& 1\\1-2(-3) &0-2(1)\end{bmatrix}A$
$\begin{bmatrix}1& 1\\0 &-1\end{bmatrix}=\begin{bmatrix}-3& 1\\7 &-2\end{bmatrix}A$
Step:5 Applying $R_2\rightarrow (-1)R_2$
$\begin{bmatrix}1& 1\\0 &1\end{bmatrix}=\begin{bmatrix}-3& 1\\-7 &2\end{bmatrix}A$
Step:6 Applying $R_1\rightarrow R_1- R_2$
$\begin{bmatrix}1-0& 1-1\\0 &1\end{bmatrix}=\begin{bmatrix}-3-(-7)& 1-2\\-7 &2\end{bmatrix}A$
$\begin{bmatrix}1& 0\\0 &1\end{bmatrix}=\begin{bmatrix}4& -1\\-7 &2\end{bmatrix}A\Rightarrow A^{-1}=\begin{bmatrix}4 & -1\\-7 & 2\end{bmatrix}$
answered Feb 16, 2013
edited Mar 18, 2013

+1 vote