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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Using elementary transformations, find the inverse of the following matrix : $ \begin{bmatrix} 2 & -1 & 4 \\ 4 & 0 & 2 \\ 3 & -2 & 7 \end{bmatrix} $

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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:
$A=\begin{bmatrix}2 & -1 & 4\\4 &0 &2\\3 &-2 & 7\end{bmatrix}$
In order to use the elementary row transformations, we may write $A=IA$
$\begin{bmatrix}2 & -1 & 4\\4 &0 &2\\3 &-2 & 7\end{bmatrix}=\begin{bmatrix}1 &0 & 0\\0 &1 & 0\\0&0 &1\end{bmatrix}A$
Step 1: Apply $R_1\leftrightarrow R_2$
$\begin{bmatrix}4 & 0 & 2\\2 &-1 &4\\3 &-2 & 7\end{bmatrix}=\begin{bmatrix}0 &1 & 0\\1 &0 & 0\\0&0 &1\end{bmatrix}A$
Step 2: Apply $R_1\rightarrow R_1-R_3$
$\begin{bmatrix}1 & 2 & -5\\2 &-1 &4\\3 &-2 & 7\end{bmatrix}=\begin{bmatrix}0 &1 & -1\\1 &0 & 0\\0&0 &1\end{bmatrix}A$
Step 3: Apply $R_2\rightarrow 6R_2-4R_3$
$\begin{bmatrix}1 & 2 & -5\\0 &2 &0\\3 &-2 & 7\end{bmatrix}=\begin{bmatrix}0 &1 & -1\\6 &0 & -4\\0&0 &1\end{bmatrix}A$
Step 4: Apply $R_2\rightarrow \frac{1}{2}R_2$
$\begin{bmatrix}1 & 2 & -5\\0 &1 &0\\3 &-2 & 7\end{bmatrix}=\begin{bmatrix}0 &1 & -1\\3 &0 & -2\\0&0 &1\end{bmatrix}A$
Step 5: Apply $R_3\rightarrow R_3+R_1$
$\begin{bmatrix}1 & 2 & -5\\0 &1 &0\\4 &0 & 2\end{bmatrix}=\begin{bmatrix}0 &1 & -1\\3 &0 & -2\\0&1 &0\end{bmatrix}A$
Step 6: Apply $R_3\rightarrow \frac{1}{2}R_3$
$\begin{bmatrix}1 & 2 & -5\\0 &1 &0\\2 &0 & 1\end{bmatrix}=\begin{bmatrix}0 &1 & -1\\3 &0 & -2\\0&\frac{1}{2} &0\end{bmatrix}A$
Step 7: Apply $R_1\rightarrow 2R_2-R_1$
$\begin{bmatrix}1 & 0 & 5\\0 &1 &0\\2 &0 & 1\end{bmatrix}=\begin{bmatrix}6 &-1 & -3\\3 &0 & -2\\0&\frac{1}{2} &0\end{bmatrix}A$
Step 8: Apply $R_3\rightarrow 2R_1-R_3$
$\begin{bmatrix}1 & 0 & 5\\0 &1 &0\\0 &0 & 9\end{bmatrix}=\begin{bmatrix}6 &-1 & -3\\3 &0 & -2\\12&\frac{-5}{2} &-6\end{bmatrix}A$
Step 9: Apply $R_3\rightarrow \frac{1}{9}R_3$
$\begin{bmatrix}1 & 0 & 5\\0 &1 &0\\0 &0 & 1\end{bmatrix}=\begin{bmatrix}6 &-1 & -3\\3 &0 & -2\\\frac{4}{3}&\frac{-5}{18} &\frac{-2}{3}\end{bmatrix}A$
Step 10: Apply $R_1\rightarrow R_1-5R_3$
$\begin{bmatrix}1 & 0 & 50\\0 &1 &0\\0 &0 & 1\end{bmatrix}=\begin{bmatrix}\frac{-2}{3} &\frac{7}{18} & \frac{1}{3}\\3 &0 & -2\\\frac{4}{3}&\frac{-5}{18} &\frac{-2}{3}\end{bmatrix}A$
Step 11: $A^{-1}=\frac{1}{3}\begin{bmatrix}-2 &\frac{7}{6} &1\\1 &0 & -6\\4 & \frac{-5}{6} & -2\end{bmatrix}$
answered Apr 4, 2013 by sharmaaparna1
 

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