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

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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
$A=\begin{bmatrix}1 & -1 & 2\\0 & 2 & -3\\3 & -2 & 4\end{bmatrix}$
In order to use column elementary transformation we write A=AI.
$\begin{bmatrix}1 & -1 & 2\\0 & 2 & -3\\3 & -2 & 4\end{bmatrix}=A\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}$
Step 1: Apply $C_2\rightarrow 2C_2+C_3$
$\begin{bmatrix}1 & 0 & 2\\0 & 1 & -3\\3 & 0 & 4\end{bmatrix}=A\begin{bmatrix}1 & 0 & 0\\0 & 2 & 0\\0 & 1 & 1\end{bmatrix}$
Step 2: Apply $C_3\rightarrow 3C_2+C_3$
$\begin{bmatrix}1 & 0 & 2\\0 & 1 & 0\\3 & 0 & 4\end{bmatrix}=A\begin{bmatrix}1 & 0 & 0\\0 & 2 & 6\\0 & 1 & 4\end{bmatrix}$
Step 3: Apply $C_3\rightarrow C_3-C_1$
$\begin{bmatrix}1 & 0 & 1\\0 & 1 & 0\\3 & 0 & 1\end{bmatrix}=A\begin{bmatrix}1 & 0 & -1\\0 & 2 & 6\\0 & 1 & 4\end{bmatrix}$
Step 4: Apply $C_3\rightarrow C_1-C_3$
$\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\3 & 0 & 2\end{bmatrix}=A\begin{bmatrix}1 & 0 & 2\\0 & 2 & -6\\0 & 1 & -4\end{bmatrix}$
Step 5: Apply $C_3\rightarrow \frac{1}{2}C_3$
$\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\3 & 0 & 1\end{bmatrix}=A\begin{bmatrix}1 & 0 & 1\\0 & 2 & -3\\0 & 1 & -2\end{bmatrix}$
Step 6: Apply $C_1\rightarrow C_1-3C_3$
$\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}=A\begin{bmatrix}-2 & 0 & 1\\9 & 2 & -3\\6 & 1 & -2\end{bmatrix}$
Step 7: $A^{-1}=\begin{bmatrix}-2 & 0 & 1\\9 & 2 & -3\\6 & 1 & -2\end{bmatrix}$
answered Apr 11, 2013 by sharmaaparna1
 

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