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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Using elementary transformations, find the inverse of the matrix $\begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \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
$A=\begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix}$
In order to find the inverse we use row elementary transformation we write A=IA.
$\begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix} =\begin{bmatrix}1 & 0 &0\\0 &1 & 0\\0 & 0 & 1\end{bmatrix}A$
Step 1: Apply $R_2\leftrightarrow R_3$
$\begin{bmatrix} 1 & 3 & -2 \\ 2 & 1 & 0 \\ -3 & 0 & -1 \end{bmatrix} =\begin{bmatrix}1 & 0 &0\\0 &0 & 1\\0 & 1 & 0\end{bmatrix}A$
Step 2: Apply $R_3\rightarrow (-1) R_3$
$\begin{bmatrix} 1 & 3 & -2 \\ 2 & 1 & 0 \\ 3 & 0 & 1 \end{bmatrix} =\begin{bmatrix}1 & 0 &0\\0 &0 & 1\\0 & -1 & 0\end{bmatrix}A$
Step 3: Apply $R_1\rightarrow 3 R_2-R_1$
$\begin{bmatrix} 5 & 0 & 2 \\ 2 & 1 & 0 \\ 3 & 0 & 1 \end{bmatrix} =\begin{bmatrix}-1 & 0 &3\\0 &0 & 1\\0 & -1 & 0\end{bmatrix}A$
Step 4: Apply $R_1\rightarrow 2 R_3-R_1$
$\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 0 & 1 \end{bmatrix} =\begin{bmatrix}1 & -2 &-3\\0 &0 & 1\\0 & -1 & 0\end{bmatrix}A$
Step 5: Apply $R_3\rightarrow R_3-3R_1$
$\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} =\begin{bmatrix}1 & -2 &-3\\0 &0 & 1\\-3 & 5 & 9\end{bmatrix}A$
Step 6: Apply $R_2\rightarrow R_2-2R_1$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} =\begin{bmatrix}1 & -2 &-3\\-2 &4 & 7\\-3 & 5 & 9\end{bmatrix}A$
Step 7: $A^{-1}=\begin{bmatrix}1 & -2 &-3\\-2 &4 & 7\\-3 & 5 & 9\end{bmatrix}$
answered Apr 9, 2013 by sharmaaparna1
 

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