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Home  >>  CBSE XII  >>  Math  >>  Matrices
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Find inverse,by elementary row operations(if possible),of the following matrices$\quad\begin{bmatrix}1 & 3\\-5 & 7\end{bmatrix}$

Note: This is part 1 of a 2 part question, split as 2 separate questions here.
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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.
Step1:
Given
$A=\begin{bmatrix}1 & 3\\-5 & 7\end{bmatrix}$
In order to use elementary row operations we may write A=IA.
$\Rightarrow \begin{bmatrix}1 & 3\\-5 & 7\end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}A$
Step2:
By Applying $R_2=R_2+5R_1$
$\Rightarrow \begin{bmatrix}1 & 3\\0 & 22\end{bmatrix}=\begin{bmatrix}1 & 0\\5 & 1\end{bmatrix}A$
Step3:
Apply $R_2=\frac{1}{22}R_2$
$\begin{bmatrix}1 & 3\\0 & 1\end{bmatrix}=\begin{bmatrix}1 & 0\\5/22 & 1/22\end{bmatrix}A$
$\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}=\frac{1}{22}\begin{bmatrix}22 & 0\\5 & 1\end{bmatrix}A$
Step4:
Apply $R_1=R_1-3R_2$
$\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}=\frac{1}{22}\begin{bmatrix}7 & -3\\5 & 1\end{bmatrix}A$
$\Rightarrow A^{-1}=\frac{1}{22}\begin{bmatrix}7 & -3\\5 & 1\end{bmatrix}$
$A^{-1}= \begin{bmatrix}7/22 & -3/22\\5/22 & 1/22\end{bmatrix}$
answered Mar 25, 2013 by sharmaaparna1
 

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