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Home  >>  CBSE XII  >>  Math  >>  Matrices
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Using elementary transformations, find the inverse of the matrix if it exists - $ \begin{bmatrix} 2 & 3 \\ 5 & 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 $A=\begin{bmatrix}2 & 3\\5 & 7\end{bmatrix}$
Step 1: In order to use the elementary row transformations, we may write $A=IA$
$\begin{bmatrix}2 & 3\\5 & 7\end{bmatrix}=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}A$
Step 2:Applying $R_1\leftrightarrow R_2$
$\begin{bmatrix}5 & 7\\2 & 3\end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}A$
Step 3:Applying $R_1\rightarrow R_1-2R_2$
$\begin{bmatrix}5-2(2) & 7-2(3)\\2& 3\end{bmatrix}=\begin{bmatrix}0-2(1) & 1-2(0)\\1 & 0\end{bmatrix}A$
$\Rightarrow\begin{bmatrix}1 & 1\\2 & 3\end{bmatrix}=\begin{bmatrix}-2 & 1\\1 & 0\end{bmatrix}A$
Step 4:Applying $R_2\rightarrow R_2-2R_1$
$\begin{bmatrix}1 & 1\\2-2(1)& 3-2(1)\end{bmatrix}=\begin{bmatrix}-2 & 1\\1-2(-2) & 0-2(1)\end{bmatrix}A$
$\Rightarrow\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}=\begin{bmatrix}-2 & 1\\5 & -2\end{bmatrix}A$
Step 5: Applying $R_1\rightarrow R_1-R_2$
$\begin{bmatrix}1-0 & (1-1)\\0& 1\end{bmatrix}=\begin{bmatrix}-2-5 & 1-(-2)\\5 & -2\end{bmatrix}A$
$\Rightarrow\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}=\begin{bmatrix}-7 & 3\\5 & -2\end{bmatrix}A$
$A^{-1}=\begin{bmatrix}-7 & 3\\5 & -2\end{bmatrix}$
answered Feb 15, 2013 by sreemathi.v
edited Mar 18, 2013 by sreemathi.v
 

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