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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} 3 & -1 \\ -4 & 2 \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:$A=\begin{bmatrix}3 & -1\\-4 &2\end{bmatrix}$
Step 1: In order to use row elementary transformation we write as A=IA.
$\begin{bmatrix}3 & -1\\-4& 2\end{bmatrix}=\begin{bmatrix}1 & 0\\0&1\end{bmatrix}A$
Step :2 Applying $R_1\rightarrow R_2$
$\begin{bmatrix}-4 & 2\\3 &-1\end{bmatrix}=\begin{bmatrix}0 & 1\\1&0\end{bmatrix}A$
Step :3 Applying $R_1= (-1)R_1$
$\begin{bmatrix}4 & -2\\3 & -1\end{bmatrix}=\begin{bmatrix}0 & -1\\0&0\end{bmatrix}A$
Step:4 Applying $R_1\Rightarrow R_1-R_2$
$\begin{bmatrix}4-3 &-2-(-1)\\3 &-1\end{bmatrix}=\begin{bmatrix}-1-0 & 0-1\\0&1\end{bmatrix}A$
$\begin{bmatrix}1 & -1\\3 &-1\end{bmatrix}=\begin{bmatrix}-1 & -1\\1&0\end{bmatrix}A$
Step:5 Applying $R_2\Rightarrow R_2-3R_1$
$\begin{bmatrix}1 &-1\\3-3(1) &-1-3(-1)\end{bmatrix}=\begin{bmatrix}-1 & -1\\1-3(-1)&0-3(-1)\end{bmatrix}A$
$\begin{bmatrix}1 & -1\\0 & 2\end{bmatrix}=\begin{bmatrix}-1 & -1\\4&3\end{bmatrix}A$
Step 6: Applying $R_2=\frac{1}{2}\times R_2$
$\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}=\begin{bmatrix}-1 & -1\\2&3/2\end{bmatrix}A$
Step 7: Applying $R_1\Rightarrow R_1+R_2$
$\begin{bmatrix}1+0 & -1+1\\0 & 1\end{bmatrix}=\begin{bmatrix}-1+2 & -1+3/2\\2&3/2\end{bmatrix}A$
$\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}=\begin{bmatrix}1 & 1/2\\2&3/2\end{bmatrix}A$
$A^{-1}=\begin{bmatrix}1 & 1/2\\2&3/2\end{bmatrix}$
answered Mar 16, 2013 by sharmaaparna1
edited Mar 18, 2013 by sreemathi.v
 

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