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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Obtain the inverse of the following matrix, using elementary operations $ A = \begin{bmatrix} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \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:
$A=\begin{bmatrix}3 & 0 & -1\\2 & 3 & 0\\0 & 4 &1\end{bmatrix}$
In order to find the inverse,using the row elementary transformation we write A=IA.
$\begin{bmatrix}3 & 0 & -1\\2 & 3 & 0\\0 & 4 &1\end{bmatrix}=\begin{bmatrix}1 & 0 & 0\\0 & 1 &0\\0&0 & 1\end{bmatrix}A$
Step 1: Apply $R_1\rightarrow R_1-R_2$
$\begin{bmatrix}1 & -1 & -1\\2 & 3 & 0\\0 & 4 &1\end{bmatrix}=\begin{bmatrix}1 & -1 & 0\\0 & 1 &0\\0&0 & 1\end{bmatrix}A$
Step 2: Apply $R_1\rightarrow R_1+R_3$
$\begin{bmatrix}1 & 3 & 0\\2 & 3 & 0\\0 & 4 &1\end{bmatrix}=\begin{bmatrix}1 & -1 & 1\\0 & 1 &0\\0&0 & 1\end{bmatrix}A$
Step 3: Apply $R_1\rightarrow R_2-R_1$
$\begin{bmatrix}1 & 0 & 0\\2 & 3 & 0\\0 & 4 &1\end{bmatrix}=\begin{bmatrix}-1 & 2 & -1\\0 & 1 &0\\0&0 & 1\end{bmatrix}A$
Step 4: Apply $R_2\rightarrow R_2-2R_1$
$\begin{bmatrix}1 & 0 & 0\\0 & 3 & 0\\0 & 4 &1\end{bmatrix}=\begin{bmatrix}-1 & 2 & -1\\2 & -3 &1\\0&0 & 1\end{bmatrix}A$
Step 5: Apply $R_2\rightarrow \frac{1}{3}R_2$
$\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 4 &1\end{bmatrix}=\begin{bmatrix}-1 & 2 & -1\\\frac{2}{3} & -1 &\frac{1}{3}\\0&0 & 1\end{bmatrix}A$
Step 6: Apply $R_3\rightarrow R_3-4R_2$
$\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 &1\end{bmatrix}=\begin{bmatrix}-1 & 2 & -1\\\frac{2}{3} & -1 &\frac{1}{3}\\\frac{8}{3}&4 & \frac{-1}{3}\end{bmatrix}A$
Step :7 $A^{-1}=\begin{bmatrix}-1 & 2 & -1\\\frac{2}{3} & -1 &\frac{1}{3}\\\frac{8}{3}&4 & \frac{-1}{3}\end{bmatrix}$
$A^{-1}=\frac{1}{3}\begin{bmatrix}-3 & 6 & -3\\2 & -3 &1\\8&12 & -1\end{bmatrix}$
answered Apr 8, 2013 by sharmaaparna1
 

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