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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 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} $

$\begin{array}{1 1} (A) Does \;not\; exist \\ A^{-1}=\begin{bmatrix} -3 & -1 & 1 \\ -15& 6 & -5 \\- 5& -2 & 2 \end{bmatrix} \\ A^{-1}=\begin{bmatrix} 3 & 3 & 1 \\ -15& 8 & -5 \\ 5& -2 & 2 \end{bmatrix} \\ A^{-1}=\begin{bmatrix} 3 & -1 & 1 \\ -15& 6 & -5 \\ 5& -2 & 2 \end{bmatrix} \end{array} $

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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 $\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} $
Step 1: In order to use row elementary transformation we write as\[A=I_3A\]
$\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} =\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}A $
Step 2: Applying $ R_1\leftrightarrow R_2$
$\begin{bmatrix}5 & 1 & 0\\2 & 0 & -1\\0 & 1 &3\end{bmatrix}=\begin{bmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & 0 & 1\end{bmatrix}A$
Step 3: Applying $R_1\rightarrow R_1-2R_2$
$\begin{bmatrix}5-2( 2) & 1-2(0) & 0-2(-1) \\ 2 & 0 & -1 \\ 0 & 1 & 3 \end{bmatrix} =\begin{bmatrix} 0-2(1) &1-2( 0 )& 0-2(0) \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}A $
$\begin{bmatrix} 1 & 1 & 2 \\ 2 & 0 & -1\\ 0 & 1 & 3 \end{bmatrix} =\begin{bmatrix} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}A $
Step 4: Applying $ R_2\Rightarrow R_2-2R_1$
$\begin{bmatrix} 1 & 1 & 2 \\ 0 & -2 & -5\\ 0 & 1 & 3 \end{bmatrix} =\begin{bmatrix} -2 & 1 & 0 \\ 5 & -2 & 0 \\ 0 & 0 & 1 \end{bmatrix}A $
Step 5: Applying $ R_2\rightarrow (-1)R_2$
$\begin{bmatrix} 1 & 1 & 2 \\ 0 & 2 & 5\\ 0 & 1 & 3 \end{bmatrix} =\begin{bmatrix} -2 & 1 & 0 \\ -5 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}A $
Step 6: Applying $ R_2\rightarrow R_2-R_3$
$\begin{bmatrix} 1 & 1 & 2 \\ 0-0 & 2-1 & 5-3\\ 0 & 1 & 3 \end{bmatrix} =\begin{bmatrix} -2 & 1 & 0 \\ -5-0 & 2-0 & 0-1 \\ 0 & 0 & 1 \end{bmatrix}A $
$\begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 2\\ 0 & 1 & 3 \end{bmatrix} =\begin{bmatrix} -2 & 1 & 0 \\ -5& 2 & -1 \\ 0 & 0 & 1 \end{bmatrix}A $
Step 7: Applying $ R_3\rightarrow R_3-R_2$ & $ R_1\rightarrow R_1-R_2$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 2\\ 0 & 1 & 3 \end{bmatrix} =\begin{bmatrix} 3 & -1 & 1 \\ -5& 2 & -1 \\ 0 & 0 & 1 \end{bmatrix}A $
Step 8: Applying $ R_3\rightarrow R_3-R_2$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 2\\ 0 & 0 & 1 \end{bmatrix} =\begin{bmatrix} 3 & -1 & 1 \\ -5& 2 & -1 \\ 5& -2 & 2 \end{bmatrix}A $
Step 9: Applying $ R_2\rightarrow R_2-2R_3$
$\begin{bmatrix} 1 & 0 & 0 \\ 0-2(0) & 1-2(0) & 2-2(1)\\ 0 & 0 & 1 \end{bmatrix} =\begin{bmatrix} 3 & -1 & 1 \\ -5-2(5)& 2-2(-2) & -1-2(2) \\ 5& -2 & 2 \end{bmatrix}A $
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} =\begin{bmatrix} 3 & -1 & 1 \\ -15& 6 & -5 \\ 5& -2 & 2 \end{bmatrix}A $
$A^{-1}=\begin{bmatrix} 3 & -1 & 1 \\ -15& 6 & -5 \\ 5& -2 & 2 \end{bmatrix} $
answered Mar 4, 2013 by sharmaaparna1
edited Mar 18, 2013 by sreemathi.v
 

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