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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} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{bmatrix} $

$\begin{array}{1 1}A^{-1}=\begin{bmatrix} 1& -10/25 & -3/25 \\ -10/25 & 4/25 & 1\\ -3/25 & 1/25 & 0 \end{bmatrix} \\ A^{-1}=\begin{bmatrix} 1& -10/25 & 0 \\ -10/25 & 4/25 & 11/25\\ -3/25 & 1/25 & 9/25 \end{bmatrix} \\ A^{-1}=\begin{bmatrix} 1& -2/5 & -3/5 \\ -2/25 & 4/25 & 11/25\\ -3/5 & 1/25 & 9/25 \end{bmatrix} \\ \text{Does not exist} \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} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{bmatrix} $
Step :1 In order to use row elementary transformation we write as\[A=I_3A\]
$ \begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} A$
Step 2: Applying $R_1\rightarrow R_2+ 3R_1$
$ \begin{bmatrix} 1 & 3 & -2 \\ -3+3(1) & 0+3(3) & -5+3(-2) \\ 2 & 5 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0+3(0) & 1+3(0) & 0+3(0)\\ 0 & 0 & 1 \end{bmatrix} A$
$ \begin{bmatrix} 1 & 3 & -2 \\ 0 & 9 & -11 \\ 2 & 5 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} A$
Step 3: Applying $R_3\rightarrow R_3- 2R_1$
$ \begin{bmatrix} 1 & 3 & -2 \\ 0 & 9 & -11 \\ 2-2(1) & 5-2(3) & 0-2(-2) \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0\\ 0-2(1) & 0-0 & 1-2(0) \end{bmatrix} A$
$ \begin{bmatrix} 1 & 3 & -2 \\ 0 & 9 & -11 \\ 0 & -1 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0\\ -2 & 0 & 1 \end{bmatrix} A$
Step :4 Applying $R_2\rightarrow R_2+ 8R_3$
$ \begin{bmatrix} 1 & 3 & -2 \\ 0+0 & 9+8(-1) & -11+8(4) \\ 0 & -1 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 3+8(-2) & 1+0 & 0+8(1)\\ -2 & 0 & 1 \end{bmatrix} A$
$ \begin{bmatrix} 1 & 3 & -2 \\ 0 & 1 & 21 \\ 0 & -1 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ -13 & 1 & 8\\ -2 & 0 & 1 \end{bmatrix} A$
Step:5 Applying $R_1\rightarrow R_1- 3R_2$
$ \begin{bmatrix} 1-3(0) & 3-3(1 & -2-3(21) \\ 0 & 1 & 21 \\ 0 & -1 & 4 \end{bmatrix} = \begin{bmatrix} \\ 1-3(-13) & 0-3(1) & 0-3(8)\\ -13 & 1 & 8\\-2 \end{bmatrix} A$
$ \begin{bmatrix} 1 & 0 & -65 \\ 0 & 1 & 21 \\ 0 & -1 & 4 \end{bmatrix} = \begin{bmatrix} 40 & -3 & -24 \\ -13 & 1 & 8\\ -2 & 0 & 1 \end{bmatrix} A$
Step :6 Applying $R_3\rightarrow R_3+R_2$
$ \begin{bmatrix} 1 & 0 & -65 \\ 0 & 1 & 21 \\ 0+0 & -1+1 & 4+21 \end{bmatrix} = \begin{bmatrix} 40 & -3 & -24 \\ -13 & 1 & 8\\ -2-13 & 0+1 & 1+8 \end{bmatrix} A$
$ \begin{bmatrix} 1 & 0 & -65 \\ 0 & 1 & 21 \\ 0 & 0 & 25\end{bmatrix} = \begin{bmatrix} 40 & -3 & -24 \\ -13 & 1 & 8\\ -15 & 1 & 9 \end{bmatrix} A$
Step :7 Applying $R_3\rightarrow \frac{1}{25} R_3$
$ \begin{bmatrix} 1 & 0 & -65 \\ 0 & 1 & 21 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix} 40 & -3 & -24 \\ -12 & 1 & 8\\ -15/25 & 1/25 & 9/25 \end{bmatrix} A$
$ \begin{bmatrix} 1 & 0 & -65 \\ 0 & 1 & 21 \\ 0 & 0 & 1\end{bmatrix} = \frac{1}{25}\begin{bmatrix} 1000 & -75 & -600 \\ -325 & 25 & 200\\ -15 & 1 & 9 \end{bmatrix} A$
Step 8: Applying $R_1\rightarrow R_1+65 R_3$
$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 21 \\ 0 & 0 & 1\end{bmatrix} = \frac{1}{25}\begin{bmatrix} 25& -10 & -15 \\ -325 & 25 & 200\\ -15 & 1 & 9 \end{bmatrix} A$
Step :9 Applying $R_2\rightarrow R_2-21 R_3$
$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \frac{1}{25}\begin{bmatrix} 25& -10 & -15 \\ -10 & 4 & 11\\ -15 & 1 & 9 \end{bmatrix} A$
$A^{-1}=\begin{bmatrix} 1& -10/25 & -15/25 \\ -10/25 & 4/25 & 11/25\\ -15/25 & 1/25 & 9/25 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 1& -2/5 & -3/5 \\ -2/5 & 4/25 & 11/25\\ -3/5 & 1/25 & 9/25 \end{bmatrix}$
answered Feb 17, 2013 by sreemathi.v
edited Mar 18, 2013 by sreemathi.v
 

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