logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Matrices
0 votes

Using elementary transformations, find the inverse of the matrix if it exists - $ \begin{bmatrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{bmatrix} $

$\begin{array}{1 1}A^{-1}=\frac{1}{5}\begin{bmatrix}-2 & 0 & 3\\-1 & 0& 0\\2 & 1 & -0\end{bmatrix} \\ \text{Does not exist} \\ A^{-1}=\frac{1}{5}\begin{bmatrix}-2 & 0 & 3\\-1 & 1& 0\\2 & 1 & -2\end{bmatrix} \\A^{-1}=\frac{1}{5}\begin{bmatrix}-9 & 0 & 3\\-1 & 0& 0\\2 & 1 & -2\end{bmatrix} \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
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:$A=\begin{bmatrix}2 & -3 & 3\\2 & 2 & 3\\3 & -2 & 2\end{bmatrix}$
Step 1: In order to use row elementary transformation we write as\[A=I_3A\]
$\begin{bmatrix}2 & -3 & 3\\2 & 2 & 3\\3 & -2 & 2\end{bmatrix}=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}A$
Step 2: Applying $R_1\leftrightarrow R_3$
$\begin{bmatrix}3 & -2 & 2\\2 & 2 & 3\\2 & -3 & 3\end{bmatrix}=\begin{bmatrix}0 & 0 & 1\\0 & 1 & 0\\1 & 0 & 0\end{bmatrix}A$
Apply $R_1\rightarrow R_1- R_2$
$\begin{bmatrix}3-2 & -2-2 & 2-3\\2 & 2 & 3\\2 & -3 & 3\end{bmatrix}=\begin{bmatrix}0-0 & 0 -1& 1-0\\0 & 1 & 0\\1 & 0 & 0\end{bmatrix}A$
$\begin{bmatrix}1 & -4 & -1\\2 & 2 & 3\\2 & -3 & 3\end{bmatrix}=\begin{bmatrix}0 & -1 & 1\\0 & 1 & 0\\1 & 0 & 0\end{bmatrix}A$
Step 3: Applying $R_2\rightarrow R_2- 2 R_1$
$\begin{bmatrix}1 & -4 & -1\\2-2(1) & 2-2(-4) & 3-2(-1)\\2 & -3 & 3\end{bmatrix}=\begin{bmatrix}0 & -1 & 1\\0-2(0) & 1-2(-1) & 0-2(1)\\1 & 0 & 0\end{bmatrix}A$
$\begin{bmatrix}1 & -4 & -1\\0 & 10 & 5\\2 & -3 & 3\end{bmatrix}=\begin{bmatrix}0 & -1 & 1\\0 & 3 & -2\\1 & 0 & 0\end{bmatrix}A$
Step 4: Applying $R_3= R_3- 2 R_1$
$\begin{bmatrix}1 & -4 & -1\\0 & 10 & 5\\2-2(1) & -3-2(-4) & 3-2(-1)\end{bmatrix}=\begin{bmatrix}0 & -1 & 1\\0 & 3 & -2\\1-2(0) & 0-2(-1) & 0-2(1)\end{bmatrix}A$
$\begin{bmatrix}1 & -4 & -1\\0 & 10 & 5\\0 & 5 & 5\end{bmatrix}=\begin{bmatrix}0 & -1 & 1\\0 & 3 & -2\\1 & 2 & -2\end{bmatrix}A$
Step 5: Applying $R_2\rightarrow R_2- R_3$
$\begin{bmatrix}1 & -4 & -1\\0 & 5 & 0\\0 & 5 & 5\end{bmatrix}=\begin{bmatrix}0 & -1 & 1\\-1 & 1 & 0\\1 & 2 & -2\end{bmatrix}A$
Step 6: Applying $R_2\rightarrow \frac{1}{5} \times R_2$
$\begin{bmatrix}1 & -4 & -1\\0 & 1 & 0\\0 & 5 & 5\end{bmatrix}=\begin{bmatrix}0 & -1 & 1\\-1/5 & 1/5 & 0\\1 & 2 & -2\end{bmatrix}A$
Step :7 Applying $R_3\rightarrow \frac{1}{5} \times R_3$
$\begin{bmatrix}1 & -4 & -1\\0 & 1 & 0\\0 & 1 & 1\end{bmatrix}=\begin{bmatrix}0 & -1 & 1\\-1/5 & 1/5 & 0\\1/5 & 2/5 & -2/5\end{bmatrix}A$
Step 8:Applying $R_1\rightarrow R_1+4 R_2$
$\begin{bmatrix}1+0 & -4+4 & -1+4(0)\\0 & 1 & 0\\0 & 1 & 1\end{bmatrix}=\begin{bmatrix}0-4/5 & -1+4(1/5) & 1+0\\-1/5 & 1/5 & 0\\1/5 & 2/5 & -2/5\end{bmatrix}A$
$\begin{bmatrix}1 & 0 & -1\\0 & 1 & 0\\0 & 1 & 1\end{bmatrix}=\begin{bmatrix}-4/5 & -1/5 & 1\\-1/5 & 1/5 & 0\\1/5 & 2/5 & -2/5\end{bmatrix}A$
Step 9: Applying $R_3\rightarrow R_3- R_2$
$\begin{bmatrix}1 & 0 & -1\\0 & 1 & 0\\0 & 1 & 1\end{bmatrix}=\begin{bmatrix}-4/5 & -1/5 & 1\\-1/5 & 1/5 & 0\\2/5 & 1/5 & -2/5\end{bmatrix}A$
$\begin{bmatrix}1 & 0 & -1\\0 & 1 & 0\\0 & 1 & 1\end{bmatrix}=\frac{1}{5}\begin{bmatrix}-4 & -1 & 1\\-1 & 1& 0\\2 & 1 & -2\end{bmatrix}A$
Step 10: Applying $R_1\rightarrow R_1+R_3$
$\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}=\frac{1}{5}\begin{bmatrix}-2 & 0 & 3\\-1 & 1& 0\\2 & 1 & -2\end{bmatrix}A$
$A^{-1}=\frac{1}{5}\begin{bmatrix}-2 & 0 & 3\\-1 & 1& 0\\2 & 1 & -2\end{bmatrix}$
answered Feb 16, 2013 by sreemathi.v
edited Mar 18, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...