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

Can you answer this question?
 
 

1 Answer

+1 vote
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 $A=\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}$
Step 1: In order to use the elementary row transformations, we may write $A=IA$
$\Rightarrow \begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}A$
Step 2: Applying $R_2=R_2-2R_1$
$\Rightarrow $$\begin{bmatrix}1 & -1\\2-2(1) & 3-2(-1)\end{bmatrix}=\begin{bmatrix}1 & 0\\0-2(1) & 1-2(0)\end{bmatrix}A$
$\Rightarrow $$\begin{bmatrix}1 & -1\\0 & 3+2\end{bmatrix}=\begin{bmatrix}1 & 0\\-2 & 1\end{bmatrix}A$
$\Rightarrow $$\begin{bmatrix}1 & -1\\0 & 5\end{bmatrix}=\begin{bmatrix}1 & 0\\-2 & 1\end{bmatrix}A$
Step 3: Applying $R_2\rightarrow\frac{1}{5}R_2$
$\Rightarrow \begin{bmatrix}1 & -1\\0/5 & 5/5\end{bmatrix}=\begin{bmatrix}1 & 0\\-2/5 & 1/5\end{bmatrix}A$
$\Rightarrow \begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}=\begin{bmatrix}1 & 0\\-2/5 & 1/5\end{bmatrix}A$
Step 4: Applying $R_1=R_1+R_2$
$\begin{bmatrix}1 & -1+1\\0 & 1\end{bmatrix}=\begin{bmatrix}1-(\frac{-2}{5} )& 0+\frac{1}{5}\\-2/5 & 1/5\end{bmatrix}A$
$\Rightarrow $ $\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}=\begin{bmatrix}1+\frac{2}{5} & 1/5\\-2/5 & 1/5\end{bmatrix}A$
$\Rightarrow A^{-1}=\begin{bmatrix}3/5 & 1/5\\-2/5 & 1/5\end{bmatrix}$
answered Feb 15, 2013 by sreemathi.v
edited Mar 17, 2013 by balaji.thirumalai
 

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
...