Browse Questions

# Using elements transformation find the inverse of $\begin{bmatrix} 2 & -6 \\ 1 & -2 \end{bmatrix}$

$\begin{array}{1 1} A^{-1}=\begin{bmatrix}1 & -3\\\frac{-1}{2} & 1\end{bmatrix} \\ A^{-1}=\begin{bmatrix}-1 & 3\\\frac{-1}{2} & 1\end{bmatrix} \\ A^{-1}=\begin{bmatrix}-1 & 3\\\frac{1}{2} & 7\end{bmatrix} \\ A^{-1}=\begin{bmatrix}-1 & -3\\\frac{-1}{2} & -1\end{bmatrix} \end{array}$

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
Step 1:Let A=$\begin{bmatrix}2 & -6\\1 & -2\end{bmatrix}$
In order to find inverse we use row elementary transformation we can write as $A=I_2A$
$\Rightarrow \begin{bmatrix}2 &-6\\1 & -2\end{bmatrix}=\begin{bmatrix}1 & 0\\0 &1\end{bmatrix}A$
Step 2:Apply $R_1\rightarrow R_1-R_2$
$\begin{bmatrix}1 &-4\\1 & -2\end{bmatrix}=\begin{bmatrix}1 & -1\\0 &1\end{bmatrix}A$
Step 3:Apply $R_2\rightarrow R_2-R_1$
$\begin{bmatrix}1 &-4\\0 & 2\end{bmatrix}=\begin{bmatrix}1 & -1\\-1 &2\end{bmatrix}A$
Step 4:Apply $R_2\rightarrow \frac{1}{2}\times R_2$
$\begin{bmatrix}1 &-4\\0 & 1\end{bmatrix}=\begin{bmatrix}1 & -1\\\frac{-1}{2} &1\end{bmatrix}A$
Step 5:Apply $R_1\rightarrow R_1+4R_2$
$\begin{bmatrix}1 &0\\0 & 1\end{bmatrix}=\begin{bmatrix}-1 & 3\\\frac{-1}{2} &1\end{bmatrix}A$
Step 6:$A^{-1}=\begin{bmatrix}-1 & 3\\\frac{-1}{2} & 1\end{bmatrix}$