Elementary row operations - Practice Questions & MCQ

Elementary Row Operations
Row transformation: The following three types of operation (Transformation) on the rows of a given matrix are known as elementary row operation (transformation).
i) Interchange of $i^{\text {th }}$ row with $j^{\text {th }}$ row, this operation is denoted by

$
R_i \leftrightarrow R_j
$

ii) The multiplication of $\mathrm{i}^{\text {th }}$ row by a constant $\mathrm{k}(\mathrm{k} \neq 0)$ is denoted by

$
\mathrm{R}_{\mathrm{i}} \leftrightarrow \mathrm{kR}_{\mathrm{i}}
$

iii) Adding of $\mathrm{i}^{\text {th }}$ row elements with of $\mathrm{j}^{\text {th }}$ row multiplied by constant $\mathrm{k}(\mathrm{k} \neq 0$ ) is denoted by

$
R_i \leftrightarrow R_i+k R_j
$

In the same way, three-column operations can also be defined.

Steps for finding the inverse of a matrix of order 2 by elementary row operations
Step 1: Write $A=I_n A$
Step II: Perform a sequence of elementary row operations successively on A on the LHS and the pre factor $I_n$ on the RHS till we obtain the result $I_n=B A$
Step III: Write $A^{-1}=B$

For example: 

Given matrix $\mathrm{A}=\left[\begin{array}{cc}a & b \\ c & \left(\frac{1+b c}{a}\right)\end{array}\right]$, then to find the inverse of matrix A

We write,

$
\begin{aligned}
& {\left[\begin{array}{cc}
a & b \\
c & \left(\frac{1+b c}{a}\right)
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \mathrm{A}} \\
& \mathrm{R}_1 \rightarrow \frac{1}{\mathrm{a}} \mathrm{R}_1 \\
& {\left[\begin{array}{ll}
1 & \frac{b}{a} \\
c & \left(\frac{1+b c}{a}\right)
\end{array}\right]=\left[\begin{array}{ll}
\frac{1}{a} & 0 \\
0 & 1
\end{array}\right] \mathrm{A}} \\
& \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{cR}_1 \\
& {\left[\begin{array}{ll}
1 & \frac{b}{q} \\
0 & \frac{1}{a}
\end{array}\right]=\left[\begin{array}{cc}
\frac{1}{a} & 0 \\
\frac{a}{a} & 1
\end{array}\right] \mathrm{A}} \\
& \mathrm{R}_2 \rightarrow \mathrm{aR}_2 \\
& {\left[\begin{array}{ll}
1 & \frac{b}{a} \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
\frac{1}{a} & 0 \\
-c & a
\end{array}\right] \mathrm{A}} \\
& \mathrm{R}_2 \rightarrow \mathrm{R}_1-\frac{\mathrm{b}}{\mathrm{a}} \mathrm{R}_2 \\
& {\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
\frac{1+b c}{a} & -b \\
-c & a
\end{array}\right] \mathrm{A}} \\
& \mathrm{~A}^{-1}=\left[\begin{array}{cc}
\frac{1+b c}{a} & -b \\
-c & a
\end{array}\right]
\end{aligned}
$

Algorithm for finding the lnverse of a Non singular 3 x 3 Matrix by Elementary Row Transformations

1. Introduce unity at the intersection of first row and first column either by interchanging two rows or by adding a constant multiple of elements of some other row to first row.
2. After introducing unity at (1,1) place introduce zeros at all other places in first column.
3. Introduce unity at the intersection of 2nd row and 2nd column with the help of 2nd and 3rd row.
4. Introduce zeros at all other places in the second column except at the intersection of 2nd row and 2nd column.
5. Introduce unity at the intersection of 3rd row and third column.
6. Finally introduce zeros at all other places in the third column except at the intersection of third row and third column.

For example, to find the inverse of matrix A

$A=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 1\end{array}\right]$

First write, A = IA

First write, $A=I A$

$
\Rightarrow\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 2 \\
3 & 1 & 1
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \mathrm{A}
$
Apply, $\mathrm{R}_3 \rightarrow \mathrm{R}_3-3 \mathrm{R}_1$

$
\Rightarrow\left[\begin{array}{ccc}
1 & 2 & 3 \\
0 & 1 & 2 \\
0 & -5 & -8
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
-3 & 0 & 1
\end{array}\right] \mathrm{A}
$
Apply, $\mathrm{R}_1 \rightarrow \mathrm{R}_1-2 \mathrm{R}_2$

$
\Rightarrow\left[\begin{array}{ccc}
1 & 0 & -1 \\
0 & 1 & 2 \\
0 & -5 & -8
\end{array}\right]=\left[\begin{array}{ccc}
1 & -2 & 0 \\
0 & 1 & 0 \\
-3 & 0 & 1
\end{array}\right] \mathrm{A}
$
Apply, $\mathrm{R}_3 \rightarrow \mathrm{R}_3+5 \mathrm{R}_2$

$
\Rightarrow\left[\begin{array}{ccc}
1 & 0 & -1 \\
0 & 1 & 2 \\
0 & 0 & 2
\end{array}\right]=\left[\begin{array}{ccc}
1 & -2 & 0 \\
0 & 1 & 0 \\
-3 & 5 & 1
\end{array}\right] \mathrm{A}
$

Apply, $\mathrm{R}_3 \rightarrow \frac{1}{2} \mathrm{R}_3$

$
\Rightarrow\left[\begin{array}{ccc}
1 & 0 & -1 \\
0 & 1 & 2 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
1 & -2 & 0 \\
0 & 1 & 0 \\
\frac{-3}{2} & \frac{5}{2} & \frac{1}{2}
\end{array}\right] \mathrm{A}
$
Apply, $\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_3$

$
\Rightarrow\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 2 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
-\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
0 & 1 & 0 \\
\frac{-3}{2} & \frac{5}{2} & \frac{1}{2}
\end{array}\right] \mathrm{A}
$
Apply, $\mathrm{R}_2 \rightarrow \mathrm{R}_2-2 \mathrm{R}_3$

$
\Rightarrow\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
-\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
3 & -4 & -1 \\
\frac{-3}{2} & \frac{5}{2} & \frac{1}{2}
\end{array}\right] \mathrm{A}
$
Hence,

$
\mathrm{A}^{-1}=\left[\begin{array}{ccc}
-\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
3 & -4 & -1 \\
\frac{-3}{2} & \frac{5}{2} & \frac{1}{2}
\end{array}\right]
$