Q: Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right)$, if exists.

Let $A=\left(\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right)$ We know that $A=I A$ Therefore, $$ \begin{array}{ll} \Rightarrow\left(\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right)=\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) A & \\ \Rightarrow\left(\begin{array}{cc} 1 & -1 \\ 0 & 5 \end{array}\right)=\left(\begin{array}{cc} 1 & 0 \\ -2 & 1 \end{array}\right) A & \left(R_{2} \rightarrow R_{2}-2 R_{1}\right) \\ \Rightarrow\left(\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right)=\left(\begin{array}{cc} 1 & 0 \\ \frac{-2}{5} & \frac{1}{5} \end{array}\right) A & \left(R_{2} \rightarrow \frac{1}{5} R_{2}\right) \\ \Rightarrow\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)=\left(\begin{array}{cc} \frac{3}{5} & \frac{1}{5} \\ \frac{-2}{5} & \frac{1}{5} \end{array}\right) A & \left(R_{1} \rightarrow R_{1}+R_{2}\right) \\ \Rightarrow A^{-1}=\left(\begin{array}{cc} \frac{3}{5} & \frac{1}{5} \\ \frac{-2}{5} & \frac{1}{5} \end{array}\right) & \end{array} $$

Q: Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ll}1 & 3 \\ 2 & 7\end{array}\right)$, if exists.

Let $A=\left(\begin{array}{ll}1 & 3 \\ 2 & 7\end{array}\right)$ We know that $A=I A$ Therefore, $$ \begin{array}{ll} \Rightarrow\left(\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right)=\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) A & \\ \Rightarrow\left(\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right)=\left(\begin{array}{cc} 1 & 0 \\ -2 & 1 \end{array}\right) A & \left(R_{2} \rightarrow R_{2}-2 R_{1}\right) \\ \Rightarrow\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)=\left(\begin{array}{cc} 7 & -3 \\ -2 & 1 \end{array}\right) A & \left(R_{1} \rightarrow R_{1}-3 R_{2}\right) \\ \Rightarrow A^{-1}=\left(\begin{array}{cc} 7 & -3 \\ -2 & 1 \end{array}\right) & \end{array} $$

Q: Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right)$, if exists.

Let $A=\left(\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right)$ We know that $A=I A$ Therefore, $$ \begin{aligned} & \Rightarrow\left(\begin{array}{ll} 2 & 3 \\ 5 & 7 \end{array}\right)=\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) A \\ & \Rightarrow\left(\begin{array}{cc} 1 & \frac{3}{2} \\ 5 & 7 \end{array}\right)=\left(\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & 1 \end{array}\right) A \\ & \Rightarrow\left(\begin{array}{cc} 1 & \frac{3}{2} \\ 0 & \frac{-1}{2} \end{array}\right)=\left(\begin{array}{cc} \frac{1}{2} & 0 \\ \frac{-5}{2} & 1 \end{array}\right) A \\ & \Rightarrow\left(\begin{array}{cc} 1 & 0 \\ 0 & \frac{-1}{2} \end{array}\right)=\left(\begin{array}{cc} -7 & 3 \\ \frac{-5}{2} & 1 \end{array}\right) A \\ & \Rightarrow\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)=\left(\begin{array}{cc} -7 & 3 \\ 5 & -2 \end{array}\right) A \\ & \Rightarrow A^{-1}=\left(\begin{array}{cc} -7 & 3 \\ 5 & -2 \end{array}\right) \end{aligned}\left(R_{1} \rightarrow R_{1}+3 R_{2}\right) $$

Q: Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right)$, if exists.

Let $A=\left(\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right)$ We know that $A=I A$ Therefore, $$ \begin{aligned} & \Rightarrow\left(\begin{array}{ll} 2 & 1 \\ 7 & 4 \end{array}\right)=\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) A \\ & \Rightarrow\left(\begin{array}{ll} 1 & \frac{1}{2} \\ 7 & 4 \end{array}\right)=\left(\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & 1 \end{array}\right) A \\ & \Rightarrow\left(\begin{array}{ll} 1 & \frac{1}{2} \\ 0 & \frac{1}{2} \end{array}\right)=\left(\begin{array}{cc} \frac{1}{2} & 0 \\ \frac{-7}{2} & 1 \end{array}\right) A \\ & \Rightarrow\left(\begin{array}{ll} 1 & 0 \\ 0 & \frac{1}{2} \end{array}\right)=\left(\begin{array}{cc} 4 & -1 \\ \frac{-7}{2} & 1 \end{array}\right) A \\ & \Rightarrow\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)=\left(\begin{array}{cc} 4 & -1 \\ -7 & 2 \end{array}\right) A \\ & \Rightarrow A^{-1}=\left(\begin{array}{cc} 4 & -1 \\ -7 & 2 \end{array}\right) \end{aligned}\left(R_{1} \rightarrow R_{2}-7 R_{1}\right) ~\left(R_{2} \rightarrow 2 R_{1}\right) ~{ } $$

Q: Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ll}2 & 5 \\ 1 & 3\end{array}\right)$, if exists.

Let $A=\left(\begin{array}{ll}2 & 5 \\ 1 & 3\end{array}\right)$ We know that $A=I A$ Therefore, $$ \begin{aligned} & \Rightarrow\left(\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right)=\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) A \\ & \Rightarrow\left(\begin{array}{ll} 1 & \frac{5}{2} \\ 1 & 3 \end{array}\right)=\left(\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & 1 \end{array}\right) A \\ & \Rightarrow\left(\begin{array}{ll} 1 & \frac{5}{2} \\ 0 & \frac{1}{2} \end{array}\right)=\left(\begin{array}{cc} \frac{1}{2} & 0 \\ \frac{-1}{2} & 1 \end{array}\right) A \\ & \Rightarrow\left(\begin{array}{ll} 1 & 0 \\ 0 & \frac{1}{2} \end{array}\right)=\left(\begin{array}{cc} 3 & -5 \\ \frac{-1}{2} & 1 \end{array}\right) A \\ & \Rightarrow\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)=\left(\begin{array}{cc} 3 & -5 \\ -1 & 2 \end{array}\right) A \\ & \Rightarrow A^{-1}=\left(\begin{array}{cc} 3 & -5 \\ -1 & 2 \end{array}\right) \end{aligned}\left(R_{1} \rightarrow R_{2}-R_{1}\right) .\left(R_{2} \rightarrow 2 R_{2}\right){ } $$

Q: Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ll}3 & 1 \\ 5 & 2\end{array}\right)$, if exists.

Let $A=\left(\begin{array}{ll}3 & 1 \\ 5 & 2\end{array}\right)$ We know that $A=I A$ Therefore, $$ \begin{aligned} & \Rightarrow\left(\begin{array}{ll} 3 & 1 \\ 5 & 2 \end{array}\right)=A\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \\ & \Rightarrow\left(\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right)=A\left(\begin{array}{cc} 1 & 0 \\ -2 & 1 \end{array}\right) \quad\left(C_{1} \rightarrow C_{1}-2 C_{2}\right) \\ & \Rightarrow\left(\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right)=A\left(\begin{array}{cc} 1 & -1 \\ -2 & 3 \end{array}\right) \quad\left(C_{2} \rightarrow C_{2}-C_{1}\right) \\ & \Rightarrow\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)=A\left(\begin{array}{cc} 2 & -1 \\ -5 & 3 \end{array}\right) \quad\left(C_{1} \rightarrow C_{1}-C_{2}\right) \\ & \Rightarrow A^{-1}=\left(\begin{array}{cc} 2 & -1 \\ -5 & 3 \end{array}\right) \end{aligned} $$

Q: Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ll}4 & 5 \\ 3 & 4\end{array}\right)$, if exists.

Let $A=\left(\begin{array}{ll}4 & 5 \\ 3 & 4\end{array}\right)$ We know that $A=I A$ Therefore, $$ \begin{array}{ll} \Rightarrow\left(\begin{array}{ll} 4 & 5 \\ 3 & 4 \end{array}\right)=\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) A & \\ \Rightarrow\left(\begin{array}{cc} 1 & 1 \\ 3 & 4 \end{array}\right)=\left(\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right) A & \left(R_{1} \rightarrow R_{1}-R_{2}\right) \\ \Rightarrow\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right)=\left(\begin{array}{cc} 1 & -1 \\ -3 & 4 \end{array}\right) A & \left(R_{2} \rightarrow R_{2}-3 R_{1}\right) \\ \Rightarrow\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)=\left(\begin{array}{cc} 4 & -5 \\ -3 & 4 \end{array}\right) A & \left(R_{1} \rightarrow R_{1}-R_{2}\right) \\ \Rightarrow A^{-1}=\left(\begin{array}{cc} 4 & -5 \\ -3 & 4 \end{array}\right) & \end{array} $$

Q: Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{cc}3 & 10 \\ 2 & 7\end{array}\right)$, if exists.

Let $A=\left(\begin{array}{cc}3 & 10 \\ 2 & 7\end{array}\right)$ We know that $A=I A$ Therefore, $$ \begin{array}{ll} \Rightarrow\left(\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}\right)=\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) A & \\ \Rightarrow\left(\begin{array}{cc} 1 & 3 \\ 2 & 7 \end{array}\right)=\left(\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right) A & \left(R_{1} \rightarrow R_{1}-R_{2}\right) \\ \Rightarrow\left(\begin{array}{cc} 1 & 3 \\ 0 & 1 \end{array}\right)=\left(\begin{array}{cc} 1 & -1 \\ -2 & 3 \end{array}\right) A & \left(R_{2} \rightarrow R_{2}-2 R_{1}\right) \\ \Rightarrow\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)=\left(\begin{array}{cc} 7 & -10 \\ -2 & 3 \end{array}\right) A & \left(R_{1} \rightarrow R_{1}-3 R_{2}\right) \\ \Rightarrow A^{-1}=\left(\begin{array}{cc} 7 & -10 \\ -2 & 3 \end{array}\right) & \end{array} $$

Q: Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{cc}3 & -1 \\ -4 & 2\end{array}\right)$, if exists.

Let $A=\left(\begin{array}{cc}3 & -1 \\ -4 & 2\end{array}\right)$ We know that $A=I A$ Therefore, $$ \begin{array}{ll} \Rightarrow\left(\begin{array}{cc} 3 & -1 \\ -4 & 2 \end{array}\right)=A\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) & \\ \Rightarrow\left(\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right)=A\left(\begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array}\right) & \left(C_{1} \rightarrow C_{1}+2 C_{2}\right) \\ \Rightarrow\left(\begin{array}{cc} 1 & 0 \\ 0 & 2 \end{array}\right)=A\left(\begin{array}{cc} 1 & 1 \\ 2 & 3 \end{array}\right) & \left(C_{2} \rightarrow C_{2}+C_{1}\right) \\ \Rightarrow\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)=A\left(\begin{array}{ll} 1 & \frac{1}{2} \\ 2 & \frac{3}{2} \end{array}\right) & \left(C_{2} \rightarrow \frac{1}{2} C_{2}\right) \\ \Rightarrow A^{-1}=\left(\begin{array}{ll} 1 & \frac{1}{2} \\ 2 & \frac{3}{2} \end{array}\right) & \end{array} $$

Q: Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ll}2 & -6 \\ 1 & -2\end{array}\right)$, if exists.

Let $A=\left(\begin{array}{ll}2 & -6 \\ 1 & -2\end{array}\right)$ We know that $A=I A$ Therefore, $$ \begin{array}{ll} \Rightarrow\left(\begin{array}{c} 2 \\ 1 \\ -2 \end{array}\right)=A\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) & \\ \Rightarrow\left(\begin{array}{ll} 2 & 0 \\ 1 & 1 \end{array}\right)=A\left(\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right) & \left(C_{2} \rightarrow C_{2}+3 C_{1}\right) \\ \Rightarrow\left(\begin{array}{ll} 2 & 0 \\ 0 & 1 \end{array}\right)=A\left(\begin{array}{ll} -2 & 3 \\ -1 & 1 \end{array}\right) & \left(C_{1} \rightarrow C_{1}-C_{2}\right) \\ \Rightarrow\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)=A\left(\begin{array}{ll} -1 & 3 \\ -\frac{1}{2} & 1 \end{array}\right) & \left(C_{1} \rightarrow \frac{1}{2} C_{1}\right) \\ \Rightarrow A^{-1}=\left(\begin{array}{ll} -1 & 3 \\ -\frac{1}{2} & 1 \end{array}\right) & \end{array} $$

Question 1

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{cc}1 & -1 \ 2 & 3\end{array}ight)$, if exists.

Accepted Answer

Let $A=\left(\begin{array}{cc}1 & -1 \ 2 & 3\end{array}ight)$

We know that $A=I A$

Therefore,

$$

\begin{array}{ll}

\Rightarrow\left(\begin{array}{cc}

1 & -1 \

2 & 3

\end{array}ight)=\left(\begin{array}{cc}

1 & 0 \

0 & 1

\end{array}ight) A & \

\Rightarrow\left(\begin{array}{cc}

1 & -1 \

0 & 5

\end{array}ight)=\left(\begin{array}{cc}

1 & 0 \

-2 & 1

\end{array}ight) A & \left(R_{2} ightarrow R_{2}-2 R_{1}ight) \

\Rightarrow\left(\begin{array}{cc}

1 & -1 \

0 & 1

\end{array}ight)=\left(\begin{array}{cc}

1 & 0 \

\frac{-2}{5} & \frac{1}{5}

\end{array}ight) A & \left(R_{2} ightarrow \frac{1}{5} R_{2}ight) \

\Rightarrow\left(\begin{array}{cc}

1 & 0 \

0 & 1

\end{array}ight)=\left(\begin{array}{cc}

\frac{3}{5} & \frac{1}{5} \

\frac{-2}{5} & \frac{1}{5}

\end{array}ight) A & \left(R_{1} ightarrow R_{1}+R_{2}ight) \

\Rightarrow A^{-1}=\left(\begin{array}{cc}

\frac{3}{5} & \frac{1}{5} \

\frac{-2}{5} & \frac{1}{5}

\end{array}ight) &

\end{array}

$$

Question 2

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ll}2 & 1 \ 1 & 1\end{array}ight)$, if exists.

Solution:

Let $A=\left(\begin{array}{ll}2 & 1 \ 1 & 1\end{array}ight)$

We know that $A=I A$

Therefore,

$$

\begin{array}{ll}

\Rightarrow\left(\begin{array}{ll}

2 & 1 \

1 & 1

\end{array}ight)=\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight) A & \

\Rightarrow\left(\begin{array}{ll}

1 & 0 \

1 & 1

\end{array}ight)=\left(\begin{array}{cc}

1 & -1 \

0 & 1

\end{array}ight) A & \left(R_{1} ightarrow R_{1}-R_{2}ight) \

\Rightarrow\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight)=\left(\begin{array}{cc}

1 & -1 \

-1 & 2

\end{array}ight) A & \left(R_{2} ightarrow R_{2}-R_{1}ight) \

\Rightarrow A^{-1}=\left(\begin{array}{cc}

1 & -1 \

-1 & 2

\end{array}ight) &

\end{array}

$$

Accepted Answer

(No solution provided.)

Question 3

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ll}1 & 3 \ 2 & 7\end{array}ight)$, if exists.

Accepted Answer

Let $A=\left(\begin{array}{ll}1 & 3 \ 2 & 7\end{array}ight)$

We know that $A=I A$

Therefore,

$$

\begin{array}{ll}

\Rightarrow\left(\begin{array}{ll}

1 & 3 \

2 & 7

\end{array}ight)=\left(\begin{array}{cc}

1 & 0 \

0 & 1

\end{array}ight) A & \

\Rightarrow\left(\begin{array}{ll}

1 & 3 \

0 & 1

\end{array}ight)=\left(\begin{array}{cc}

1 & 0 \

-2 & 1

\end{array}ight) A & \left(R_{2} ightarrow R_{2}-2 R_{1}ight) \

\Rightarrow\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight)=\left(\begin{array}{cc}

7 & -3 \

-2 & 1

\end{array}ight) A & \left(R_{1} ightarrow R_{1}-3 R_{2}ight) \

\Rightarrow A^{-1}=\left(\begin{array}{cc}

7 & -3 \

-2 & 1

\end{array}ight) &

\end{array}

$$

Question 4

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ll}2 & 3 \ 5 & 7\end{array}ight)$, if exists.

Accepted Answer

Let $A=\left(\begin{array}{ll}2 & 3 \ 5 & 7\end{array}ight)$

We know that $A=I A$

Therefore,

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{ll}

2 & 3 \

5 & 7

\end{array}ight)=\left(\begin{array}{cc}

1 & 0 \

0 & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{cc}

1 & \frac{3}{2} \

5 & 7

\end{array}ight)=\left(\begin{array}{cc}

\frac{1}{2} & 0 \

0 & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{cc}

1 & \frac{3}{2} \

0 & \frac{-1}{2}

\end{array}ight)=\left(\begin{array}{cc}

\frac{1}{2} & 0 \

\frac{-5}{2} & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{cc}

1 & 0 \

0 & \frac{-1}{2}

\end{array}ight)=\left(\begin{array}{cc}

-7 & 3 \

\frac{-5}{2} & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{cc}

1 & 0 \

0 & 1

\end{array}ight)=\left(\begin{array}{cc}

-7 & 3 \

5 & -2

\end{array}ight) A \

& \Rightarrow A^{-1}=\left(\begin{array}{cc}

-7 & 3 \

5 & -2

\end{array}ight)

\end{aligned}\left(R_{1} ightarrow R_{1}+3 R_{2}ight)

$$

Question 5

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ll}2 & 1 \ 7 & 4\end{array}ight)$, if exists.

Accepted Answer

Let $A=\left(\begin{array}{ll}2 & 1 \ 7 & 4\end{array}ight)$

We know that $A=I A$

Therefore,

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{ll}

2 & 1 \

7 & 4

\end{array}ight)=\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{ll}

1 & \frac{1}{2} \

7 & 4

\end{array}ight)=\left(\begin{array}{cc}

\frac{1}{2} & 0 \

0 & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{ll}

1 & \frac{1}{2} \

0 & \frac{1}{2}

\end{array}ight)=\left(\begin{array}{cc}

\frac{1}{2} & 0 \

\frac{-7}{2} & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{ll}

1 & 0 \

0 & \frac{1}{2}

\end{array}ight)=\left(\begin{array}{cc}

4 & -1 \

\frac{-7}{2} & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight)=\left(\begin{array}{cc}

4 & -1 \

-7 & 2

\end{array}ight) A \

& \Rightarrow A^{-1}=\left(\begin{array}{cc}

4 & -1 \

-7 & 2

\end{array}ight)

\end{aligned}\left(R_{1} ightarrow R_{2}-7 R_{1}ight) ~\left(R_{2} ightarrow 2 R_{1}ight) ~{ }

$$

Question 6

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ll}2 & 5 \ 1 & 3\end{array}ight)$, if exists.

Accepted Answer

Let $A=\left(\begin{array}{ll}2 & 5 \ 1 & 3\end{array}ight)$

We know that $A=I A$

Therefore,

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{ll}

2 & 5 \

1 & 3

\end{array}ight)=\left(\begin{array}{cc}

1 & 0 \

0 & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{ll}

1 & \frac{5}{2} \

1 & 3

\end{array}ight)=\left(\begin{array}{cc}

\frac{1}{2} & 0 \

0 & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{ll}

1 & \frac{5}{2} \

0 & \frac{1}{2}

\end{array}ight)=\left(\begin{array}{cc}

\frac{1}{2} & 0 \

\frac{-1}{2} & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{ll}

1 & 0 \

0 & \frac{1}{2}

\end{array}ight)=\left(\begin{array}{cc}

3 & -5 \

\frac{-1}{2} & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{cc}

1 & 0 \

0 & 1

\end{array}ight)=\left(\begin{array}{cc}

3 & -5 \

-1 & 2

\end{array}ight) A \

& \Rightarrow A^{-1}=\left(\begin{array}{cc}

3 & -5 \

-1 & 2

\end{array}ight)

\end{aligned}\left(R_{1} ightarrow R_{2}-R_{1}ight) .\left(R_{2} ightarrow 2 R_{2}ight){ }

$$

Question 7

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ll}3 & 1 \ 5 & 2\end{array}ight)$, if exists.

Accepted Answer

Let $A=\left(\begin{array}{ll}3 & 1 \ 5 & 2\end{array}ight)$

We know that $A=I A$

Therefore,

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{ll}

3 & 1 \

5 & 2

\end{array}ight)=A\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ll}

1 & 1 \

1 & 2

\end{array}ight)=A\left(\begin{array}{cc}

1 & 0 \

-2 & 1

\end{array}ight) \quad\left(C_{1} ightarrow C_{1}-2 C_{2}ight) \

& \Rightarrow\left(\begin{array}{ll}

1 & 0 \

1 & 1

\end{array}ight)=A\left(\begin{array}{cc}

1 & -1 \

-2 & 3

\end{array}ight) \quad\left(C_{2} ightarrow C_{2}-C_{1}ight) \

& \Rightarrow\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight)=A\left(\begin{array}{cc}

2 & -1 \

-5 & 3

\end{array}ight) \quad\left(C_{1} ightarrow C_{1}-C_{2}ight) \

& \Rightarrow A^{-1}=\left(\begin{array}{cc}

2 & -1 \

-5 & 3

\end{array}ight)

\end{aligned}

$$

Question 8

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ll}4 & 5 \ 3 & 4\end{array}ight)$, if exists.

Accepted Answer

Let $A=\left(\begin{array}{ll}4 & 5 \ 3 & 4\end{array}ight)$

We know that $A=I A$

Therefore,

$$

\begin{array}{ll}

\Rightarrow\left(\begin{array}{ll}

4 & 5 \

3 & 4

\end{array}ight)=\left(\begin{array}{cc}

1 & 0 \

0 & 1

\end{array}ight) A & \

\Rightarrow\left(\begin{array}{cc}

1 & 1 \

3 & 4

\end{array}ight)=\left(\begin{array}{cc}

1 & -1 \

0 & 1

\end{array}ight) A & \left(R_{1} ightarrow R_{1}-R_{2}ight) \

\Rightarrow\left(\begin{array}{cc}

1 & 1 \

0 & 1

\end{array}ight)=\left(\begin{array}{cc}

1 & -1 \

-3 & 4

\end{array}ight) A & \left(R_{2} ightarrow R_{2}-3 R_{1}ight) \

\Rightarrow\left(\begin{array}{cc}

1 & 0 \

0 & 1

\end{array}ight)=\left(\begin{array}{cc}

4 & -5 \

-3 & 4

\end{array}ight) A & \left(R_{1} ightarrow R_{1}-R_{2}ight) \

\Rightarrow A^{-1}=\left(\begin{array}{cc}

4 & -5 \

-3 & 4

\end{array}ight) &

\end{array}

$$

Question 9

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{cc}3 & 10 \ 2 & 7\end{array}ight)$, if exists.

Accepted Answer

Let $A=\left(\begin{array}{cc}3 & 10 \ 2 & 7\end{array}ight)$

We know that $A=I A$

Therefore,

$$

\begin{array}{ll}

\Rightarrow\left(\begin{array}{cc}

3 & 10 \

2 & 7

\end{array}ight)=\left(\begin{array}{cc}

1 & 0 \

0 & 1

\end{array}ight) A & \

\Rightarrow\left(\begin{array}{cc}

1 & 3 \

2 & 7

\end{array}ight)=\left(\begin{array}{cc}

1 & -1 \

0 & 1

\end{array}ight) A & \left(R_{1} ightarrow R_{1}-R_{2}ight) \

\Rightarrow\left(\begin{array}{cc}

1 & 3 \

0 & 1

\end{array}ight)=\left(\begin{array}{cc}

1 & -1 \

-2 & 3

\end{array}ight) A & \left(R_{2} ightarrow R_{2}-2 R_{1}ight) \

\Rightarrow\left(\begin{array}{cc}

1 & 0 \

0 & 1

\end{array}ight)=\left(\begin{array}{cc}

7 & -10 \

-2 & 3

\end{array}ight) A & \left(R_{1} ightarrow R_{1}-3 R_{2}ight) \

\Rightarrow A^{-1}=\left(\begin{array}{cc}

7 & -10 \

-2 & 3

\end{array}ight) &

\end{array}

$$

Question 10

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{cc}3 & -1 \ -4 & 2\end{array}ight)$, if exists.

Accepted Answer

Let $A=\left(\begin{array}{cc}3 & -1 \ -4 & 2\end{array}ight)$

We know that $A=I A$

Therefore,

$$

\begin{array}{ll}

\Rightarrow\left(\begin{array}{cc}

3 & -1 \

-4 & 2

\end{array}ight)=A\left(\begin{array}{cc}

1 & 0 \

0 & 1

\end{array}ight) & \

\Rightarrow\left(\begin{array}{cc}

1 & -1 \

0 & 2

\end{array}ight)=A\left(\begin{array}{cc}

1 & 0 \

2 & 1

\end{array}ight) & \left(C_{1} ightarrow C_{1}+2 C_{2}ight) \

\Rightarrow\left(\begin{array}{cc}

1 & 0 \

0 & 2

\end{array}ight)=A\left(\begin{array}{cc}

1 & 1 \

2 & 3

\end{array}ight) & \left(C_{2} ightarrow C_{2}+C_{1}ight) \

\Rightarrow\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight)=A\left(\begin{array}{ll}

1 & \frac{1}{2} \

2 & \frac{3}{2}

\end{array}ight) & \left(C_{2} ightarrow \frac{1}{2} C_{2}ight) \

\Rightarrow A^{-1}=\left(\begin{array}{ll}

1 & \frac{1}{2} \

2 & \frac{3}{2}

\end{array}ight) &

\end{array}

$$

Question 11

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ll}2 & -6 \ 1 & -2\end{array}ight)$, if exists.

Accepted Answer

Let $A=\left(\begin{array}{ll}2 & -6 \ 1 & -2\end{array}ight)$

We know that $A=I A$

Therefore,

$$

\begin{array}{ll}

\Rightarrow\left(\begin{array}{c}

2 \

1 \

-2

\end{array}ight)=A\left(\begin{array}{cc}

1 & 0 \

0 & 1

\end{array}ight) & \

\Rightarrow\left(\begin{array}{ll}

2 & 0 \

1 & 1

\end{array}ight)=A\left(\begin{array}{ll}

1 & 3 \

0 & 1

\end{array}ight) & \left(C_{2} ightarrow C_{2}+3 C_{1}ight) \

\Rightarrow\left(\begin{array}{ll}

2 & 0 \

0 & 1

\end{array}ight)=A\left(\begin{array}{ll}

-2 & 3 \

-1 & 1

\end{array}ight) & \left(C_{1} ightarrow C_{1}-C_{2}ight) \

\Rightarrow\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight)=A\left(\begin{array}{ll}

-1 & 3 \

-\frac{1}{2} & 1

\end{array}ight) & \left(C_{1} ightarrow \frac{1}{2} C_{1}ight) \

\Rightarrow A^{-1}=\left(\begin{array}{ll}

-1 & 3 \

-\frac{1}{2} & 1

\end{array}ight) &

\end{array}

$$

Question 12

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{cc}6 & -3 \ -2 & 1\end{array}ight)$, if exists.

Accepted Answer

Let $A=\left(\begin{array}{cc}6 & -3 \ -2 & 1\end{array}ight)$

We know that $A=I A$

Therefore,

$$

\begin{array}{ll}

\Rightarrow\left(\begin{array}{cc}

6 & -3 \

-2 & 1

\end{array}ight)=\left(\begin{array}{cc}

1 & 0 \

0 & 1

\end{array}ight) A & \

\Rightarrow\left(\begin{array}{cc}

1 & \frac{-1}{2} \

-2 & 1

\end{array}ight)=\left(\begin{array}{cc}

\frac{1}{6} & 0 \

0 & 1

\end{array}ight) A & \left(R_{1} ightarrow \frac{1}{6} R_{1}ight) \

\Rightarrow\left(\begin{array}{cc}

1 & \frac{-1}{2} \

0 & 0

\end{array}ight)=\left(\begin{array}{cc}

\frac{1}{6} & 0 \

\frac{1}{3} & 1

\end{array}ight) A & \left(R_{2} ightarrow R_{2}+2 R_{1}ight)

\end{array}

$$

In the above equation, we can see all the zeros in the second row of the matrix on the L.H.S. Thus, $A^{-1}$ does not exist.

Question 13

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{cc}2 & -3 \ -1 & 2\end{array}ight)$, if exists.

Accepted Answer

Let $A=\left(\begin{array}{cc}2 & -3 \ -1 & 2\end{array}ight)$

We know that $A=I A$

Therefore,

$$

\begin{array}{ll}

\Rightarrow\left(\begin{array}{cc}

2 & -3 \

-1 & 2

\end{array}ight)=\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight) A & \

\Rightarrow\left(\begin{array}{cc}

1 & -1 \

-1 & 2

\end{array}ight)=\left(\begin{array}{ll}

1 & 1 \

0 & 1

\end{array}ight) A & \left(R_{1} ightarrow R_{1}+R_{2}ight) \

\Rightarrow\left(\begin{array}{cc}

1 & -1 \

0 & 1

\end{array}ight)=\left(\begin{array}{ll}

1 & 1 \

1 & 2

\end{array}ight) A & \left(R_{2} ightarrow R_{2}+R_{1}ight) \

\Rightarrow\left(\begin{array}{cc}

1 & 0 \

0 & 1

\end{array}ight)=\left(\begin{array}{ll}

2 & 3 \

1 & 2

\end{array}ight) A & \left(R_{1} ightarrow R_{1}+R_{2}ight) \

\Rightarrow A^{-1}=\left(\begin{array}{ll}

2 & 3 \

1 & 2

\end{array}ight) &

\end{array}

$$

Question 14

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ll}2 & 1 \ 4 & 2\end{array}ight)$, if exists.

Accepted Answer

Let $A=\left(\begin{array}{ll}2 & 1 \ 4 & 2\end{array}ight)$

We know that $A=I A$

Therefore,

$$

\begin{array}{ll}

\Rightarrow\left(\begin{array}{ll}

2 & 1 \

4 & 2

\end{array}ight)=\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight) A & \

\Rightarrow\left(\begin{array}{ll}

0 & 0 \

4 & 2

\end{array}ight)=\left(\begin{array}{ll}

1 & \frac{-1}{2} \

0 & 1

\end{array}ight) A & \left(R_{1} ightarrow R_{1}-\frac{1}{2} R_{2}ight)

\end{array}

$$

In the above equation, we can see all the zeros in the first row of the matrix on the L.H.S. Thus, $A^{-1}$ does not exist.

Question 15

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ccc}2 & -3 & 3 \ 2 & 2 & 3 \ 3 & -2 & 2\end{array}ight)$, if exists.

Accepted Answer

Let $A=\left(\begin{array}{ccc}2 & -3 & 3 \ 2 & 2 & 3 \ 3 & -2 & 2\end{array}ight)$

We know that $A=I A$

Therefore,

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{ccc}

2 & -3 & 3 \

2 & 2 & 3 \

3 & -2 & 2

\end{array}ight)=\left(\begin{array}{ccc}

1 & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{ccc}

2 & -3 & 3 \

0 & 5 & 0 \

3 & -2 & 2

\end{array}ight)=\left(\begin{array}{ccc}

1 & 0 & 0 \

-1 & 1 & 0 \

0 & 0 & 1

\end{array}ight) A \quad\left(R_{2} ightarrow R_{2}-R_{1}ight) \

& \Rightarrow\left(\begin{array}{ccc}

2 & -3 & 3 \

0 & 1 & 0 \

3 & -2 & 2

\end{array}ight)=\left(\begin{array}{ccc}

1 & 0 & 0 \

\frac{-1}{5} & \frac{1}{5} & 0 \

0 & 0 & 1

\end{array}ight) A \quad\left(R_{2} ightarrow \frac{1}{5} R_{2}ight) \

& \Rightarrow\left(\begin{array}{ccc}

-1 & -1 & 1 \

0 & 1 & 0 \

3 & -2 & 2

\end{array}ight)=\left(\begin{array}{ccc}

1 & 0 & -1 \

\frac{-1}{5} & \frac{1}{5} & 0 \

0 & 0 & 1

\end{array}ight) A \quad\left(R_{1} ightarrow R_{1}-R_{3}ight) \

& \Rightarrow\left(\begin{array}{ccc}

-1 & 0 & 1 \

0 & 1 & 0 \

3 & 0 & 2

\end{array}ight)=\left(\begin{array}{ccc}

\frac{4}{5} & \frac{1}{5} & -1 \

\frac{-1}{5} & \frac{1}{5} & 0 \

\frac{-2}{5} & \frac{2}{5} & 1

\end{array}ight) A \quad\left(R_{1} ightarrow R_{1}+R_{2} 	ext { and } R_{3} ightarrow R_{3}+2 R_{2}ight) \

& \Rightarrow\left(\begin{array}{ccc}

-1 & 0 & 1 \

0 & 1 & 0 \

0 & 0 & 5

\end{array}ight)=\left(\begin{array}{ccc}

\frac{4}{5} & \frac{1}{5} & -1 \

\frac{-1}{5} & \frac{1}{5} & 0 \

2 & 1 & -2

\end{array}ight) A \quad\left(R_{3} ightarrow R_{3}+3 R_{1}ight)

\end{aligned}

$$

\begin{array}{ll}

\Rightarrow\left(\begin{array}{ccc}

-1 & 0 & 1 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight)=\left(\begin{array}{ccc}

\frac{4}{5} & \frac{1}{5} & -1 \

\frac{-1}{5} & \frac{1}{5} & 0 \

\frac{2}{5} & \frac{1}{5} & \frac{-2}{5}

\end{array}ight) A & \left(R_{3} ightarrow \frac{1}{5} R_{3}ight) \

\Rightarrow\left(\begin{array}{ccc}

-1 & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight)=\left(\begin{array}{ccc}

\frac{2}{5} & 0 & \frac{-3}{5} \

\frac{-1}{5} & \frac{1}{5} & 0 \

\frac{2}{5} & \frac{1}{5} & \frac{-2}{5}

\end{array}ight) A & \left(R_{1} ightarrow R_{1}-R_{3}ight) \

\Rightarrow\left(\begin{array}{lll}

1 & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight)=\left(\begin{array}{ccc}

\frac{-2}{5} & 0 & \frac{3}{5} \

\frac{-1}{5} & \frac{1}{5} & 0 \

\frac{2}{5} & \frac{1}{5} & \frac{-2}{5}

\end{array}ight) A & \left(R_{1} ightarrow(-1) R_{1}ight) \

\Rightarrow A^{-1}=\left(\begin{array}{ccc}

\frac{-2}{5} & 0 & \frac{3}{5} \

\frac{-1}{5} & \frac{1}{5} & 0 \

\frac{2}{5} & \frac{1}{5} & \frac{-2}{5}

\end{array}ight) &

\end{array}

$$

Question 16

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ccc}1 & 3 & -2 \ -3 & 0 & -5 \ 2 & 5 & 0\end{array}ight)$, if exists.

Accepted Answer

Let

$$

A=\left(\begin{array}{ccc}

1 & 3 & -2 \

-3 & 0 & -5 \

2 & 5 & 0

\end{array}ight)

$$

We know that $A=I A$

Therefore,

$\Rightarrow\left(\begin{array}{ccc}1 & 3 & -2 \ -3 & 0 & -5 \ 2 & 5 & 0\end{array}ight)=\left(\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}ight) A$

$\Rightarrow\left(\begin{array}{ccc}1 & 3 & -2 \ 0 & 9 & -11 \ 0 & -1 & 4\end{array}ight)=\left(\begin{array}{ccc}1 & 0 & 0 \ 3 & 1 & 0 \ -2 & 0 & 1\end{array}ight) A \quad\left(R_{2} ightarrow R_{2}+3 R_{1}ight.$ and $\left.R_{3} ightarrow R_{3}-2 R_{1}ight)$

$\Rightarrow\left(\begin{array}{ccc}1 & 0 & 10 \ 0 & 1 & 21 \ 0 & -1 & 4\end{array}ight)=\left(\begin{array}{ccc}-5 & 0 & 3 \ -13 & 1 & 8 \ -2 & 0 & 1\end{array}ight) A \quad\left(R_{1} ightarrow R_{1}+3 R_{3}ight.$ and $\left.R_{2} ightarrow R_{2}+8 R_{3}ight)$

$\Rightarrow\left(\begin{array}{lll}1 & 0 & 10 \ 0 & 1 & 21 \ 0 & 0 & 25\end{array}ight)=\left(\begin{array}{ccc}-5 & 0 & 3 \ -13 & 1 & 8 \ -15 & 1 & 9\end{array}ight) A \quad\left(R_{3} ightarrow R_{3}+R_{2}ight)$

$\Rightarrow\left(\begin{array}{ccc}1 & 0 & 10 \ 0 & 1 & 21 \ 0 & 0 & 1\end{array}ight)=\left(\begin{array}{ccc}-5 & 0 & 3 \ -13 & 1 & 8 \ \frac{-3}{5} & \frac{1}{25} & \frac{9}{25}\end{array}ight) A \quad\left(R_{3} ightarrow \frac{1}{25} R_{3}ight)$

$\Rightarrow\left(\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}ight)=\left(\begin{array}{ccc}1 & \frac{-2}{5} & \frac{-3}{5} \ \frac{-2}{5} & \frac{4}{25} & \frac{11}{25} \ \frac{-3}{5} & \frac{1}{25} & \frac{9}{25}\end{array}ight) A \quad\left(R_{1} ightarrow R_{1}-10 R_{3}ight.$ and $\left.R_{2} ightarrow R_{2}-21 R_{3}ight)$

$\Rightarrow A^{-1}=\left(\begin{array}{ccc}1 & \frac{-2}{5} & \frac{-3}{5} \ \frac{-2}{5} & \frac{4}{25} & \frac{11}{25} \ \frac{-3}{5} & \frac{1}{25} & \frac{9}{25}\end{array}ight)$

Question 17

Using elementary transformation, Find the inverse of the matrix $\left(\begin{array}{ccc}2 & 0 & -1 \ 5 & 1 & 0 \ 0 & 1 & 3\end{array}ight)$, if exists.

Accepted Answer

Let $A=\left(\begin{array}{ccc}2 & 0 & -1 \ 5 & 1 & 0 \ 0 & 1 & 3\end{array}ight)$

We know that $A=I A$

Therefore,

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{ccc}

2 & 0 & -1 \

5 & 1 & 0 \

0 & 1 & 3

\end{array}ight)=\left(\begin{array}{ccc}

1 & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{lll}

1 & 0 & \frac{-1}{2} \

5 & 1 & 0 \

0 & 1 & 3

\end{array}ight)=\left(\begin{array}{ccc}

\frac{1}{2} & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{lll}

1 & 0 & \frac{-1}{2} \

0 & 1 & \frac{5}{2} \

0 & 1 & 3

\end{array}ight)=\left(\begin{array}{ccc}

\frac{1}{2} & 0 & 0 \

\frac{-5}{2} & 1 & 0 \

0 & 0 & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{lll}

1 & 0 & \frac{-1}{2} \

0 & 1 & \frac{5}{2} \

0 & 0 & \frac{1}{2}

\end{array}ight)=\left(\begin{array}{ccc}

\frac{1}{2} & 0 & 0 \

\frac{-5}{2} & 1 & 0 \

\frac{5}{2} & -1 & 1

\end{array}ight) A \

& \Rightarrow\left(\begin{array}{lll}

1 & 0 & \frac{-1}{2} \

0 & 1 & \frac{5}{2} \

0 & 0 & 1 \

1 & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight)=\left(\begin{array}{ccc}

\frac{1}{2} & 0 & 0 \

\frac{-5}{2} & 1 & 0 \

5 & -2 & 2

\end{array}ight) A \

& \Rightarrow A^{-1}=\left(\begin{array}{ccc}

3 & -1 & 1 \

-15 & 6 & -5 \

5 & -2 & 2

\end{array}ight) A \

& \left.\left(\begin{array}{cc}

3 & 1 \

5 & -1

\end{array}ight) A R_{2}-R_{2}ight) \

& \left(R_{3} ightarrow 2 R_{3}ight) \

& \left(R_{1} ightarrow R_{1}+\frac{1}{2} R_{3} 	ext { and } R_{2} ightarrow R_{2}-\frac{5}{2} R_{3}ight)

\end{aligned}

$$

Question 18

Matrices $A$ and $B$ will be the inverse of each other only if:

(A) $A B=B A$

(B) $A B=B A=0$

(C) $A B=0, B A=I$

(D) $A B=B A=I$

Accepted Answer

We know that if $A$ is a square matrix of order $m$, and if there exists another square matrix $B$ of the same order $m$, such that $A B=B A=I$, then $B$ is said to be the inverse of $A$.

In this case, it is clear that $A$ is the inverse of $B$.

Thus, matrices $A$ and $B$ will be inverses of each other only if $A B=B A=I$.

The correct option is D.

\section*{MISCELLANEOUS EXERCISE}