Ncert Class 12 Maths Exercise 3.4 Solutions

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |A|=(1)(3)-(-1)(2)=5 \)
So the matrix is invertible.
Now to find the inverse of the matrix,
We know \( \mathrm{AA}^{-1}=\mathrm{I} \)
Let's make augmented matrix-
\( \rightarrow[ \) A: I]
\( \rightarrow\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}: \begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-2 \mathrm{R}_{1} \)
\( \rightarrow\left[\begin{array}{cc}1 & -1 \\ 0 & 5\end{array}: \begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow\frac{ \mathrm{R}_{2} }{ 5 } \)
\( \rightarrow\left[\begin{array}{cc:cc}1 & -1 \\ 0 & 1\end{array}: \begin{array}{cc}1 & 0 \\ -\frac{2}{5} & \frac{1}{5}\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2} \)
\( \rightarrow\left[\begin{array}{cc:cc}1 & 0 \\ 0 & 1\end{array}: \begin{array}{cc}\frac{3}{5} & \frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5}\end{array}\right] \)
The matrix so obtained is of the form -
\(\rightarrow\left[\mathrm{I}: \mathrm{A}^{-1}\right]\)
Hence inverse of the given matrix-
\(\rightarrow\left[\begin{array}{cc}
\frac{3}{5} & \frac{1}{5} \\
-\frac{2}{5} & \frac{1}{5}
\end{array}\right]\)

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |\mathrm{A}|=(2)(1)-(1)(1)=1 \)
So the matrix is invertible.
Now to find the inverse of the matrix,
We know \( \mathrm{AA}^{-1}=\mathrm{I} \)
Let's make augmented matrix-
\( \rightarrow[ \) A: I]
\( \rightarrow\left[\begin{array}{cc} 2 & 1 \\ 1 & 1 \end{array}: \begin{array}{ll} 1 & 0 \\ 0 & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2} \)
\( \rightarrow\left[\begin{array}{cc}1 & 0 \\ 1 & 1\end{array}: \begin{array}{ll}1 & -1 \\ 0 & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \)
\( \rightarrow\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}: \begin{array}{ll}1 & -1 \\ -1 & 2\end{array}\right] \)
The matrix so obtained is of the form -
\( \rightarrow\left[\right. \) I: \( \left.\mathrm{A}^{-1}\right] \)
Hence inverse of the given matrix-
\( \rightarrow\left[\begin{array}{cc}1 & -1 \\ -1 & 2\end{array}\right] \)

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |\mathrm{A}|=(1)(7)-(2)(3)=1 \)
So the matrix is invertible.
Now to find the inverse of the matrix,
We know \( \mathrm{AA}^{-1}=\mathrm{I} \)
Let's make augmented matrix-
\(\rightarrow\left[\mathrm{A}: \mathrm{I}\right]\)
\( \rightarrow\left[\begin{array}{cc}1 & 3 \\ 2 & 7\end{array}: \begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-2 \mathrm{R}_{1} \)
\( \rightarrow\left[\begin{array}{cc}1 & 3 \\ 0 & 1\end{array}: \begin{array}{ll}1 & 0 \\ -2 & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-3 \mathrm{R}_{2} \)
\( \rightarrow\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}: \begin{array}{ll}7 & -3 \\ 2 & 0\end{array}\right] \)
The matrix so obtained is of the form -
\(\rightarrow\left[\mathrm{I}: \mathrm{A}^{-1}\right]\)
Hence inverse of the given matrix-
\(\rightarrow\left[\begin{array}{cc}
7 & -3 \\
-2 & 1
\end{array}\right]\)

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
\(\begin{array}{l}
\rightarrow \text { [A: I] }
\end{array}\)
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |\mathrm{A}|=(2)(7)-(5)(3)=-1 \)
So the matrix is invertible.
Now to find the inverse of the matrix,
We know \( \mathrm{AA}^{-1}=\mathrm{I} \)
Let's make augmented matrix-
\( \rightarrow \) [A: I]
\( \rightarrow\left[\begin{array}{llll}2 & 3 & 1 & 0 \\ 5 & 7 & 0 & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\frac{5}{2} \mathrm{R} 1 \)
\( \rightarrow\left[\begin{array}{cccc}2 & 3 & 1 & 0 \\ 0 & -\frac{1}{2} & -\frac{5}{2} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \frac{ \mathrm{R}_{1} }{ 2 } \)
\( \rightarrow\left[\begin{array}{cccc}1 & \frac{3}{2} & \frac{1}{2} & 0 \\ 0 & -\frac{1}{2} & -\frac{5}{2} & 1\end{array}\right] \)
Apply row operation- \( R_{1} \rightarrow R_{1}+3 R_{2} \)
\( \rightarrow\left[\begin{array}{cccc}1 & 0 & -7 & 3 \\ 0 & -\frac{1}{2} & -\frac{5}{2} & 1\end{array}\right] \)
Apply row operation- \( R_{2} \rightarrow-2 R_{2} \)
\( \rightarrow\left[\begin{array}{cccc}1 & 0 & -7 & 3 \\ 0 & 1 & 5 & -2\end{array}\right] \)
The matrix so obtained is of the form -
\( \rightarrow\left[\mathrm{I}: \mathrm{A}^{-1}\right] \)
Hence inverse of the given matrix-
\(\rightarrow\left[\begin{array}{cc}
-7 & 3 \\
5 & -2
\end{array}\right]\)

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |A|=(2)(4)-(1)(7)=1 \)
So the matrix is invertible.
Now to find the inverse of the matrix,
We know \( \mathrm{AA}^{-1}=\mathrm{I} \)
Let's make augmented matrix-
\( \rightarrow \) [A: I]
\( \rightarrow\left[\begin{array}{llll}2 & 1 & 1 & 0 \\ 7 & 4 & 0 & 1\end{array}\right] \)
Apply row operation- \( R_{2} \rightarrow R_{2}-\frac{7}{2} R_{1} \)
\( \rightarrow\left[\begin{array}{cccc}2 & 1 & 1 & 0 \\ 0 & \frac{1}{2} & -\frac{7}{2} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow 2 \mathrm{R}_{2} \)
\( \rightarrow\left[\begin{array}{llcc}2 & 1 & 1 & 0 \\ 0 & 1 & -7 & 2\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2} \)
\( \rightarrow\left[\begin{array}{llcc}2 & 0 & 8 & -2 \\ 0 & 1 & -7 & 2\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \frac{1}{2} \mathrm{R}_{1} \)
\( \rightarrow\left[\begin{array}{cccc}1 & 0 & 4 & -1 \\ 0 & 1 & -7 & 2\end{array}\right] \)
The matrix so obtained is of the form -
\( \rightarrow\left[\mathrm{I}: \mathrm{A}^{-1}\right] \)
Hence inverse of the given matrix-
\( \rightarrow\left[\begin{array}{cc}4 & -1 \\ -7 & 2\end{array}\right] \)

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |A|=(2)(3)-(1)(5)=1 \)
So the matrix is invertible.
Now to find the inverse of the matrix,
We know \( \mathrm{AA}^{-1}=\mathrm{I} \)
Let's make augmented matrix-
\( \rightarrow[ \) A: I]
\( \rightarrow\left[\begin{array}{llll}2 & 5 & 1 & 0 \\ 1 & 3 & 0 & 1\end{array}\right] \)
Apply row operation- \( R_{2} \rightarrow R_{2}-\frac{1}{2} R_{1} \)
\( \rightarrow\left[\begin{array}{cccc}2 & 5 & 1 & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \frac{ \mathrm{R}_{1} }{ 2 } \)
\( \rightarrow\left[\begin{array}{cccc}1 & \frac{5}{2} & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} & 1\end{array}\right] \)
Apply row operation- \( R_{1} \rightarrow R_{1}-5 R_{2} \)
\( \rightarrow\left[\begin{array}{cccc}1 & 0 & 3 & -5 \\ 0 & \frac{1}{2} & \frac{-1}{2} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow 2 \mathrm{R}_{2} \)
\( \rightarrow\left[\begin{array}{cccc}1 & 0 & 3 & -5 \\ 0 & 1 & -1 & 2\end{array}\right] \)
The matrix so obtained is of the form -
\( \rightarrow\left[\right. \) I: \( \left.\mathrm{A}^{-1}\right] \)
Hence inverse of the given matrix-
\( \rightarrow\left[\begin{array}{cc}3 & -5 \\ -1 & 2\end{array}\right] \)

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |A|=(3)(2)-(5)(1)=1 \)
So the matrix is invertible.
Now to find the inverse of the matrix,
We know \( \mathrm{AA}^{-1}=\mathrm{I} \)
Let's make augmented matrix-
\( \rightarrow[ \) A: I]
\( \rightarrow\left[\begin{array}{cccc}3 & 1 & 1 & 0 \\ 5 & 2 & 0 & 1\end{array}\right] \)
Apply row operation- \( R_{2} \rightarrow R_{2}-\frac{5}{3} R_{1} \)
\( \rightarrow\left[\begin{array}{cccc}3 & 1 & 1 & 0 \\ 0 & \frac{1}{3} & -\frac{5}{3} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow 3 \mathrm{R}_{2} \)
\( \rightarrow\left[\begin{array}{cccc}3 & 1 & 1 & 0 \\ 0 & 1 & -5 & 3\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2} \)
\( \rightarrow\left[\begin{array}{cccc}3 & 1 & 6 & -3 \\ 0 & 1 & -5 & 3\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \frac{1}{3} \mathrm{R}_{2} \)
\(\rightarrow\left[\begin{array}{cccc}
1 & 0 & 2 & -1 \\
0 & 1 & -5 & 3
\end{array}\right]\)
The matrix so obtained is of the form -
\(\rightarrow\left[\mathrm{I}: \mathrm{A}^{-1}\right]\)
Hence inverse of the given matrix-
\(\rightarrow\left[\begin{array}{cc}
2 & -1 \\
-5 & 3
\end{array}\right]\)

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |A|=(4)(4)-(5)(3)=1 \)
So the matrix is invertible.
Now to find the inverse of the matrix,
We know \( \mathrm{AA}^{-1}=\mathrm{I} \)
Let's make augmented matrix-
\( \rightarrow[ \) A: I]
\( \rightarrow\left[\begin{array}{llll}4 & 5 & 1 & 0 \\ 0 & 4 & 0 & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\frac{3}{4} \mathrm{R}_{1} \)
\( \rightarrow\left[\begin{array}{cccc}4 & 5 & 1 & 0 \\ 0 & \frac{1}{4} & -\frac{3}{4} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \frac{ \mathrm{R}_{1} }{ 4 } \)
\( \rightarrow\left[\begin{array}{cccc}1 & \frac{5}{4} & \frac{1}{4} & 0 \\ 0 & \frac{1}{4} & -\frac{3}{4} & 1\end{array}\right] \)
Apply row operation- \( R_{1} \rightarrow R_{1}-5 R_{2} \)
\( \rightarrow\left[\begin{array}{llcc}1 & 0 & 4 & -5 \\ 0 & \frac{1}{4} \vdots & -\frac{3}{4} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow 4 \mathrm{R}_{2} \)
\( \rightarrow\left[\begin{array}{cccc}1 & 0 & 4 & -5 \\ 0 & 1 & -3 & 4\end{array}\right] \)
The matrix so obtained is of the form -
\( \rightarrow\left[\right. \) I: \( \left.\mathrm{A}^{-1}\right] \)
Hence inverse of the given matrix -
\( \rightarrow\left[\begin{array}{cc}4 & -5 \\ -3 & 4\end{array}\right] \)

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |A|=(3)(7)-(2)(10)=1 \)
So the matrix is invertible.
Now to find the inverse of the matrix,
We know \( \mathrm{AA}^{-1}=\mathrm{I} \)
Let's make augmented matrix-
\( \rightarrow[\mathrm{A}: \mathrm{I}] \)
\( \rightarrow\left[\begin{array}{cccc}3 & 10 & 1 & 0 \\ 2 & 7 & 0 & 1\end{array}\right] \)
Apply row operation- \( R_{2} \rightarrow R_{2}-\frac{2}{3} R_{1} \)
\( \rightarrow\left[\begin{array}{cccc}3 & 10 & 1 & 0 \\ 0 & \frac{1}{3} & -\frac{2}{3} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \frac{ \mathrm{R}_{1} }{ 3 } \)
\( \rightarrow\left[\begin{array}{cccc}1 & \frac{10}{3} & \frac{1}{3} & 0 \\ 0 & \frac{1}{3} & -\frac{2}{3} & 1\end{array}\right] \)
Apply row operation- \( R_{1} \rightarrow R_{1}-10 R_{2} \)
\( \rightarrow\left[\begin{array}{ccrc}1 & 0 & 7 & -10 \\ 0 & \frac{1}{3} & -\frac{2}{3} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow 3 \mathrm{R}_{2} \)
\( \rightarrow\left[\begin{array}{cccc}1 & 0 & 7 & -10 \\ 0 & 1 & -2 & 3\end{array}\right] \)
The matrix so obtained is of the form -
\(\rightarrow\left[\mathrm{I}: \mathrm{A}^{-1}\right]\)
Hence inverse of the given matrix-
\(\rightarrow\left[\begin{array}{cc}
7 & -10 \\
-2 & 3
\end{array}\right]\)

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |\mathrm{A}|=(3)(2)-(-1)(-4)=2 \)
So the matrix is invertible.
Now to find the inverse of the matrix,
We know \( \mathrm{AA}^{-1}=\mathrm{I} \)
Let's make augmented matrix-
\(\begin{array}{l}
\rightarrow \text { [A: I] }
\end{array}\)
\( \rightarrow\left[\begin{array}{cc}3 & -1 \\ -4 & 2\end{array} \begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+\frac{4}{3} \mathrm{R}_{1} \)
\( \rightarrow\left[\begin{array}{cccc}3 & -1 & 1 & 0 \\ 0 & \frac{2}{3} & \frac{4}{3} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \frac{ \mathrm{R}_{1} }{ 3 } \)
\( \rightarrow\left[\begin{array}{cccc}1 & -\frac{1}{3} & \frac{1}{3} & 0 \\ 0 & \frac{2}{3} & \frac{4}{3} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\frac{1}{2} \mathrm{R}_{2} \)
\( \rightarrow\left[\begin{array}{llll}1 & 0 & 1 & \frac{1}{2} \\ 0 & \frac{2}{4} & \frac{4}{3} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow \frac{3}{2} \mathrm{R}_{2} \)
\(\rightarrow\left[\begin{array}{llll}
1 & 0 & 1 & \frac{1}{2} \\
0 & 1 & 2 & \frac{3}{2}
\end{array}\right]\)
The matrix so obtained is of the form -
\(\rightarrow\left[\mathrm{I}: \mathrm{A}^{-1}\right]\)
Hence inverse of the given matrix-
\(\rightarrow\left[\begin{array}{llll}
1 & 0 & 1 & \frac{1}{2} \\
0 & 1 & 2 & \frac{3}{2}
\end{array}\right]\)

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |\mathrm{A}|=(2)(-2)-(-6)(1)=2 \)
So the matrix is invertible.
Now to find the inverse of the matrix,
We know \( \mathrm{AA}^{-1}=\mathrm{I} \)
Let's make augmented matrix-
\( \rightarrow[\mathrm{A}: \mathrm{I}] \)
\( \rightarrow\left[\begin{array}{ll}2 & -6 & 1 & 0 \\1 & -2 & 0 & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\frac{1}{2} \mathrm{R}_{1} \)
\( \rightarrow\left[\begin{array}{cccc}2 & -6 & 1 & 0 \\ 0 & 1 & -\frac{1}{2} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \frac{ \mathrm{R}_{1} }{ 2 } \)
\( \rightarrow\left[\begin{array}{cccc}1 & -3 & \frac{1}{2} & 0 \\ 0 & 1 & -\frac{1}{2} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+3 \mathrm{R}_{2} \)
\( \rightarrow\left[\begin{array}{cccc}1 & 0 & -1 & 3 \\ 0 & 1 & -\frac{1}{2} & 1\end{array}\right] \)
The matrix so obtained is of the form -
\( \rightarrow\left[\right. \) I: \( \left.\mathrm{A}^{-1}\right] \)
Hence inverse of the given matrix-
\(\rightarrow\left[\begin{array}{ll}
-1 & 3 \\
-\frac{1}{2} & 1
\end{array}\right]\)

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |A|=(6)(1)-(-2)(-3)=0 \)
So the matrix is not invertible.
\( \therefore \) the inverse of the given matrix does not exist.

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |A|=(2)(2)-(-1)(-3)=1 \)
So the matrix is invertible.
Now to find the inverse of the matrix,
We know \( \mathrm{AA}^{-1}=\mathrm{I} \)
Let's make augmented matrix-
\( \rightarrow \) A: I]
\( \rightarrow\left[\begin{array}{cccc}2 & -1 & 1 & 0 \\ -3 & 0 & 0 & 1\end{array}\right] \)
Apply row operation- \( R_{2} \rightarrow R_{2}+\frac{3}{2} R_{1} \)
\( \rightarrow\left[\begin{array}{cccc}2 & -1 & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & \frac{3}{2} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \frac{ \mathrm{R}_{1} }{ 2 } \)
\( \rightarrow\left[\begin{array}{cccc}1 & -\frac{1}{2} & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & \frac{3}{2} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2} \)
\( \rightarrow\left[\begin{array}{cccc}1 & 0 & 2 & 1 \\ 0 & \frac{1}{2} & \frac{3}{2} & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow 2 \mathrm{R}_{2} \)
\( \rightarrow\left[\begin{array}{cccc}1 & 0 & 2 & 1 \\ 0 & 1 & 3 & 2\end{array}\right] \)
The matrix so obtained is of the form -
\( \rightarrow\left[\right. \) I: \( \left.\mathrm{A}^{-1}\right] \)
Hence inverse of the given matrix-
\( \rightarrow\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right] \)

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |\mathrm{A}|=(2)(2)-(1)(4)=0 \)
So the matrix is not invertible.
\( \therefore \) the inverse of the given matrix does not exist.

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |\mathrm{A}|=[(2)\{2 \times 2-3 \times(-2)\}-(-3)\{2 \times 2-3 \times 3\}+(3)\{2 \times(-2)- \)
\(2 \times 3\}]\)
\(=[2\{4-(-6)\}+3\{4-9\}+3\{-4-6\}]\)
\(=[2(10)+3(-5)+3(-10)]\)
\(=[20-15-30]\)
\(=-25\)
So the matrix is invertible.
Now to find the inverse of the matrix,
We know \( \mathrm{AA}^{-1}=\mathrm{I} \)
Let's make augmented matrix-
\( \rightarrow[\mathrm{A}: \mathrm{I}] \)
\( \rightarrow\left[\begin{array}{cccccc}2 & -3 & 3 & 1 & 0 & 0 \\ 2 & 2 & 3 & 0 & 1 & 0 \\ 3 & -2 & 2 & 0 & 0 & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \frac{1}{2} \mathrm{R}_{1} \)
\(\rightarrow\left[\begin{array}{cccccc}
1 & -\frac{3}{2} & \frac{3}{2} & \frac{1}{2} & 0 & 0 \\
2 & 2 & 3 & 0 & 1 & 0 \\
3 & -2 & 2 & 0 & 0 & 1
\end{array}\right]\)
Apply row operation- \( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-2 \mathrm{R}_{1} \)
\( \rightarrow\left[\begin{array}{cccccc}1 & -\frac{3}{2} & \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 5 & 0 & -1 & 1 & 0 \\ 3 & -2 & 2 & 0 & 0 & 1\end{array}\right] \)
Apply row operation- \( R_{3} \rightarrow R_{3}-3 R_{1} \)
\( \rightarrow\left[\begin{array}{cccccc}1 & -\frac{3}{2} & \frac{3}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 5 & 0 & -1 & 1 & 0\\ 0 & \frac{5}{2} & -\frac{5}{2} & -\frac{3}{2} & 0 & 1 \end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow \frac{1}{5} \mathrm{R}_{2} \)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\frac{3}{2} \mathrm{R}_{2} \)
\(\rightarrow\left[\begin{array}{cccccc}
1 & 0 & \frac{3}{2} & \frac{1}{5} & \frac{3}{10} & 0 \\
0 & 1 & 0 & -\frac{1}{5} & \frac{1}{5} & 0 \\
0 & \frac{5}{2} & -\frac{5}{2} & -\frac{3}{2} & 0 & 1
\end{array}\right]\)
Apply row operation- \( R_{3} \rightarrow R_{3}-\frac{5}{2} R_{2} \)
\(\rightarrow\left[\begin{array}{cccccc}
1 & 0 & \frac{3}{2} & \frac{1}{5} & \frac{3}{10} & 0 \\
0 & 1 & 0 & -\frac{1}{5} & \frac{1}{5} & 0 \\
0 & 0 & -\frac{5}{2} & -1 & -\frac{1}{2} & 1
\end{array}\right]\)
Apply row operation- \( \mathrm{R}_{3} \rightarrow-\frac{2}{5} \mathrm{R}_{3} \)
\(\rightarrow\left[\begin{array}{cccccc}
1 & 0 & \frac{3}{2} & \frac{1}{5} & \frac{3}{10} & 0 \\
0 & 1 & 0 & -\frac{1}{5} & \frac{1}{5} & 0 \\
0 & 0 & 1 & \frac{2}{5} & \frac{1}{5} & -\frac{2}{5}
\end{array}\right]\)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\frac{3}{2} \mathrm{R}_{3} \)
\(\rightarrow\left[\begin{array}{cccccc}
1 & 0 & 0 & -\frac{2}{5} & 0 & \frac{3}{5} \\
0 & 1 & 0 & -\frac{1}{5} & \frac{1}{5} & 0 \\
0 & 0 & 1 & \frac{2}{5} & \frac{1}{5} & -\frac{2}{5}
\end{array}\right]\)
The matrix so obtained is of the form -
\(\rightarrow\left[\mathrm{I}: \mathrm{A}^{-1}\right]\)
Hence inverse of the given matrix-
\(\rightarrow\left[\begin{array}{ccc}
-\frac{2}{5} & 0 & \frac{3}{5} \\
-\frac{1}{5} & \frac{1}{5} & 0 \\
\frac{2}{5} & \frac{1}{5} & -\frac{2}{5}
\end{array}\right]\)

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |\mathrm{A}|=[(1)\{0-5 \times(-5)\}-(3)\{(-3) \times 0-(-5) \times 2\}+(-2)\{5 \times(-3) \)
\(\begin{array}{l}
-2 \times 0\}] \\
=[1\{25\}-3\{0+10\}-2\{-15\}] \\
=[1(25)-3(10)-2(-15)]
\end{array}\)
\(\begin{array}{l}
=[25-30+30] \\
=25
\end{array}\)
So the matrix is invertible.
Now to find the inverse of the matrix,
We know \( \mathrm{AA}^{-1}=\mathrm{I} \)
Let's make augmented matrix-
\( \rightarrow[\mathrm{A}: \mathrm{I}] \)
\( \rightarrow\left[\begin{array}{cccccc}1 & 3 & -2 & 1 & 0 & 0 \\ -3 & 0 & -5 & 0 & 1 & 0 \\ 2 & 5 & 0 & 0 & 0 & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+3 \mathrm{R}_{1} \)
\(\rightarrow\left[\begin{array}{cccccc}
1 & 3 & -2 & 1 & 0 & 0 \\
0 & 9 & -11 & 3 & 1 & 0 \\
2 & 5 & 0 & 0 & 0 & 1
\end{array}\right]\)
Apply row operation- \( \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-2 \mathrm{R}_{1} \)
\(\rightarrow\left[\begin{array}{cccccc}
1 & 3 & -2 & 1 & 0 & 0 \\
0 & 9 & -11 & 3 & 1 & 0 \\
0 & -1 & 4 & -2 & 0 & 1
\end{array}\right]\)
Apply row operation- \( \mathrm{R}_{2} \rightarrow \frac{1}{9} \mathrm{R}_{2} \)
\(\rightarrow\left[\begin{array}{cccccc}
1 & 3 & -2 & 1 & 0 & 0 \\
0 & 1 & -\frac{11}{9} & \frac{1}{3} & \frac{1}{9} & 0 \\
0 & -1 & 4 & -2 & 0 & 1
\end{array}\right]\)
Apply row operation- \( R_{1} \rightarrow R_{1}-3 R_{2} \)
\(\rightarrow\left[\begin{array}{cccccc}
1 & 0 & \frac{5}{3} & 0 & -\frac{1}{3} & 0 \\
0 & 1 & -\frac{11}{9} & \frac{1}{3} & \frac{1}{9} & 0 \\
0 & -1 & 4 & -2 & 0 & 1
\end{array}\right]\)
Apply row operation- \( \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}+\mathrm{R}_{2} \)
\(\rightarrow\left[\begin{array}{cccccc}
1 & 0 & \frac{5}{3} & 0 & -\frac{1}{3} & 0 \\
0 & 1 & -\frac{11}{9} & \frac{1}{3} & \frac{1}{9} & 0 \\
0 & 0 & \frac{25}{9} & -\frac{5}{3} & \frac{1}{9} & 1
\end{array}\right]\)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3} \)
\(\rightarrow\left[\begin{array}{cccccc}
1 & 0 & 0 & 1 & -\frac{2}{5} & -\frac{3}{5} \\
0 & 1 & -\frac{11}{9} & \frac{1}{3} & \frac{1}{9} & 0 \\
0 & 0 & \frac{25}{9} & -\frac{5}{3} & \frac{1}{9} & 1
\end{array}\right]\)
Apply row operation- \( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+\frac{11}{25} \mathrm{R}_{3} \)
\(\rightarrow\left[\begin{array}{cccccc}
1 & 0 & 0 & 1 & -\frac{2}{5} & -\frac{3}{5} \\
0 & 1 & 0 & -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\
0 & 0 & \frac{25}{9} & -\frac{5}{3} & \frac{1}{9} & 1
\end{array}\right]\)
Apply row operation- \( \mathrm{R}_{3} \rightarrow \frac{9}{25} \mathrm{R}_{3} \)
\(\rightarrow\left[\begin{array}{cclccc}
1 & 0 & 0 & 1 & -\frac{2}{5} & -\frac{3}{5} \\
0 & 1 & 0 & -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\
0 & 0 & 1 & -\frac{3}{5} & \frac{1}{25} & \frac{9}{25}
\end{array}\right]\)
The matrix so obtained is of the form -
\( \rightarrow\left[\mathrm{I}: \mathrm{A}^{-1}\right] \)
Hence inverse of the given matrix-
\(\rightarrow\left[\begin{array}{ccc}
1 & -\frac{2}{5} & -\frac{3}{5} \\
-\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\
-\frac{3}{5} & \frac{1}{25} & \frac{9}{25}
\end{array}\right]\)

First of all, we need to check whether the matrix is invertible or not. For that-
For the inverse of a matrix A to exist,
Determinant of \( \mathrm{A} \neq 0 \)
Here \( |\mathrm{A}|=[(2)\{1 \times 3-1 \times(0)\}-(0)\{3 \times 5-0 \times 0\}+(-1)\{5 \times 1-1 \times \) \( 0\}] \)
\(=[2\{3\}-0\{15\}-1\{5\}]\)
\(=[6-0-5]\)
\(=1\)
So the matrix is invertible.
Now to find the inverse of the matrix,
We know \( \mathrm{AA}^{-1}=\mathrm{I} \)
Let's make augmented matrix-
\( \rightarrow[\mathrm{A}: \mathrm{I}] \)
\(\rightarrow\left[\begin{array}{cccccc}
2 & 0 & -1 & & 1 & 0 & 0 \\
5 & 1 & 0 & \vdots & 0 & 1 & 0 \\
0 & 1 & 3 & & 0 & 0 & 1 \\
\end{array}\right]\)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \frac{1}{2} \mathrm{R}_{1} \)
\( \rightarrow\left[\begin{array}{cccccc}1 & 0 & -\frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 5 & 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 3 & 0 & 0 & 1\end{array}\right] \)
Apply row operation- \( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-5 \mathrm{R}_{1} \)
\(\rightarrow\left[\begin{array}{cccccc}
1 & 0 & -\frac{1}{2} & \frac{1}{2} & 0 & 0 \\
0 & 1 & \frac{5}{2} & \vdots-\frac{5}{2} & 1 & 0 \\
0 & 0 & 3 & 0 & 0 & 1
\end{array}\right]\)
Apply row operation- \( R_{3} \rightarrow R_{3}-R_{2} \)
\(\rightarrow\left[\begin{array}{cccccc}
1 & 0 & -\frac{1}{2} & \frac{1}{2} & 0 & 0 \\
0 & 1 & \frac{5}{2} & \vdots-\frac{5}{2} & 1 & 0 \\
0 & 0 & \frac{1}{2} & \frac{5}{2} & -1 & 1
\end{array}\right]\)
Apply row operation- \( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{3} \)
\(\rightarrow\left[\begin{array}{cclccc}
1 & 0 & 0 & 3 & -1 & 1 \\
0 & 1 & \frac{5}{2} & -\frac{5}{2} & 1 & 0 \\
0 & 0 & \frac{1}{2} & \frac{5}{2} & -1 & 1
\end{array}\right]\)
Apply row operation- \( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-5 \mathrm{R}_{3} \)
\(\rightarrow\left[\begin{array}{cclccc}
1 & 0 & 0 & 3 & -1 & 1 \\
0 & 1 & 0 & -15 & 6 & -5 \\
0 & 0 & \frac{1}{2} & \frac{5}{2} & -1 & 1
\end{array}\right]\)
Apply row operation- \( \mathrm{R}_{3} \rightarrow 2 \mathrm{R}_{3} \)
\(\rightarrow\left[\begin{array}{cccccc}
1 & 0 & 0 & 3 & -1 & 0 \\
0 & 1 & \frac{5}{2} & \vdots-15 & 6 &-5 \\
0 & 0 & 1 & 5 & -2 & 2
\end{array}\right]\)
The matrix so obtained is of the form -
\(\rightarrow\left[\mathrm{I}: \mathrm{A}^{-1}\right]\)
Hence inverse of the given matrix-
\(\rightarrow\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\)

Here it is given that A & B are inverse of each other.
\(\therefore \mathrm{A}-1=\mathrm{B}\ldots(i)\)
Also \(B-1 = A \ldots(ii)\)
From definition of inverse matrix, we know that-
\(\begin{array}{l}
\rightarrow \mathrm{AA}^{-1}=\mathrm{I} \\
\because \mathrm{A}-1=\mathrm{B}\{\text { from eq. (i) }\} \\
\rightarrow \mathrm{AB}=\mathrm{I} \ldots(iii)\end{array}\)
Similarly, \( \mathrm{BB}^{-1}=\mathrm{I} \)
\( \because \mathrm{B}-1=\mathrm{A}\{ \) from eq. (ii) \( \} \)
\( \rightarrow \mathrm{BA}=\mathrm{I} \ldots(iv)\)
So \( \mathrm{AB}=\mathrm{BA}=\mathrm{I} \) {from eq. (iii) and eq. (iv)
Hence option D is the correct answer.