Question

Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 &  3\\ 2 & 7\end{bmatrix}\)

Answer 1

Let \(A=\)\(\begin{bmatrix} 1 &  3\\ 2 & 7\end{bmatrix}\)

We know that \(A = IA\) 

so\(\begin{bmatrix} 1 &  3\\ 2 & 7\end{bmatrix}\)= \(\begin{bmatrix} 1 &  0\\ 0 & 1 \end{bmatrix}A\) 

⇒ \(\begin{bmatrix} 1 &  3\\ 0 & 1\end{bmatrix}\)= \(\begin{bmatrix} 1 &  0\\ -2 & 1\end{bmatrix}A\) \((R_2\rightarrow R_2-2R_1) \)

⇒ \(\begin{bmatrix} 1 &  0\\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 7 &  -3\\ -2 & 1 \end{bmatrix}A\)  \((R_1\rightarrow R_1-3R_2) \)

\(\therefore A^{-1}\) =\(\begin{bmatrix} 7 &  -3\\ -2 & 1 \end{bmatrix}\)

Answer 2

\(\mathbf{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\)

The inverse \(\mathbf{A}^{-1}\) is:

\(\mathbf{A}^{-1} = \frac{1}{\det(\mathbf{A})} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\)

where \(\det(\mathbf{A})\) (the determinant of \(\mathbf{A}\)) is given by:

\(\det(\mathbf{A}) = ad - bc\)

Given the matrix:

\(\mathbf{A} = \begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix}\)

First, we compute the determinant \(\det(\mathbf{A})\):

\(\det(\mathbf{A}) = (1)(7) - (3)(2) = 7 - 6 = 1\)

Since the determinant is not zero, the inverse exists. Now, we apply the formula for the inverse:

\(\mathbf{A}^{-1} = \frac{1}{\det(\mathbf{A})} \begin{bmatrix} 7 & -3 \\ -2 & 1 \end{bmatrix}\)

Substituting \(\det(\mathbf{A}) = 1\):

\(\mathbf{A}^{-1} = \frac{1}{1} \begin{bmatrix} 7 & -3 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 7 & -3 \\ -2 & 1 \end{bmatrix}\)

So, the answer is:  \(\mathbf{A}^{-1} = \begin{bmatrix} 7 & -3 \\ -2 & 1 \end{bmatrix}\)