Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)
Approach Solution - 1
Let \(A=\)\(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)
We know that \(A = IA\)
so\(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)= \(\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}A\)
⇒ \(\begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix}\)= \(\begin{bmatrix} 1 & 0\\ -2 & 1\end{bmatrix}A\) \((R_2\rightarrow R_2-2R_1) \)
⇒ \(\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 7 & -3\\ -2 & 1 \end{bmatrix}A\) \((R_1\rightarrow R_1-3R_2) \)
\(\therefore A^{-1}\) =\(\begin{bmatrix} 7 & -3\\ -2 & 1 \end{bmatrix}\)
Approach Solution -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: \(\mathbf{A}^{-1} = \begin{bmatrix} 7 & -3 \\ -2 & 1 \end{bmatrix}\)
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Concepts Used:
Invertible matrices
A matrix for which matrix inversion operation exists, given that it satisfies the requisite conditions is known as an invertible matrix. Any given square matrix A of order n × n is called invertible if and only if there exists, another n × n square matrix B such that, AB = BA = In, where In is an identity matrix of order n × n.
For example,
It can be observed that the determinant of the following matrices is non-zero.
