Let A=\(\begin{bmatrix} 2 & 3\\ 5 & 7 \end{bmatrix}\)
We know that \(A = IA\)
⇒ \(\begin{bmatrix} 2 & 3\\ 5 & 7 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}A\)
⇒ \(\begin{bmatrix} 1 & \frac32\\ 5 & 7 \end{bmatrix}\)=\(\begin{bmatrix} \frac12 & 0\\ 0 & 1 \end{bmatrix}\)\(A\) \((R_1 \rightarrow (\frac{1}{2}R_1) )\)
⇒ \(\begin{bmatrix} 1 & \frac32\\ 0 & \frac{-1}{2} \end{bmatrix}\)= \(\begin{bmatrix} \frac12 & 0\\ -\frac{5}{2} & 1 \end{bmatrix}A\) \( (R_2→ R_2-5R_1) \)
⇒ \(\begin{bmatrix} 1 & 0\\ 0 & \frac{-1}{2} \end{bmatrix}\)\(=\begin{bmatrix} -7 & 3\\ -\frac{5}{2} & 1 \end{bmatrix}A\) \((R_1→ R_1+3R_2)\)
⇒ \(\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\) = \(\begin{bmatrix} -7 & 3\\ 5 & -2 \end{bmatrix}A\) \((R_2\rightarrow-2R_1)\)
\(A^{-1}\)= \(\begin{bmatrix} -7 & 3\\ 5 & -2 \end{bmatrix}\)
\(\mathbf{A} = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}\)
we use the formula for the inverse of a \(2 \times 2\) matrix:
\(\mathbf{A}^{-1} = \frac{1}{\det(\mathbf{A})} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\)
where \(\mathbf{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\) and \(\det(\mathbf{A}) = ad - bc\).
First, we calculate the determinant \(\det(\mathbf{A})\):
\(\det(\mathbf{A}) = (2)(7) - (3)(5) = 14 - 15 = -1\)
Since the determinant is not zero, the inverse exists. Now, we apply the formula for the inverse:
\(\mathbf{A}^{-1} = \frac{1}{\det(\mathbf{A})} \begin{bmatrix} 7 & -3 \\ -5 & 2 \end{bmatrix}\)
Substituting \(\det(\mathbf{A}) = -1\):
\(\mathbf{A}^{-1} = \frac{1}{-1} \begin{bmatrix} 7 & -3 \\ -5 & 2 \end{bmatrix} = \begin{bmatrix} -7 & 3 \\ 5 & -2 \end{bmatrix}\)
So, the answer is \(\begin{bmatrix} -7 & 3 \\ 5 & -2 \end{bmatrix}\).
What is the Planning Process?
Evaluate \(\begin{vmatrix} cos\alpha cos\beta &cos\alpha sin\beta &-sin\alpha \\ -sin\beta&cos\beta &0 \\ sin\alpha cos\beta&sin\alpha\sin\beta &cos\alpha \end{vmatrix}\)
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.
It can be observed that the determinant of the following matrices is non-zero.