Question

Find the inverse of each of the matrices,if it exists. \(\begin{bmatrix} 2 &  3\\ 5 & 7 \end{bmatrix}\)

Answer 1

Let A=\(\begin{bmatrix} 2 &  3\\ 5 & 7 \end{bmatrix}\)  

We know that \(A = IA\) 

⇒ \(\begin{bmatrix} 2 &  3\\ 5 & 7 \end{bmatrix}\) = \(\begin{bmatrix} 1 &  0\\ 0 & 1 \end{bmatrix}A\)  

⇒ \(\begin{bmatrix} 1 &  \frac32\\ 5 & 7 \end{bmatrix}\)=\(\begin{bmatrix} \frac12 &  0\\ 0 & 1 \end{bmatrix}\)\(A\)          \((R_1 \rightarrow  (\frac{1}{2}R_1)  )\) 

⇒ \(\begin{bmatrix} 1 &  \frac32\\ 0 & \frac{-1}{2} \end{bmatrix}\)= \(\begin{bmatrix} \frac12 &  0\\ -\frac{5}{2} & 1 \end{bmatrix}A\)       \( (R_2→ R_2-5R_1)  \)

⇒ \(\begin{bmatrix} 1 &  0\\ 0 & \frac{-1}{2} \end{bmatrix}\)\(=\begin{bmatrix} -7 &  3\\ -\frac{5}{2} & 1 \end{bmatrix}A\)        \((R_1→ R_1+3R_2)\)

⇒ \(\begin{bmatrix} 1 &  0\\ 0 & 1 \end{bmatrix}\) = \(\begin{bmatrix} -7 &  3\\ 5 & -2 \end{bmatrix}A\)        \((R_2\rightarrow-2R_1)\) 

\(A^{-1}\)= \(\begin{bmatrix} -7 &  3\\ 5 & -2 \end{bmatrix}\)

Answer 2

\(\mathbf{A} = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}\)

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 \(\begin{bmatrix} -7 & 3 \\ 5 & -2 \end{bmatrix}\).