Question

Find the inverse of each of the matrices,if it exists \(\begin{bmatrix} 2 & 1 \\ 7 & 4  \end{bmatrix}\)

Answer 1

Let A=\(\begin{bmatrix} 2 & 1 \\ 7 & 4  \end{bmatrix}\)

We know that\(A = IA\)  

\(\therefore\) \(\begin{bmatrix} 2 & 1 \\ 7 & 4  \end{bmatrix}\)= \(\begin{bmatrix} 1 & 0 \\ 0 & 1  \end{bmatrix}\) A 

⇒ \(\begin{bmatrix} 1 & \frac12 \\ 7 & 4  \end{bmatrix}\)=\(\begin{bmatrix} \frac12 & 0 \\ 0 & 1  \end{bmatrix}\)A            (R₁\(→\) \(\frac{1}{2}R_1\)) 

⇒\(\begin{bmatrix} 1 & \frac12 \\ 0 & \frac12  \end{bmatrix}\)=\(\begin{bmatrix} \frac12 &0\\  -\frac72  & 1  \end{bmatrix}\)A           (R₂→R₂-7R₁) 

⇒ \(\begin{bmatrix} 1 & 0 \\ 0 & \frac12  \end{bmatrix}\)=\(\begin{bmatrix} 4 & -1 \\ -\frac72 & 1  \end{bmatrix}\)A        (R₁->R₁-R₂) 

⇒ \(\begin{bmatrix} 1 & 0 \\ 0 & 2  \end{bmatrix}\) = \(\begin{bmatrix} 4 & -1 \\ -7 & 2  \end{bmatrix}\)A  (R₂->2R₂) 

\(\therefore\) A^-1=\(\begin{bmatrix} 4 & -1 \\ -7 & 2  \end{bmatrix}\)