If \(B = \begin{pmatrix}  2 & 5  \\  1 & 3  \end{pmatrix}\), find \(B^{-1}\).

If \(B = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\), find \(B^{-1}\).

\(B.B^{-1} = 1\), let \(B^{-1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\)

\(\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\)\(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) = \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)

Multiplying \(B \times B^{-1}\), we have the following equations:

\(2a+5c = 1......... (1)\); \(a+3c = 0 ........(2)\)

\(2b+5d = 0 ..........(3)\); \(b+3d = 1..........(4)\)

Solving the equations simultaneously, we have

\(a = 3; b = -5; c = -1; d = 2 \implies B^{-1} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\)

{{ settings.no_comment_msg ? settings.no_comment_msg : 'There are no comments' }}