If \(B = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\), find \(B^{-1}\).
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Correct Answer: Option C
Explanation:
\(B.B^{-1} = 1\), 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
\(a = 3; b = -5; c = -1; d = 2 \implies B^{-1} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\)
\(B.B^{-1} = 1\), 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
\(a = 3; b = -5; c = -1; d = 2 \implies B^{-1} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\)