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

  • A \(A = \begin{pmatrix} -3 & -5 \\ 1 & 2 \end{pmatrix}\)
  • B \(A = \begin{pmatrix} 3 & -5 \\ 1 & 2 \end{pmatrix}\)
  • C \(A = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\)
  • D \(A = \begin{pmatrix} -3 & 5 \\ 1 & -2 \end{pmatrix}\)
waec 2017furthermathematics

The correct answer is C. \(A = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\)

\(B.B^{-1} = 1\), 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

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

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