If \(T = \begin{pmatrix} -2 & -5 \\ 3 & 8 \end{pmatrix}\), find \(T^{-1}\), the

FURTHER MATHEMATICS
WAEC 2014

If \(T = \begin{pmatrix} -2 & -5 \\ 3 & 8 \end{pmatrix}\), find \(T^{-1}\), the inverse of T.

  • A. \(\begin{pmatrix} -8 & -5 \\ 3 & 2 \end{pmatrix}\)
  • B. \(\begin{pmatrix} -8 & -5 \\ 3 & -2 \end{pmatrix}\)
  • C. \(\begin{pmatrix} -8 & -5 \\ -3 & 2 \end{pmatrix}\)
  • D. \(\begin{pmatrix} -8 & -5 \\ -3 & -2 \end{pmatrix}\)

Correct Answer: A. \(\begin{pmatrix} -8 & -5 \\ 3 & 2 \end{pmatrix}\)

Explanation

Let \(\begin{pmatrix} a & b \\ c & d \end{pmatrix} = T^{-1}\)

\(T . T^{-1} = I\)

\(\begin{pmatrix} -2 & -5 \\ 3 & 8 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)

\(\implies -2a - 5c = 1\)

\(-2b - 5d = 0 \implies b = \frac{-5d}{2}\)

\(3a + 8c = 0 \implies a = \frac{-8c}{3}\)

\(3b + 8d = 1\)

\(-2(\frac{-8c}{3}) - 5c = \frac{16c}{3} - 5c = \frac{c}{3} = 1 \implies c = 3\)

\(3(\frac{-5d}{2}) + 8d = \frac{-15d}{2} + 8d = \frac{d}{2} = 1 \implies d = 2\)

\(b = \frac{-5 \times 2}{2} = -5\)

\(a = \frac{-8 \times 3}{3} = -8\)

\(\therefore T^{-1} = \begin{pmatrix} -8 & -5 \\ 3 & 2 \end{pmatrix}\)



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