If \(T = \begin{pmatrix} -2 & -5 \\ 3 & 8 \end{pmatrix}\), find \(T^{-1}\), the inverse of T.
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\(\begin{pmatrix} -8 & -5 \\ 3 & 2 \end{pmatrix}\)
- \(\begin{pmatrix} -8 & -5 \\ 3 & -2 \end{pmatrix}\)
- \(\begin{pmatrix} -8 & -5 \\ -3 & 2 \end{pmatrix}\)
- \(\begin{pmatrix} -8 & -5 \\ -3 & -2 \end{pmatrix}\)
The correct answer is: A
Explanation
Let \(\begin{pmatrix} a & b \\ c & d \end{pmatrix} = T^{-1}\)
\(T . T^{-1} = I\)
\(\begin{pmatrix} -2 & -5 \\ 3 & 8 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
\(\implies -2a - 5c = 1\)
\(-2b - 5d = 0 \implies b = \frac{-5d}{2}\)
\(3a + 8c = 0 \implies a = \frac{-8c}{3}\)
\(3b + 8d = 1\)
\(-2(\frac{-8c}{3}) - 5c = \frac{16c}{3} - 5c = \frac{c}{3} = 1 \implies c = 3\)
\(3(\frac{-5d}{2}) + 8d = \frac{-15d}{2} + 8d = \frac{d}{2} = 1 \implies d = 2\)
\(b = \frac{-5 \times 2}{2} = -5\)
\(a = \frac{-8 \times 3}{3} = -8\)
\(\therefore T^{-1} = \begin{pmatrix} -8 & -5 \\ 3 & 2 \end{pmatrix}\)