Inverse matrices Problem set A Answers

1. (a) \(\begin{pmatrix}3 & -1 \\ -5 & 2\end{pmatrix}\)
2. (b) \(\begin{pmatrix}1 & -1.5 \\ -1 & 2\end{pmatrix}\)
3. (c) No inverse: the determinant is 0.
4. (d) \(\begin{pmatrix}-2 & 1 \\ -1.5 & 0.5\end{pmatrix}\)
5. (e) \(\begin{pmatrix}2.5 & -1 \\ 0.5 & 0\end{pmatrix}\)

1. (a) \(B = A^{-1}C = \begin{pmatrix}3 & -3 \\ -2 & 5\end{pmatrix}\)
2. (b) \(B = CA^{-1} = \begin{pmatrix}17 & -6 \\ 27 & -9\end{pmatrix}\)

\(\begin{pmatrix}\frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\ -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & -\frac{1}{3}\end{pmatrix}\)

\(k = 3\) or \(k = -2\). Inverse: \(\dfrac{1}{k^2 - k - 6}\begin{pmatrix}k-1 & -2 \\ -3 & k\end{pmatrix}\).

\(\begin{pmatrix}3 & -1 \\ -5 & 2\end{pmatrix}\begin{pmatrix}28 & 26 \\ 76 & 73\end{pmatrix} = \begin{pmatrix}8 & 5 \\ 12 & 16\end{pmatrix}\), which is H, E, L, P.

1. (a) \(\begin{pmatrix}39 & 32 \\ 98 & 82\end{pmatrix}\)
2. (b) \(K^{-1} = \begin{pmatrix}3 & -1 \\ -5 & 2\end{pmatrix}\)
3. (c) HOME

The determinant is \(3 + 2c^2\), which is at least 3, so it is never zero.

\(Q = P\begin{pmatrix}3 & 0 \\ 0 & -1\end{pmatrix}P^{-1} = \begin{pmatrix}11 & -8 \\ 12 & -9\end{pmatrix}\)

\((AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I\). The inverses come in the opposite order.