Examples

Given that \(\mathbf{ABC=I}\), prove that \(\mathbf{B^{-1}}=CA\)
- \(\mathbf{ABC=I}\)
- \(\mathbf{ABCB^{-1}=IB^{-1}}\)
- \(\mathbf{ACI=IB^{-1}}\)
- \(\mathbf{AC=B^{-1}}\)

Given that \(\mathbf{A}=\begin{pmatrix}0 & 1 \\ -1 & -6\end{pmatrix}\) and \(\mathbf{C}=\begin{pmatrix}2 & 1 \\ -3 & -1\end{pmatrix}\), find \(\mathbf{B}\)
- \(\mathbf{CA}=\begin{pmatrix}2 & 1 \\ -3 & -1\end{pmatrix}\begin{pmatrix}0 & 1 \\ -1 & -6\end{pmatrix}=\begin{pmatrix}-1 & -4 \\1 & 3\end{pmatrix}\)
- invert this matrix to find \(\mathbf{B}\)

• \(x+y+z=2000\)
• \(x-y+0z=250\)
• \(0.99x+0.98y+1.04z=2040\)
• \(\mathbf{A}=\begin{pmatrix}1 & 1 & 1 \\ 1 & -1 & 0 \\ 0.99 & 0.98 & 1.04\end{pmatrix}\)
• \(\begin{pmatrix}1 & 1 & 1 \\ 1 & -1 & 0 \\ 0.99 & 0.98 & 1.04\end{pmatrix}\times \begin{pmatrix}x \\ y \\ z\end{pmatrix}=\begin{pmatrix}2000 \\ 250 \\ 2040\end{pmatrix}\)
• \(\mathbf{A^{-1}}=\begin{pmatrix}9.45 & 0.54 & -9.09 \\ 9.45 & -0.45 & -9.09 \\ -17.9 & -0.09 & 18.18\end{pmatrix}\)
• \(\mathbf{A^{-1}B}=\begin{pmatrix}500 \\ 250 \\ 1250\end{pmatrix}\)

A: \(x+ay+2z=a\)
B: \(x-y-z=a\)
C: \(x+4y+4z=0\)
- Given that \(\det \mathbf{A}=0\): \(a=2\)

A: \(x+2y+2z=2\)
B: \(x-y-z=2\)
C: \(x+4y+4z=0\)

1. \(-18--8=-10\) therefore the scale factor is \(-10\)
2. Inverse of Matrix = \(\begin{pmatrix} \frac{3}{5} & -\frac{1}{5} \\ \frac{2}{5} & -\frac{3}{10}\end{pmatrix}\)
3. To find the point mapped to (9, 2): multiply the inverted matrix by (9, 2).
   \(\begin{pmatrix} \frac{3}{5} & -\frac{1}{5} \\ \frac{2}{5} & -\frac{3}{10}\end{pmatrix}\times\begin{pmatrix}9 \\2\end{pmatrix}=\begin{pmatrix}5 \\ 3\end{pmatrix}\)

1. The determinant has to be 0, therefore \(12+3a=0\) which means \(a=-4\)
2. \(\begin{pmatrix}\end{pmatrix}\)