Exam-Style Problem

9231 P22 - Nov 2023 - Q6 - 10 marks

5964

The matrix \(\mathbf{P}\) is given by
\(\mathbf{P}=\left(\begin{array}{rrr} 1 & -1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & -1 \end{array}\right) .\)
(a) State the eigenvalues of \(\mathbf{P}\).

(b) Use the characteristic equation of \(\mathbf{P}\) to find \(\mathbf{P}^{-1}\).

The \(3 \times 3\) matrix \(\mathbf{A}\) has distinct non-zero eigenvalues \(a, \frac{1}{2}, 2\) with corresponding eigenvectors
\(\left(\begin{array}{l} 1 \\ 0 \\ \end{array}\right), \quad\left(\begin{array}{r} -1 \\ 2 \\ \end{array}\right), \quad\left(\begin{array}{r} 1 \\ 1 \\ -1 \end{array}\right),\)
respectively.
(c) Find \(\mathbf{A}^{-1}\) in terms of \(a\).

Solution Checked by expert

Answer: (a) The eigenvalues of \(\mathbf{P}\) are \(1, 2, -1\).

(b) \(\mathbf{P}^{-1}=\left(\begin{array}{ccc}1 & \frac{1}{2} & \frac{3}{2} \\ 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & -1\end{array}\right)\).

(c) \(\mathbf{A}^{-1}=\left(\begin{array}{ccc}\frac{1}{a} & \frac{1-2a}{2a} & \frac{3-3a}{2a} \\ 0 & 2 & \frac{3}{2} \\ 0 & 0 & \frac{1}{2}\end{array}\right)\).

(a) Since \(\mathbf{P}\) is upper triangular, its eigenvalues are the diagonal entries.

Hence the eigenvalues are \(1, 2, -1\).

(b) The characteristic polynomial is

\((\lambda-1)(\lambda-2)(\lambda+1)=\lambda^3-2\lambda^2-\lambda+2.\)

So \(\mathbf{P}\) satisfies

\(\mathbf{P}^3-2\mathbf{P}^2-\mathbf{P}+2\mathbf{I}=\mathbf{0}.\)

Since \(\mathbf{P}\) is invertible, multiply by \(\mathbf{P}^{-1}\):

\(\mathbf{P}^2-2\mathbf{P}-\mathbf{I}+2\mathbf{P}^{-1}=\mathbf{0},\)

\(2\mathbf{P}^{-1}=\mathbf{I}+2\mathbf{P}-\mathbf{P}^2.\)

Now

\(\mathbf{P}^2=\left(\begin{array}{ccc}1 & -3 & -1 \\ 0 & 4 & 1 \\ 0 & 0 & 1\end{array}\right).\)

Also,

\(\mathbf{I}+2\mathbf{P}=\left(\begin{array}{ccc}3 & -2 & 2 \\ 0 & 5 & 2 \\ 0 & 0 & -1\end{array}\right).\)

Therefore

\(2\mathbf{P}^{-1}=\left(\begin{array}{ccc}2 & 1 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & -2\end{array}\right),\)

hence

\(\mathbf{P}^{-1}=\left(\begin{array}{ccc}1 & \frac{1}{2} & \frac{3}{2} \\ 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & -1\end{array}\right).\)

\(\mathbf{P}=\left(\begin{array}{ccc}1 & -1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & -1\end{array}\right).\)

Then

\(\mathbf{A}=\mathbf{P}\left(\begin{array}{ccc}a & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & 2\end{array}\right)\mathbf{P}^{-1},\)

\(\mathbf{A}^{-1}=\mathbf{P}\left(\begin{array}{ccc}\frac{1}{a} & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & \frac{1}{2}\end{array}\right)\mathbf{P}^{-1}.\)

First,

\(\mathbf{P}\left(\begin{array}{ccc}\frac{1}{a} & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & \frac{1}{2}\end{array}\right)=\left(\begin{array}{ccc}\frac{1}{a} & -2 & \frac{1}{2} \\ 0 & 4 & \frac{1}{2} \\ 0 & 0 & -\frac{1}{2}\end{array}\right).\)

Then

\(\mathbf{A}^{-1}=\left(\begin{array}{ccc}\frac{1}{a} & -2 & \frac{1}{2} \\ 0 & 4 & \frac{1}{2} \\ 0 & 0 & -\frac{1}{2}\end{array}\right)\left(\begin{array}{ccc}1 & \frac{1}{2} & \frac{3}{2} \\ 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & -1\end{array}\right).\)

Multiplying gives

\(\mathbf{A}^{-1}=\left(\begin{array}{ccc}\frac{1}{a} & \frac{1-2a}{2a} & \frac{3-3a}{2a} \\ 0 & 2 & \frac{3}{2} \\ 0 & 0 & \frac{1}{2}\end{array}\right).\)