Homework 7

1. Find the determinant of the matrices using Gaussian elimination.

(a) $$\begin{bmatrix} 2 & 0 & 3 \\ 1 & 3 & 1 \\ 0 & 1 & 1\end{bmatrix}$$

$$\begin{align*} \det\begin{bmatrix} 2 & 0 & 3 \\ 1 & 3 & 1 \\ 0 & 1 & 1 \end{bmatrix} &\overset{R_2-R_3}{=} \begin{vmatrix} 2 & 0 & 3 \\ 1 & 2 & 0 \\ 0 & 1 & 1 \end{vmatrix} \\ &\overset{R_1\leftrightarrow R_2}{=} (-1)\begin{vmatrix} 1 & 2 & 0 \\ 2 & 0 & 3 \\ 0 & 1 & 1 \end{vmatrix} \\ &\overset{R_2 - 2R_1}{=} (-1)\begin{vmatrix} 1 & 2 & 0 \\ 0 & -4 & 3 \\ 0 & 1 & 1 \end{vmatrix} \\ &\overset{R_2 + 5R_1}{=} (-1)\begin{vmatrix} 1 & 2 & 0 \\ 0 & 1 & 8 \\ 0 & 1 & 1 \end{vmatrix} \\ &\overset{R_3 - R_2}{=} (-1)\begin{vmatrix} 1 & 2 & 0 \\ 0 & 1 & 8 \\ 0 & 0 & -7 \end{vmatrix} \\ &= (-1)(1\cdot1\cdot-7) \\ &= 7 \end{align*}$$

(b) $$\begin{bmatrix} 0 & 1 & 2 \\ −1 & 1 & 3 \\ 2 & −2 & 0\end{bmatrix}$$

$$\begin{align*} \det\begin{bmatrix} 0 & 1 & 2 \\ −1 & 1 & 3 \\ 2 & −2 & 0 \end{bmatrix} &\overset{R_1\leftrightarrow R_2}{=} (-1) \begin{vmatrix} −1 & 1 & 3 \\ 0 & 1 & 2 \\ 2 & −2 & 0 \end{vmatrix} \\ &\overset{R_3 + 2R_2}{=} (-1) \begin{vmatrix} −1 & 1 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 6 \end{vmatrix} \\ &= (-1) (-1\cdot1\cdot6) \\ &= 6 \end{align*}$$

(c) $$\begin{bmatrix}1 & −1 & 1 & 1\\1 & −1 & −1 & 0\\1 & 2 & 0 & −2\\2 & 0 & 2 & 1\end{bmatrix}$$

$$\begin{align*} \det\begin{bmatrix} 1 & −1 & 1 & 1 \\ 1 & −1 & −1 & 0 \\ 1 & 2 & 0 & −2 \\ 2 & 0 & 2 & 1 \end{bmatrix} &\overset{R_2 - R_1}{=} \begin{vmatrix} 1 & −1 & 1 & 1 \\ 0 & 0 & -2 & -1 \\ 1 & 2 & 0 & −2 \\ 2 & 0 & 2 & 1 \end{vmatrix} \\ &\overset{R_3 - R_1}{=} \begin{vmatrix} 1 & −1 & 1 & 1 \\ 0 & 0 & -2 & -1 \\ 0 & 3 & -1 & −3 \\ 2 & 0 & 2 & 1 \end{vmatrix} \\ &\overset{R_4 - 2R_1}{=} \begin{vmatrix} 1 & −1 & 1 & 1 \\ 0 & 0 & -2 & -1 \\ 0 & 3 & -1 & −3 \\ 0 & 2 & 0 & -1 \end{vmatrix} \\ &\overset{R_2\leftrightarrow R_3}{=} (-1) \begin{vmatrix} 1 & −1 & 1 & 1 \\ 0 & 3 & -1 & −3 \\ 0 & 0 & -2 & -1 \\ 0 & 2 & 0 & -1 \end{vmatrix} \\ &\overset{\frac{2}{3}R_2}{=} \left(-\frac{3}{2}\right) \begin{vmatrix} 1 & −1 & 1 & 1 \\ 0 & 2 & -\frac{2}{3} & -2 \\ 0 & 0 & -2 & -1 \\ 0 & 2 & 0 & -1 \end{vmatrix} \\ &\overset{R_4 - R_2}{=} \left(-\frac{3}{2}\right) \begin{vmatrix} 1 & −1 & 1 & 1 \\ 0 & 2 & -\frac{2}{3} & -2 \\ 0 & 0 & -2 & -1 \\ 0 & 0 & \frac{2}{3} & 1 \end{vmatrix} \\ &\overset{\frac{1}{3}R_3}{=} \left(-\frac{9}{2}\right) \begin{vmatrix} 1 & −1 & 1 & 1 \\[.5em] 0 & 2 & -\frac{2}{3} & -2 \\[.5em] 0 & 0 & -\frac{2}{3} & -\frac{1}{3} \\[.5em] 0 & 0 & \frac{2}{3} & 1 \end{vmatrix} \\ &\overset{R_4 + R_3}{=} \left(-\frac{9}{2}\right) \begin{vmatrix} 1 & −1 & 1 & 1 \\[.5em] 0 & 2 & -\frac{2}{3} & -2 \\[.5em] 0 & 0 & -\frac{2}{3} & -\frac{1}{3} \\[.5em] 0 & 0 & 0 & \frac{2}{3} \end{vmatrix} \\ &= \left(-\frac{9}{2}\right)\left(1\cdot2\cdot-\frac{2}{3}\cdot\frac{2}{3}\right) \\ &= 4 \end{align*}$$

(d)$$\begin{bmatrix}2 & 1 & 1 & 1 & 1 \\1 & 2 & 1 & 1 & 1 \\1 & 1 & 2 & 1 & 1 \\1 & 1 & 1 & 2 & 1 \\1 & 1 & 1 & 1 & 2\end{bmatrix}$$

$$\begin{align*} \det\begin{bmatrix} 2 & 1 & 1 & 1 & 1 \\ 1 & 2 & 1 & 1 & 1 \\ 1 & 1 & 2 & 1 & 1 \\ 1 & 1 & 1 & 2 & 1 \\ 1 & 1 & 1 & 1 & 2 \end{bmatrix} &\overset{R_1\leftrightarrow R_2}{=} (-1)\begin{vmatrix} 1 & 2 & 1 & 1 & 1 \\ 2 & 1 & 1 & 1 & 1 \\ 1 & 1 & 2 & 1 & 1 \\ 1 & 1 & 1 & 2 & 1 \\ 1 & 1 & 1 & 1 & 2 \end{vmatrix} \\ &\overset{R_2 - 2R_1}{=} (-1)\begin{vmatrix} 1 & 2 & 1 & 1 & 1 \\ 0 & -3 & -1 & -1 & -1 \\ 1 & 1 & 2 & 1 & 1 \\ 1 & 1 & 1 & 2 & 1 \\ 1 & 1 & 1 & 1 & 2 \end{vmatrix} \\ &\overset{\substack{R_3 - R_1 \\ R_4 - R_1 \\ R_5 - R_1}}{=} (-1)\begin{vmatrix} 1 & 2 & 1 & 1 & 1 \\ 0 & -3 & -1 & -1 & -1 \\ 0 & -1 & 1 & 0 & 0 \\ 0 & -1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 & 1 \end{vmatrix} \\ &\overset{R_2\leftrightarrow R_3}{=} \begin{vmatrix} 1 & 2 & 1 & 1 & 1 \\ 0 & -1 & 1 & 0 & 0 \\ 0 & -3 & -1 & -1 & -1 \\ 0 & -1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 & 1 \end{vmatrix} \\ &\overset{R_3 - 3R_2}{=} \begin{vmatrix} 1 & 2 & 1 & 1 & 1 \\ 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & -4 & -1 & -1 \\ 0 & -1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 & 1 \end{vmatrix} \\ &\overset{\substack{R_4 - R_2 \\ R_5 - R_2}}{=} \begin{vmatrix} 1 & 2 & 1 & 1 & 1 \\ 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & -4 & -1 & -1 \\ 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 0 & 1 \end{vmatrix} \\ &\overset{R_3 \leftrightarrow R_4}{=} (-1)\begin{vmatrix} 1 & 2 & 1 & 1 & 1 \\ 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & -4 & -1 & -1 \\ 0 & 0 & -1 & 0 & 1 \end{vmatrix} \\ &\overset{R_4 - 4R_3}{=} (-1)\begin{vmatrix} 1 & 2 & 1 & 1 & 1 \\ 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & -5 & -1 \\ 0 & 0 & -1 & 0 & 1 \end{vmatrix} \\ &\overset{R_5 - R_3}{=} (-1)\begin{vmatrix} 1 & 2 & 1 & 1 & 1 \\ 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & -5 & -1 \\ 0 & 0 & 0 & -1 & 1 \end{vmatrix} \\ &\overset{R_4 \leftrightarrow R_5}{=} \begin{vmatrix} 1 & 2 & 1 & 1 & 1 \\ 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & -5 & -1 \\ \end{vmatrix} \\ &\overset{R_5 - 5R_4}{=} \begin{vmatrix} 1 & 2 & 1 & 1 & 1 \\ 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & 0 & -6 \\ \end{vmatrix} \\ &= 1\cdot-1\cdot-1\cdot-1\cdot-6 \\ &= 6 \end{align*}$$

2. Given $5\times5$ matrices $A$, $B$, $Q$. Suppose that $\det A = 3$, $\det B = 2$ and $Q$ is an invertible matrix. Find the determinant of $A^TB$, $A^3$, $2A$, $ABA$ and $Q^{−1}AQ$.

$\det A^T B$

$$\begin{align*} \det A^T B &= \det A^T \det B \\ &= \det A \det B \\ &= 3\cdot 2 \\ &= 6 \end{align*}$$

$\det A^3$

$$\begin{align*} \det A^3 &= \det AAA \\ &= \det A \det A \det A \\ &= 3\cdot3\cdot3 \\ &= 27 \end{align*}$$

$\det 2A$

$$\begin{align*} \det 2A &= \det 2\begin{pmatrix} a_{11} & & \\ &\ddots& \\ & & a_{55} \end{pmatrix}, & a_{ij} \in \R \\ &= \det \begin{pmatrix} 2a_{11} & & \\ &\ddots& \\ & & 2a_{55} \end{pmatrix}, & a_{ij} \in \R \\ &= 2^5 \det \begin{pmatrix} a_{11} & & \\ &\ddots& \\ & & a_{55} \end{pmatrix}, & a_{ij} \in \R \\ &= 2^5 \det A \\ &= 32 \cdot 3 \\ &= 96 \end{align*}$$

$\det ABA$

$$\begin{align*} \det ABA &= \det A \det B \det A \\ &= 3 \cdot 2 \cdot 3 \\ &= 18 \end{align*}$$

$\det Q^{-1}AQ$

$$\begin{align*} \det Q^{-1}AQ &= \det Q^{-1} \det A \det Q \\ &= \frac{1}{\det Q} \det A \det Q \\ &= \det A \\ &= 3 \end{align*}$$

3. Consider the following system of linear equations: $$\begin{cases} px + y + z = 6, \\ 3x − y + 11z = 6, \\ 2x + y + 4z = q,\end{cases}$$

(a) Find the condition on $p$ so that the system has unique solution (Hint: $\det(A) \neq 0$).

$$\begin{cases} px + y + z \\ 3x − y + 11z \\ 2x + y + 4z \end{cases} \iff \begin{pmatrix} p & 1 & 1 \\ 3 & -1 & 11 \\ 2 & 1 & 4 \end{pmatrix} \\ \begin{align*} \det \begin{pmatrix} p & 1 & 1 \\ 3 & -1 & 11 \\ 2 & 1 & 4 \end{pmatrix} &= p \begin{vmatrix} -1 & 11 \\ 1 & 4 \end{vmatrix} - \begin{vmatrix} 3 & 11 \\ 2 & 4 \end{vmatrix} + \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} \\ &= p(-4-11) - (12-22) + (3-(-2)) \\ &= -15p + 15 \end{align*} \\ \therefore \det\begin{pmatrix} p & 1 & 1 \\ 3 & -1 & 11 \\ 2 & 1 & 4 \end{pmatrix} \neq 0 \iff p \neq 1$$

As such, the system has unique solution for all $p\neq1$.

(b) Find the condition on $p$ and $q$ so that the system has infinitely many solutions (Hint: $\det(A) = 0$ and no inconsistent equations). Describe the solution set.

$$\begin{cases} px + y + z = 6 \\ 3x − y + 11z = 6 \\ 2x + y + 4z = q \end{cases} \iff \left( \begin{array}{ccc|c} p & 1 & 1 & 6 \\ 3 & -1 & 11 & 6 \\ 2 & 1 & 4 & q \end{array} \right)$$

From (a), we know that $\det\begin{pmatrix} p & 1 & 1 \\ 3 & -1 & 11 \\ 2 & 1 & 4 \end{pmatrix} = 0 \iff p = 1$. So, we can just Gaussian with $p=1$.

$$\begin{array}{c} \left( \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 3 & -1 & 11 & 6\\ 2 & 1 & 4 & q \end{array} \right) &\xrightarrow[R_3 - 2R_1]{R_2 - 3R_1} &\left( \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & -4 & 8 & -12 \\ 0 & -1 & 2 & q-12 \end{array} \right) \\ &\xrightarrow{-\frac{1}{4}R_2} &\left( \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & -2 & 3 \\ 0 & -1 & 2 & q-12 \end{array} \right) \\ &\xrightarrow{R_3 + R_2} &\left( \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & -2 & 3 \\ 0 & 0 & 0 & q-9 \end{array} \right) \end{array}$$

Here, we see that the last row is $0 = q-9$. As such, the system will have a consistent solution if and only if $q=9$.

To get the solution set, we continue to Gaussian to obtain the reduced-row echelon form.

$$\begin{array}{c} \left( \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & -2 & 3 \\ 0 & 0 & 0 & q-9 \end{array} \right) &\xrightarrow{R_1 - R_2} &\left( \begin{array}{ccc|c} 1 & 0 & 3 & 3 \\ 0 & 1 & -2 & 3 \\ 0 & 0 & 0 & q-9 \end{array} \right) \end{array} \\ \therefore y = 2z + 3 \\ x = -3z + 3$$

Therefore, the system

$$\begin{cases} px + y + z = 6 \\ 3x − y + 11z = 6 \\ 2x + y + 4z = q \end{cases}$$

contains infinitely many solutions if and only if $p=1$ and $q=9$, for which its solution set is

$$\Set{ (x,y,z): \begin{pmatrix} -3z + 3 \\ 2z + 3 \\ z \end{pmatrix}: z\in\R }.$$

4. Show that if $A$ is an $n \times n$ skew-symmetric matrix (i.e. $A^T = −A$) and $n$ is an odd number, then $\det A = 0$.

Given that $\det A^T = \det A$. Then if $A^T = -A$ ($A$ is skew-symmetric), we have that:

$$\det A^T = \det A = \det (-A)$$

If $n$ is odd, then $n=k+1$ for $k\in\N$. As such:

$$\begin{align*} \det(-A) &= \det -1\cdot\begin{pmatrix} a_{11} & & \\ & \ddots & \\ & & a_{nn} \end{pmatrix}, &a_{ij}\in\R \\ &= \det\begin{pmatrix} (-1)a_{11} & & \\ & \ddots & \\ & & (-1)a_{nn} \end{pmatrix}, &a_{ij}\in\R \\ &= (-1)^n\det\begin{pmatrix} a_{11} & & \\ & \ddots & \\ & & a_{nn} \end{pmatrix}, &a_{ij}\in\R \\ &= (-1)^{k+1}\det A & \boxed{n=k+1}\\ &= -\det A & \boxed{(-1)^{k+1} = -1\quad\forall k\in\N} \end{align*}$$

Since $\det A^T = \det A = \det(-A) = -\det A$. Then, $\det A = -\det A \iff \det A = 0$.

5. Let $$A = \begin{bmatrix} 1 & 3 \\ 4 & 2\end{bmatrix},\quad B =\begin{bmatrix} 0 & −2 & −3 \\ −1 & 1 & −1 \\ 2 & 2 & 5\end{bmatrix}.$$

(i) Find the eigenvalues and eigenvectors of both $A$ and $B$.

Eigenvalues and eigenvectors for $A$

$$\begin{align*} \det (A-\lambda I) &= \begin{vmatrix} 1-\lambda & 3 \\ 4 & 2-\lambda \end{vmatrix} &= 0 \\ &= (1-\lambda)(2-\lambda) - 12 &= 0 \\ &= \lambda^2 - 3\lambda - 10 &= 0 \\ &= (\lambda + 2)(\lambda - 5) &= 0 \end{align*} \\ \therefore\lambda = -2, 5$$

The eigenvalues of $A$ are $-2$ and $5$.

$$\lambda = -2 \implies A+2I \\ A + 2I = \begin{bmatrix} 3 & 3 \\ 4 & 4 \end{bmatrix} \implies\operatorname{rref}(A+2I) = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \\ \therefore x_1 = -x_2 \\ x_2\in\R \\ \therefore\ker(A+2I) = \Set{ \begin{pmatrix} -x_2 \\ x_2 \end{pmatrix}: x_2\in\R } = \Set{ x\begin{pmatrix} -1 \\ 1 \end{pmatrix}: x\in\R }$$

An eigenvector corresponding to $\lambda=-2$ is $\begin{pmatrix} -1 \\ 1 \end{pmatrix}$.

$$\lambda = 5 \implies A - 5I \\ A - 5I = \begin{bmatrix} -4 & 3 \\ 4 & -3 \end{bmatrix} \implies \operatorname{rref}(A-5I) = \begin{bmatrix} 1 & -\frac{3}{4} \\ 0 & 0 \end{bmatrix} \\ \therefore x_1 = \frac{3}{4}x_2 \\ x_2 \in\R \\ \therefore\ker(A-5I) = \Set{ \begin{pmatrix} \frac{3}{4}x_2 \\ x_2 \end{pmatrix}: x_2 \in\R } = \Set{ x\begin{pmatrix} \frac{3}{4} \\[.2em] 1 \end{pmatrix}: x\in\R }$$

An eigenvector corresponding to $\lambda=5$ is $\begin{pmatrix} \frac{3}{4} \\[.2em] 1 \end{pmatrix}$.

Eigenvalues and eigenvectors for $B$

$$\begin{align*} \det(B-\lambda I) &= \begin{vmatrix} -\lambda & −2 & −3 \\ −1 & 1-\lambda & −1 \\ 2 & 2 & 5-\lambda \end{vmatrix} &= 0 \\ &= -\lambda \begin{vmatrix} 1-\lambda & −1 \\ 2 & 5-\lambda \end{vmatrix} - (-2) \begin{vmatrix} −1 & −1 \\ 2 & 5-\lambda \end{vmatrix} + (-3) \begin{vmatrix} −1 & 1-\lambda \\ 2 & 2 \end{vmatrix} &= 0 \\ &= -\lambda(\lambda^2 - 6\lambda + 7) -(-2)(\lambda-3) + (-3)(2\lambda-4) &= 0 \\ &= -\lambda^3 + 6\lambda^2 - 11\lambda + 6 &= 0 \\ &= -(\lambda-3)(\lambda-2)(\lambda-1) &= 0 \end{align*} \\ \therefore \lambda = 1,2,3$$

The eigenvalues of $B$ are $1$, $2$, and $3$.

$$\lambda=1 \implies B-I \\ B-I = \begin{bmatrix} -1 & −2 & −3 \\ −1 & 0 & −1 \\ 2 & 2 & 4 \end{bmatrix} \implies\operatorname{rref}(B-I) = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \\ \therefore x_2 = -x_3 \\ x_1 = -x_3 \\ x_3 \in\R \\ \therefore\ker(B-I) = \Set{ \begin{pmatrix} -x_3 \\ -x_3 \\ x_3 \end{pmatrix}: x_3 \in\R } = \Set{ x\begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix}: x\in\R }$$

An eigenvector corresponding to $\lambda=1$ is $\begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix}$.

$$\lambda=2 \implies B-2I \\ B-2I = \begin{bmatrix} -2 & −2 & −3 \\ −1 & -1 & −1 \\ 2 & 2 & 3 \end{bmatrix} \implies\operatorname{rref}(B-2I) = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \\ \therefore x_3 = 0 \\ x_1 = -x_2 \\ x_2 \in\R \\ \therefore\ker(B-2I) = \Set{ \begin{pmatrix} -x_2 \\ x_2 \\ 0 \end{pmatrix}: x_2 \in\R } = \Set{ x\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}: x\in\R }$$

An eigenvector corresponding to $\lambda=2$ is $\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$.
$$\lambda=3 \implies B-3I \\ B-3I = \begin{bmatrix} -3 & −2 & −3 \\ −1 & -2 & −1 \\ 2 & 2 & 2 \end{bmatrix} \implies\operatorname{rref}(B-3I) = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ \therefore x_2 = 0 \\ x_1 = -x_3 \\ x_3\in\R \\ \therefore\ker(B-3I) = \Set{ \begin{pmatrix} -x_3 \\ 0 \\ x_3 \end{pmatrix}: x_2 \in\R } = \Set{ x\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}: x\in\R }$$

An eigenvector corresponding to $\lambda=3$ is $\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$.

(ii) Diagonalize $A$ and $B$.

Diagonalizing $A$

For eigenvalues $\lambda_{A1}=-2$ and $\lambda_{A2}=5$, the diagonalization of $A= \begin{bmatrix} 1 & 3 \\ 4 & 2 \end{bmatrix}$ is

$$P^{-1}AP = \begin{pmatrix} -2 & 0 \\ 0 & 5 \end{pmatrix}$$

where $P$ is a matrix composed of the corresponding eigenvectors of $A$ such that

$$P = \begin{pmatrix} -1 & \frac{3}{4} \\[.2em] 1 & 1 \end{pmatrix}.$$

Diagonalizing $B$

For eigenvalues $\lambda_{B1}=1$, $\lambda_{B2}=2$, and $\lambda_{B3}=3$, the diagonalization of $B=\begin{bmatrix} 0 & −2 & −3 \\ −1 & 1 & −1 \\ 2 & 2 & 5 \end{bmatrix}$ is

$$Q^{-1}BQ = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$$

where $Q$ is a matrix composed of the corresponding eigenvectors of $B$ such that

$$Q = \begin{pmatrix} -1 & -1 & -1 \\\ -1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}.$$

(iii) Find $A^{10}$ and $B^3$.

Finding $A^{10}$

$$P = \begin{pmatrix} -1 & \frac{3}{4} \\[.2em] 1 & 1 \end{pmatrix} \implies P^{-1} = \frac{1}{7} \begin{pmatrix} -4 & 3 \\ 4 & 4 \end{pmatrix} \\ P^{-1}A^{10}P = \begin{pmatrix} -2 & 0 \\ 0 & 5 \end{pmatrix}^{10} \\ \begin{align*} \therefore A^{10} &= P\begin{pmatrix} -2 & 0 \\ 0 & 5 \end{pmatrix}^{10} P^{-1} \\ &= \begin{pmatrix} -1 & \frac{3}{4} \\ 1 & 1 \end{pmatrix} \begin{pmatrix} -2 & 0 \\ 0 & 5 \end{pmatrix}^{10} \frac{1}{7} \begin{pmatrix} -4 & 3 \\ 4 & 4 \end{pmatrix} \\ &= \begin{pmatrix} 4\ 185\ 853 & 4\ 184\ 829 \\ 5\ 579\ 772 & 5\ 580\ 796 \end{pmatrix} \end{align*}$$

Finding $B^3$

$$Q = \begin{pmatrix} -1 & -1 & -1 \\ -1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} \implies Q^{-1} = \begin{pmatrix} -1 & -1 & -1 \\ -1 & 0 & -1 \\ 1 & 1 & 2 \end{pmatrix} \\ Q^{-1}B^3 Q = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}^3 \\ \begin{align*} \therefore B^3 &= Q\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}^3Q^{-1} \\ &= \begin{pmatrix} -1 & -1 & -1 \\ -1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}^3 \begin{pmatrix} -1 & -1 & -1 \\ -1 & 0 & -1 \\ 1 & 1 & 2 \end{pmatrix} \\ &= \begin{pmatrix} -18 & -26 & -45 \\ -7 & 1 & -7 \\ 26 & 26 & 53 \end{pmatrix} \end{align*}$$