Tutorial 6: Matrix Algebra
Q1: Let \(A=\begin{bmatrix}5 & -2 & 0\\-2 & 6 & 2\\0 & 2 & 7\end{bmatrix}\), \(X=\begin{bmatrix}-x\\0\\x\end{bmatrix}\) and \(B=\begin{bmatrix}2 & 2 & -1\\2 & -1 & 2\\-1 & 2 & 2\end{bmatrix}\)
(a) Find the value of \(x\) such that \(X^{T} AX=144\)
(b) Show that \(B^{T} AB=27\begin{bmatrix}1 & 0 & 0\\0 & 2 & 0\\0 & 0 & 3\end{bmatrix}\)
Solution
Question (a)
\[\begin{aligned} X^{T} AX & =\begin{bmatrix} -x & 0 & x \end{bmatrix}\begin{bmatrix} 5 & -2 & 0\\ -2 & 6 & 2\\ 0 & 2 & 7 \end{bmatrix}\begin{bmatrix} -x\\ 0\\ x \end{bmatrix}\\ & =\begin{bmatrix} -5x+0+0 & 2x+0+2x & 0+0+7x \end{bmatrix}\begin{bmatrix} -x\\ 0\\ x \end{bmatrix}\\ & =12x^{2} \end{aligned}\] \[\therefore X^{T} AX=144\Longrightarrow 12x^{2} =144\Longrightarrow x=\pm 2\sqrt{3}\]Question (b)
\[\begin{aligned} B^{T} AB & =\begin{bmatrix} 2 & 2 & -1\\ 2 & -1 & 2\\ -1 & 2 & 2 \end{bmatrix}\begin{bmatrix} 5 & -2 & 0\\ -2 & 6 & 2\\ 0 & 2 & 7 \end{bmatrix}\begin{bmatrix} 2 & 2 & -1\\ 2 & -1 & 2\\ -1 & 2 & 2 \end{bmatrix}\\ & =\begin{bmatrix} 6 & 6 & -3\\ 12 & -6 & 12\\ -9 & 18 & 18 \end{bmatrix}\begin{bmatrix} 2 & 2 & -1\\ 2 & -1 & 2\\ -1 & 2 & 2 \end{bmatrix}\\ & =3\begin{bmatrix} 2 & 2 & -1\\ 4 & -2 & 4\\ -3 & 6 & 6 \end{bmatrix}\begin{bmatrix} 2 & 2 & -1\\ 2 & -1 & 2\\ -1 & 2 & 2 \end{bmatrix}\\ & =3\begin{bmatrix} 9 & 0 & 0\\ 0 & 18 & 0\\ 0 & 0 & 27 \end{bmatrix}\\ & =27\begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3 \end{bmatrix} \end{aligned}\]Q2: Let \(A=\begin{bmatrix} 6 & -1 & 1\\ 0 & 13 & -16\\ 0 & 8 & -11 \end{bmatrix}\) and \(x=\begin{bmatrix} 10.5\\ 21.0\\ 10.5 \end{bmatrix}\)
(a) Determine a scalar \(r\) such that \(Ax=rx\)
(b) Is it true \(A^{T} x=rx\) for the value of \(r\) determined in part (a)
Solution
\[\begin{aligned} \begin{bmatrix} 6 & -1 & 1\\ 0 & 13 & -16\\ 0 & 8 & -11 \end{bmatrix}\begin{bmatrix} 10.5\\ 21.0\\ 10.5 \end{bmatrix} & =r\begin{bmatrix} 10.5\\ 21.0\\ 10.5 \end{bmatrix}\\ \begin{bmatrix} 525\\ 1050\\ 525 \end{bmatrix} & =r\begin{bmatrix} 10.5\\ 21.0\\ 10.5 \end{bmatrix}\\ \mathbf{r} & \mathbf{=} 5 \end{aligned}\](a) \(r=5\)
(b) No.
Q3: Solve each of the following systems of linear equations using Gaussian Elimination technique.
(a) \(\begin{aligned} x+2y+3z & =9\\2x-y+z & =8\\3x-z & =3\end{aligned}\)
(b) \(\begin{aligned} -3x+2y-6z & =6\\ 5x+7y-5z & =6\\ x+4y-2z & =8 \end{aligned}\)
(c) \(\begin{aligned} 2x+y+3z & =1\\ 2x+6y+8z & =3\\ 6x+8y+18z & =5 \end{aligned}\)
(d) \(\begin{aligned} 2x-3y+z & =5\\ 3x+2y-z & =7\\ x+4y-5z & =3 \end{aligned}\)
(e) \(\begin{aligned} x+y+z & =6\\ 2x-y+z & =3\\ 3x-z & =0 \end{aligned}\)
Solution
Question (a)
\[\begin{bmatrix} 1 & 2 & 3 & 9\\ 2 & -1 & 1 & 8\\ 1 & 0 & -1 & 3 \end{bmatrix}\xrightarrow{ \begin{array}{l} R_{2}\rightarrow R_{2} -2R_{1}\\ R_{3}\rightarrow R_{3} -3R_{1} \end{array}}\begin{bmatrix} 1 & 2 & 3 & 9\\ 0 & -5 & -5 & -10\\ 0 & -6 & -10 & -24 \end{bmatrix}\xrightarrow{R_{2}\rightarrow -1/5R_{2}}\begin{bmatrix} 1 & 2 & 3 & 9\\ 0 & 1 & 1 & 2\\ 0 & -6 & -10 & -24 \end{bmatrix}\\ \\ \xrightarrow{R_{3}\rightarrow R_{3} +6R_{2}}\begin{bmatrix} 1 & 2 & 3 & 9\\ 0 & 1 & 1 & 2\\ 0 & 0 & -4 & -12 \end{bmatrix}\\ \\ \begin{aligned} -4z & =-12 & \quad y+z & =2 & \quad x+2y+3z & =9\\ \mathbf{z} & \mathbf{=3} & y+3 & =2 & x+2( -1) +3( 3) & =9\\ & & \mathbf{y} & \mathbf{=-1} & \mathbf{x} & \mathbf{=2} \end{aligned}\]Question (b)
\[\begin{bmatrix} -3 & 2 & -6 & 6\\ 5 & 7 & -5 & 6\\ 1 & 4 & -2 & 8 \end{bmatrix}\xrightarrow{R_{1}\rightarrow 3R_{3} +1}\begin{bmatrix} 0 & 14 & -12 & 30\\ 5 & 7 & -5 & 6\\ 1 & 4 & -2 & 8 \end{bmatrix}\xrightarrow{ \begin{array}{l} R_{1}\rightarrow 1/2R_{1}\\ R_{2}\rightarrow -5R_{3} +R_{2} \end{array}}\begin{bmatrix} 0 & 7 & -6 & 15\\ 0 & -13 & -5 & -34\\ 1 & 4 & -2 & 8 \end{bmatrix}\\ \\ \xrightarrow{ \begin{array}{l} R_{1}\rightarrow 7R_{2} +R_{1}\\ R_{2}\rightarrow 2R_{1} +R_{2} \end{array}}\begin{bmatrix} 0 & 0 & 43 & 43\\ 0 & 1 & -7 & -4\\ 1 & 4 & -2 & 8 \end{bmatrix}\\ \\ \begin{aligned} 43z & =43 & \quad y-7z & =-4 & \quad x+4y-2z & =8\\ \mathbf{z} & \mathbf{= 1} & y-7 & =-4 & x+4( 3) -2( 1) & =8\\ & & \mathbf{y} & \mathbf{= 3} & \mathbf{x} & \mathbf{=-2} \end{aligned}\]Question (c)
\[\begin{bmatrix} 2 & 1 & 3 & 1\\ 2 & 6 & 8 & 3\\ 6 & 8 & 18 & 5 \end{bmatrix}\xrightarrow{ \begin{array}{l} R_{2}\rightarrow -R_{1} +R_{2}\\ R_{3}\rightarrow -3R_{1} +R_{3} \end{array}}\begin{bmatrix} 2 & 1 & 3 & 1\\ 0 & 5 & 5 & 2\\ 0 & 5 & 9 & 2 \end{bmatrix}\xrightarrow{R_{3}\rightarrow -R_{2} +R_{3}}\begin{bmatrix} 2 & 1 & 3 & 1\\ 0 & 5 & 5 & 2\\ 0 & 0 & 4 & 0 \end{bmatrix}\\ \\ \begin{aligned} 4z & =0 & \quad 5y+5z & =2 & \quad 2x+y+3z & =1\\ \mathbf{z} & \mathbf{= 0} & 5y & =2 & 2x+( 2/5) +3( 0) & =1\\ & & \mathbf{y} & \mathbf{= 2/5} & \mathbf{x} & \mathbf{= 3/5} \end{aligned}\]Question (d)
\[\begin{bmatrix} 2 & -3 & 1 & -5\\ 3 & 2 & -1 & 7\\ 1 & 4 & -5 & 3 \end{bmatrix}\xrightarrow{ \begin{array}{l} R_{2}\rightarrow R_{2} -3R_{3}\\ R_{3}\rightarrow 2R_{3} -R_{1} \end{array}}\begin{bmatrix} 2 & -3 & 1 & -5\\ 0 & -10 & 14 & -2\\ 0 & 11 & -11 & 11 \end{bmatrix}\xrightarrow{R_{3}\rightarrow \frac{1}{11} R_{3}}\begin{bmatrix} 2 & -3 & 1 & -5\\ 0 & -10 & 14 & -2\\ 0 & 1 & -1 & 1 \end{bmatrix}\\ \\ \xrightarrow{R_{3}\rightarrow 10R_{3} +R_{2}}\begin{bmatrix} 2 & -3 & 1 & -5\\ 0 & -10 & 14 & -2\\ 0 & 0 & 4 & 8 \end{bmatrix}\\ \begin{aligned} 4z & =8 & \quad -10y+14z & =-2 & \quad 2x-3y+z & =-5\\ \mathbf{z} & \mathbf{= 2} & -10y+14( 2) & =-2 & 2x-3( 3) +2 & =-5\\ & & \mathbf{y} & \mathbf{= 3} & \mathbf{x} & \mathbf{= 1} \end{aligned}\]Question (e)
\[\begin{bmatrix} 1 & 1 & 1 & 6\\ 2 & -1 & 1 & 3\\ 3 & 0 & -1 & 0 \end{bmatrix}\xrightarrow{ \begin{array}{l} R_{2}\rightarrow R_{2} -2R_{1}\\ R_{3}\rightarrow R_{3} -3R_{1} \end{array}}\begin{bmatrix} 1 & 1 & 1 & 6\\ 0 & -3 & -1 & -9\\ 0 & -3 & -4 & -18 \end{bmatrix}\xrightarrow{R_{3}\rightarrow R_{3} -R_{2}}\begin{bmatrix} 1 & 1 & 1 & 6\\ 0 & -3 & -1 & -9\\ 0 & 0 & -3 & -9 \end{bmatrix}\\ \begin{aligned} -3z & =-9 & \quad -3y-z & =-9 & \quad x+y+z & =6\\ \mathbf{z} & \mathbf{= 3} & -3y-3 & =-9 & x+2+3 & =6\\ & & \mathbf{y} & \mathbf{=2} & \mathbf{x} & \mathbf{= 1} \end{aligned}\]Q4: Find the eigenvalues and their associated eigenvectors
(a) \(A=\begin{bmatrix} 7 & 0 & -3\\ -9 & -2 & 3\\ 18 & 0 & -8 \end{bmatrix}\)
(b) \(A=\begin{bmatrix} -5 & 0 & 0\\ 3 & 7 & 0\\ 4 & -2 & 3 \end{bmatrix}\)
Solution
Question (a)
Compute \(\det(\mathbf{A} \ −\lambda \mathbf{I})\) via a cofactor expansion along the second column:
\[\begin{aligned} \det\begin{vmatrix} 7-\lambda & 0 & -3\\ -9 & -2-\lambda & 3\\ 18 & 0 & -8-\lambda \end{vmatrix} & =( -2-\lambda )\begin{vmatrix} 7-\lambda & -3\\ 18 & -8-\lambda \end{vmatrix}\\ & =( -2-\lambda )[( 7-\lambda )( -8-\lambda ) -18( -3)]\\ & =( -2-\lambda )\left[ -56-7\lambda +8\lambda +\lambda ^{2} +54\right]\\ & =( -2-\lambda )\left[ \lambda ^{2} +\lambda -2\right]\\ & =-( \lambda +2)( \lambda +2)( \lambda -1)\\ & =-( \lambda +2)^{2}( \lambda -1) \end{aligned}\]Thus \(A\) has two distinct eigenvalues, \(\lambda _{1} =−2\) and \(\lambda _{2} =1\).
When \(\lambda _{1} =1\),
\[\begin{aligned} \begin{bmatrix} 6 & 0 & -3\\ -9 & -3 & 3\\ 18 & 0 & -9 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix} & =\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}\\ \begin{bmatrix} 6x_{1} -3x_{2}\\ -9x_{1} -3x_{2} +3x_{3}\\ 18x_{1} -9x_{3} \end{bmatrix} & =\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \end{aligned}\] \[\Longrightarrow \ x_{3} =2x_{1} \ \ \text{and} \ \ \ x_{2} =x_{3} -3x_{1}\] \[\Longrightarrow \ x_{3} =2x_{1} \ \ \ \text{and} \ \ \ x_{2} =-x_{1}\]When \(\lambda _{2} =−2\),
\[\begin{aligned} \begin{bmatrix} 9 & 0 & -3\\ -9 & 0 & 3\\ 18 & 0 & -6 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix} & =\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}\\ \begin{bmatrix} 9x_{1} -3x_{3}\\ -9x_{1} +3x_{3}\\ 18x_{1} -6x_{3} \end{bmatrix} & =\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \end{aligned}\]Question (b)
Compute \(\det(\mathbf{A} \ −\lambda \mathbf{I})\) via a cofactor expansion along the first row
\[\begin{aligned} \det\begin{vmatrix} -5-\lambda & 0 & 0\\ 3 & 7-\lambda & 0\\ 4 & -2 & 3-\lambda \end{vmatrix} & =( -5-\lambda )\begin{vmatrix} 7-\lambda & 0\\ -2 & 3-\lambda \end{vmatrix}\\ & =( -5-\lambda )[( 7-\lambda )( 3-\lambda ) -2( 0)]\\ & =( -5-\lambda )\left[ 21-7\lambda -3\lambda +\lambda ^{2}\right]\\ & =( -5-\lambda )\left[ \lambda ^{2} -10\lambda -21\right]\\ & =( -\lambda -5)( \lambda -7)( \lambda -3) \end{aligned}\]Thus \(A\) has three eigenvalues, \(\lambda _{1} =−5\), \(\lambda _{2} =7\), \(\lambda _{3} =3\).
When \(\lambda _{1} =-5\),
\[\begin{aligned} \begin{bmatrix} 0 & 0 & 0\\ 3 & 12 & 0\\ 4 & -2 & 8 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix} & =\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}\\ \begin{bmatrix} 0\\ 3x_{1} +12x_{2}\\ 4x_{1} -2x_{2} +8x_{3} \end{bmatrix} & =\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \end{aligned}\] \[\Longrightarrow \ x_{1} =-4x_{2} \ \ \text{and} \ \ \ x_{2} =x_{3} -3x_{1}\] \[\Longrightarrow \ x_{3} =2x_{1} \ \ \ \text{and} \ \ \ x_{2} =-x_{1}\]When \(\lambda _{1} =−2\),
\[\begin{aligned} \begin{bmatrix} -3 & 0 & 0\\ 3 & 5 & 0\\ 4 & -2 & 1 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix} & =\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}\\ \begin{bmatrix} -3x_{1}\\ 3x_{1} +5x_{2}\\ 4x_{1} -2x_{2} +x_{3} \end{bmatrix} & =\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \end{aligned}\] \[\Longrightarrow \ x_{3} =3x_{1}\]Q5: Using Cayley-Hamilthon approach, find $$A^{-1}$ for the following matrix:
(a) \(\begin{bmatrix} 7 & 2 & -2\\ -6 & -1 & 2\\ 6 & 2 & -1 \end{bmatrix}\)
(b) \(\begin{bmatrix} 2 & 1 & 1\\ 0 & 1 & 0\\ 1 & 1 & 2 \end{bmatrix}\)
Solution
Question (a)
\[\begin{aligned} | A-\lambda I| & =0\\ \left| \begin{bmatrix} 7 & 2 & -2\\ -6 & -1 & 2\\ 6 & 2 & -1 \end{bmatrix} -\lambda \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\right| & =0\\ ( 7-\lambda )[( -1-\lambda )( -1-\lambda ) -4] -2[ -6( -1-A) -12] -2[ -12-6( -1-\lambda )] & =0\\ ( 7-\lambda )\left[ 1+\lambda +\lambda +\lambda ^{2} -4\right] -2[ 6+6\lambda -12] -2[ -12+6+6\lambda ] & =0\\ ( 7-\lambda )\left( \lambda ^{2} +2\lambda -3\right) -\lambda \left( \lambda ^{2} +2\lambda -3\right) -12\lambda +12-12\lambda +12 & =0\\ 7\lambda ^{2} +14\lambda -21-\lambda ^{3} -2\lambda ^{2} +3\lambda -12\lambda +12-12\lambda +12 & =0\\ -\lambda ^{3} +5\lambda ^{2} -7\lambda +3 & =0\\ \lambda ^{3} -5\lambda ^{2} +7\lambda -3 & =0 \end{aligned}\]Replacing \(\lambda =A\),
\[\begin{aligned} A^{3} -5A^{2} +7A-3I & =0\\ A^{2} -5A+7I-3A^{-1} & =0\\ A^{-1} & =\frac{1}{3}\left[ A^{2} -5A+7I\right] \end{aligned}\] \[A^{2} =\begin{bmatrix} 7 & 2 & -2\\ -6 & -1 & 2\\ 6 & 2 & -1 \end{bmatrix} \cdot \begin{bmatrix} 7 & 2 & -2\\ -6 & -1 & 2\\ 6 & 2 & -1 \end{bmatrix} =\begin{bmatrix} 25 & 8 & -8\\ -24 & -7 & 8\\ 24 & 8 & -7 \end{bmatrix}\] \[\begin{aligned} A^{-1} & =\frac{1}{3}\begin{bmatrix} 25 & 8 & -8\\ -24 & -7 & 8\\ 24 & 8 & -7 \end{bmatrix} -5\begin{bmatrix} 7 & 2 & -2\\ -6 & -1 & 2\\ 6 & 2 & -1 \end{bmatrix} +7\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\\ & =\mathbf{\frac{1}{3}\begin{bmatrix} -3 & -2 & 2\\ 6 & 5 & -2\\ -6 & -2 & 5 \end{bmatrix}} \end{aligned}\]Question (b)
\[\begin{aligned} | A-\lambda I| & =0\\ \left| \begin{bmatrix} 2 & 1 & 1\\ 0 & 1 & 0\\ 1 & 1 & 2 \end{bmatrix} -\lambda \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\right| & =0\\ ( 2-\lambda )[( 1-\lambda )( 2-\lambda ) -0] -0+1[ -( 1-\lambda )] & =0\\ 2\left( \lambda ^{2} -3\lambda +2\right) -\lambda \left( \lambda ^{2} +3\lambda +2\right) -1+\lambda & =0\\ 2\lambda ^{2} -6\lambda +4-\lambda ^{3} -3\lambda ^{2} -2\lambda -1+\lambda & =0\\ -\lambda ^{3} -5\lambda ^{2} -7\lambda +3 & =0\\ \lambda ^{3} -5\lambda ^{2} +7\lambda -3 & =0 \end{aligned}\]Replacing \(\lambda =A\),
\[\begin{aligned} A^{3} +5A^{2} +7A-3I & =0\\ A^{2} -5A+7I-3A^{-1} & =0\\ A^{-1} & =\frac{1}{3}\left[ A^{2} -5A+7I\right] \end{aligned}\] \[A^{2} =\begin{bmatrix} 2 & 1 & 1\\ 0 & 1 & 0\\ 1 & 1 & 2 \end{bmatrix} \cdot \begin{bmatrix} 2 & 1 & 1\\ 0 & 1 & 0\\ 1 & 1 & 2 \end{bmatrix} =\begin{bmatrix} 5 & 4 & 4\\ 0 & 1 & 0\\ 4 & 4 & 5 \end{bmatrix}\] \[\begin{aligned} A^{-1} & =\frac{1}{3}\begin{bmatrix} 5 & 4 & 4\\ 0 & 1 & 0\\ 4 & 4 & 5 \end{bmatrix} -5\begin{bmatrix} 2 & 1 & 1\\ 0 & 1 & 0\\ 1 & 1 & 2 \end{bmatrix} +7\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\\ & =\mathbf{\frac{1}{3}\begin{bmatrix} 2 & -1 & -1\\ 0 & 3 & 0\\ -1 & -1 & 2 \end{bmatrix}} \end{aligned}\]Q6: Diagonalize the following matrix, if possible
(a) \(\begin{bmatrix} 2 & 0 & 0\\ 1 & 2 & 1\\ -1 & 0 & 1 \end{bmatrix}\)
(b) \(\begin{bmatrix} 2 & 4 & 6\\ 0 & 2 & 2\\ 0 & 0 & 4 \end{bmatrix}\)
Solution
Question (a)
Step 1: Find the eigenvalue,
\[\det( A-\lambda I) =\det\begin{bmatrix} 2-\lambda & 0 & 0\\ 1 & 2-\lambda & 1\\ -1 & 0 & 1-\lambda \end{bmatrix} =( 2-\lambda )^{2}( 1-\lambda ) =0\]Eigenvalue: \(\lambda =1\) and \(\lambda =2\).
Step 2: Find three linearly independent eigenvector.
Basis of \(\lambda =1\), \(v_{1} =\begin{bmatrix} 0\\ -1\\ 1 \end{bmatrix}\)
Basis of \(\lambda =2\), \(v_{2} =\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}\), \(v_{2} =\begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}\)
Step 3: Construct \(P\) from Step 2:
\[P=\begin{bmatrix} 0 & 0 & -1\\ -1 & 1 & 0\\ 1 & 0 & 1 \end{bmatrix}\]Step 4: Construct \(D\) from corresponding eigenvalue:
\[D=\begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{bmatrix}\]Step 5: Verify \(AP=PD\):
\[\begin{aligned} AP & =\begin{bmatrix} 2 & 0 & 0\\ 1 & 2 & 1\\ -1 & 0 & 1 \end{bmatrix}\begin{bmatrix} 0 & 0 & -1\\ -1 & 1 & 0\\ 1 & 0 & 1 \end{bmatrix} =\begin{bmatrix} 0 & 0 & -2\\ -1 & 2 & 0\\ 1 & 0 & 2 \end{bmatrix}\\ PD & =\begin{bmatrix} 0 & 0 & -1\\ -1 & 1 & 0\\ 1 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{bmatrix} =\begin{bmatrix} 0 & 0 & -2\\ -1 & 2 & 0\\ 1 & 0 & 2 \end{bmatrix} \end{aligned}\]Question (b)
Step 1: Find the eigenvalue,
Since the matrix is triangle,
Eigenvalue: \(\lambda =2\) and \(\lambda =4\).
Step 2: Find three linearly independent eigenvector.
Basis of \(\lambda =2\), \(v_{1} =\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}\)
Basis of \(\lambda =4\), \(v_{2} =\begin{bmatrix} 5\\ 1\\ 1 \end{bmatrix}\)
Every eigenvector of \(A\) is a multiple of \(v_{1}\) or \(v_{2}\). There are not three linearly independent eigenvector of \(A\). \(A\) is not diagonalizable.