Question

1) M = [ [-2,1,0], [-11,4,1],[-18,6,1]]
a) solve the equation M [ [x] , [y], [z] ] = 3 [ [x] , [y] , [z] ]
b) solve the equation M [ [x], [y], [z] ] = 4 [ [x] , [y], [z] ]

2) A matrix S has eigenvalues 2, -3 and 5 with corresponding eigenvectors v1 = [ [1] , [-3] , [-2] ] , v2 = [ [-2], [7], [5] ] and v3 = [ [0], [0], [1]] respectively.
a) write down the values of Sv1, Sv2 and Sv3.
b) evaluate S[ [0], [1], [0]]
d) hence or otherwise find matrix S

Answer 1

Answer on Question #45231 - Math - Matrix - Tensor Analysis

1) $M = [[-2,1,0], [-11,4,1], [-18,6,1]]$

a) solve the equation $M([x], [y], [z]) = 3([x], [y], [z])$

b) solve the equation $M([x], [y], [z]) = 4([x], [y], [z])$

Solution

a)

$$\left\{ \begin{array}{l} (-2 - 3)x + y + 0 \cdot z = 0 \\ -11x + (4 - 3)y + 1 \cdot z = 0 \\ -18x + 6y + (1 - 3) \cdot z = 0 \end{array} \right. \longrightarrow \left\{ \begin{array}{l} (-5)x + y = 0 \\ -11x + y + 1 \cdot z = 0 \\ -18x + 6y + (-2) \cdot z = 0 \end{array} \right.$$ $$y = 5x$$ $$-11x + 5x + 1 \cdot z = 0 \rightarrow z = 6x$$ $$-18x + 6 \cdot 5x + (-2) \cdot 6x = 0 \rightarrow 0 \cdot x = 0 \rightarrow x = t, t \in \mathbb{R}.$$

So

$$\left( \begin{array}{c} x \\ y \\ z \end{array} \right) = t \left( \begin{array}{c} 1 \\ 5 \\ 6 \end{array} \right).$$

b)

$$\left\{ \begin{array}{l} (-2 - 4)x + y + 0 \cdot z = 0 \\ -11x + (4 - 4)y + 1 \cdot z = 0 \\ -18x + 6y + (1 - 4) \cdot z = 0 \end{array} \right. \rightarrow \left\{ \begin{array}{l} (-6)x + y = 0 \\ -11x + 1 \cdot z = 0 \\ -18x + 6y + (-3) \cdot z = 0 \end{array} \right.$$ $$y = 6x$$ $$z = 11x$$ $$-18x + 6 \cdot 6x + (-3) \cdot 11x = 0 \rightarrow -15x = 0 \rightarrow x = 0.$$ $$\left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right).$$

2) A matrix $S$ has eigenvalues 2, -3 and 5 with corresponding eigenvectors $v1 = [[1], [-3], [-2]], v2 = [[-2], [7], [5]]$ and $v3 = [[0], [0], [1]]$ respectively.

a) write down the values of Sv1, Sv2 and Sv3.

b) evaluate $S[[0], [1], [0]]$

d) hence or otherwise find matrix S

Solution

a)

$$Sv_1 = 2v_1 = 2 \left( \begin{array}{c} 1 \\ -3 \\ -2 \end{array} \right) = \left( \begin{array}{c} 2 \\ -6 \\ -4 \end{array} \right)$$ $$\begin{array}{l} \mathrm{S v}_{2} = -3 v_{2} = -3 \binom{-2}{7} = \binom{6}{-21} \\ \mathrm{S v}_{3} = 5 v_{3} = 5 \binom{0}{0} = \binom{0}{0} \\ \end{array}$$

b)

$$\binom{0}{1} = a \binom{1}{-3} + b \binom{-2}{7} + c \binom{0}{0} \rightarrow$$ $$2a = b; \quad -3a + 7 \cdot 2a = 1 \rightarrow a = \frac{1}{11}, \quad b = \frac{2}{11}; \quad -2 \cdot \frac{1}{11} + \frac{2}{11} \cdot 5 + c = 0 \rightarrow c = -\frac{4}{11}.$$

So

$$\binom{0}{1} = \frac{1}{11} \binom{1}{-3} + \frac{2}{11} \binom{-2}{7} - \frac{4}{11} \binom{0}{0} = \frac{1}{11} v_{1} + \frac{2}{11} v_{2} - \frac{4}{11} v_{3}.$$ $$S \binom{0}{1} = S \left( \frac{1}{11} v_{1} + \frac{2}{11} v_{2} - \frac{4}{11} v_{3} \right) = \frac{1}{11} \mathrm{S v}_{1} + \frac{2}{11} \mathrm{S v}_{2} - \frac{4}{11} \mathrm{S v}_{3} = \frac{1}{11} \binom{2}{-6} + \frac{2}{11} \binom{6}{-21} - \frac{4}{11} \binom{0}{0}.$$ $$S \binom{0}{1} = \frac{1}{11} \binom{2 + 2 \cdot 6 - 4 \cdot 0}{-6 + 2 \cdot (-21) - 4 \cdot 0} = \frac{1}{11} \binom{14}{-48}.$$

d) We need to find $S \binom{1}{0}$.

$$\binom{1}{0} = d \binom{1}{-3} + e \binom{-2}{7} + f \binom{0}{0} \rightarrow$$ $$7e = 3d \rightarrow e = \frac{3}{7} d; \quad d - 2 \cdot \frac{3}{7} d = 1 \rightarrow d = 7, \quad e = 3; \quad -2 \cdot 7 + 5 \cdot 3 + f = 0 \rightarrow f = -1.$$ $$\binom{1}{0} = 7 v_{1} + 3 v_{2} - v_{3}.$$ $$S \binom{1}{0} = S (7 v_{1} + 3 v_{2} - v_{3}) = 7 S v_{1} + 3 S v_{2} - S v_{3} = 7 \binom{2}{-6} + 3 \binom{6}{-21} - \binom{0}{0} = \binom{32}{-105} - 78.$$

The rows of $S$ are transposed vectors $S \binom{1}{0}, S \binom{0}{1}$ and $S \binom{0}{0}$.

The matrix $S$ is

$$S = \begin{pmatrix} 32 & -105 & -78 \\ \frac{14}{11} & -\frac{48}{11} & -\frac{54}{11} \\ 0 & 0 & 5 \end{pmatrix}.$$

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