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Question 6, Exercise 2.2
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matrix \begin{align} & A^{2}+\alpha I=\beta A\\ \implies &\begin{bmatrix} 2 & 1 \\ 3 & -3 \end{bmatrix} \b... \begin{bmatrix} 2 & 1 \\ 3 & -3 \end{bmatrix}\\ \implies & \begin{bmatrix} 4 + 3 & 2 - 3 \\ 6 - 9 & 3 + 9 ... beta & \beta \\ 3\beta & -3\beta \end{bmatrix}\\ \implies &\begin{bmatrix} 7 + \alpha & -1 \\ -3 & 12 + \al... 1), we get: \begin{align} & 7 + \alpha = 2(-1)\\ \implies & \alpha = -2 - 7\\ \implies & \alpha = -9\end{al
Question 6, Exercise 2.6
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2} & \frac{7}{22} & \frac{1}{22} \end{bmatrix}\\ \implies A^{-1} &= \begin{bmatrix} \frac{1}{11} & \frac{5}... end{align*} Since \begin{align*} X &= A^{-1}B \\ \implies X &= A^{-1}B = \begin{bmatrix} \frac{1}{11} & \fr... x} \begin{bmatrix} 6 \\ 19 \\ 25 \end{bmatrix}\\ \implies X &= \begin{bmatrix} \frac{6 + 95 - 100}{11} \\ \... 22} \\ \frac{-18 + 133 + 25}{22} \end{bmatrix}\\ \implies X &= \begin{bmatrix} \frac{1}{11} \\ \frac{119}{2
Question 1, Exercise 2.3
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end{array}\right]\\ |A|&=2(-2-2)-3(2-8)+1(1+4)\\ \implies |A|&=-8+18+5\\ \implies |A|&=15 \end{align*} =====Question 1(ii)===== Evaluate the determinant of the ... a (cos \theta -0)+ sin \theta(sin \theta -0)+0\\ \implies |A| &= cos^2 \theta + sin^2 \theta\\ \implies |A| &= 1 \end{align*} =====Question 1(iii)===== Evaluate t
Question 4, Exercise 2.2
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s we know \begin{align*} &(x+y)^2=x^2+y^2+2xy \\ \implies & x^2+y^2 = (x+y)^2-2xy \\ \implies & x^2+y^2 = 6^2-2(8) =20 \end{align*} Hence, we conclude $z=4$, $t=0
Question 1, Exercise 2.6
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_3&=0 \\ \hline &x_2&-2x_3 &=0\\ \end{array} \\ \implies &x_2=2x_3\\ \end{align*} Put the value of $x_3$ ... &=0 \\ \hline &-5x_2&+2x_3 &=0\\ \end{array} \\ \implies & x_2 = \frac{2}{5}x_3 \end{align*} Put the value
Question 8, Exercise 2.2
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21} & b_{22} \\ b_{31} & b_{32} \end{bmatrix}\\ \implies AB &= \begin{bmatrix} a_{11}b_{11} + a_{12}b_{21
Question 4, Exercise 2.3
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)\lambda &= 1\\ \lambda &= \dfrac{1}{-14 + 2i}\\ \implies \lambda &= \dfrac{1}{-14 + 2i} \cdot \dfrac{-14 -
Question 2, Exercise 2.6
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e &-\frac{26}{11}x_2&+2x_3 &=0\\ \end{array} \\ \implies &x_2=\frac{11}{13}x_3\\ \end{align*} Put the val