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- Question 7 and 8, Exercise 2.6
- & -11 \end{bmatrix}\\ A^{-1} &= \begin{bmatrix} \dfrac{-3}{62} & \dfrac{9}{62} & \dfrac{5}{62} \\ \dfrac{26}{62} & \dfrac{-16}{62} & \dfrac{-2}{62} \\ \dfrac{19}{62} & \dfrac{-5}{62} & \dfrac{
- Question 1, Exercise 2.2
- ht]$ of order $2 \times 2$ for which is $a_{i j}=\dfrac{i+3 j}{2}$ ** Solution. ** Given \( a_{ij} = \dfrac{i + 3j}{2} \). For \( i = 1, j = 1 \): \[ a_{11} = \dfrac{1 + 3 \cdot 1}{2} = \dfrac{1 + 3}{2} = \dfrac{4}{2} = 2 \] For \( i = 1, j = 2 \): \[ a_{12} = \dfrac{
- Question 5, Exercise 2.3
- ay}\right] \end{align*} \begin{align*} A^{-1} &= \dfrac{1}{-9} \left[\begin{array}{ccc} -3 & -3 & 0 \\ 1 ... e of $A$ is: $$ A^{-1} =\left[\begin{array}{ccc} \dfrac{1}{3} & \dfrac{1}{3} & 0 \\ -\dfrac{1}{9} & \dfrac{2}{9} & -\dfrac{1}{3} \\ \dfrac{5}{9} & -\dfrac{1}{9} & -\dfrac{1}{3
- Question 4, Exercise 2.2
- } 1 & 3 \\ 2 & 4 \end{array}\right] \\ C^{-1} &= \dfrac{1}{-2} \left[\begin{array}{cc} 4 & -3 \\ -2 & 1 \... array}\right]\\ & = \left[\begin{array}{cc} -2 & \dfrac{3}{2} \\ 1 & -\dfrac{1}{2} \end{array}\right]\end{align*} \begin{align*} A &= B^{-1} C^{-1} \\ &= \left[... \end{array}\right] \left[\begin{array}{cc} -2 & \dfrac{3}{2} \\ 1 & -\dfrac{1}{2} \end{array}\right]\\ A
- Question 4, Exercise 2.3
- mbda + 16 &= 0\\ -23\lambda &= -16 \\ \lambda &= \dfrac{16}{23} \end{align*} Thus, $\lambda = \dfrac{16}{23}$. =====Question 4(ii)===== Find the value of $\lamb... a - 1 &= 0\\ (-14 + 2i)\lambda &= 1\\ \lambda &= \dfrac{1}{-14 + 2i}\\ \implies \lambda &= \dfrac{1}{-14 + 2i} \cdot \dfrac{-14 - 2i}{-14 - 2i}\\ & = \dfrac{-14
- Question 1, Exercise 2.5
- \\ \sim & \text{R}\left[\begin{array}{ccc} 1 & -\dfrac{1}{2} & 0 \\ 0 & 1 & \frac{8}{9} \\ 0 & 0 & 1 \en... \sim & \text{R}\left[\begin{array}{ccc} 1 & -2 & \dfrac{3}{2} \\ 4 & 1 & 8 \\ 7 & 3 & 0 \end{array}\right... \sim & \text{R}\left[\begin{array}{ccc} 1 & -2 & \dfrac{3}{2} \\ 0 & 9 & 2 \\ 7 & 3 & 0 \end{array}\right... sim & \text{R} \left[\begin{array}{ccc} 1 & -2 & \dfrac{3}{2} \\ 0 & 9 & 2 \\ 0 & 17 & -\dfrac{21}{2} \en
- Question 6, Exercise 2.6
- f linear equation by matrix inversion method.\\ $\dfrac{2}{x}+\dfrac{3}{y}+\dfrac{10}{z}=4$\\ $\dfrac{4}{x}-\dfrac{6}{y}+\dfrac{5}{z}=1$\\ $\dfrac{6}{x}+\dfrac{9}{y}-\dfrac{20}{z}=2$ **
- Question 7, Exercise 2.2
- rs $n, A^{n}=\left[\begin{array}{cc}x^{n} & 0 \\ \dfrac{y\left(x^{n}-1\right)}{x-1} & 1\end{array}\right]... 1$ \begin{align}A^1 =\begin{bmatrix} x^1 & 0 \\ \dfrac{y(x^1 - 1)}{x - 1} & 1 \end{bmatrix} \\ = \begin{bmatrix} x & 0 \\ \dfrac{y(x - 1)}{x - 1} & 1 \end{bmatrix}\\ = \begin{bma... } &= A^k \cdot A\\ &= \begin{bmatrix} x^k & 0 \\ \dfrac{y(x^k - 1)}{x - 1} & 1 \end{bmatrix}\begin{bmatri
- Question 7, Exercise 2.3
- array}\right] \\ |A|& = 12 - 8 = 4\\ A^{-1} &= \dfrac{1}{4} \left[\begin{array}{ll}6 & -1 \\ -8 & 2\end... |B| &= (3 \cdot 2) - (2 \cdot 0) = 6\\ B^{-1} &= \dfrac{1}{6} \left[\begin{array}{ll} 2 & -2 \\ 0 & 3 \en
- Question 2, Exercise 2.6
- end{align*} Hence S.S$=\left[ \begin{array}{c} - \dfrac{10}{11} x_{3} \\ \dfrac{11}{13}x_3 \\ x_3 \end{array} \right]$ =====Question 2(ii)===== Find the
- Question 13, Exercise 2.2
- } 7 & 21 & -7 \\ 7 & 0 & 21 \end{pmatrix}\\ X &= \dfrac{1}{7} \begin{pmatrix} 7 & 21 & -7 \\ 7 & 0 & 21 \
- Question 4, Exercise 2.6
- 7 \\ 4 & 2 & -5 & \vert & 10 \end{bmatrix}\quad \dfrac{1}{2}\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{