Search
You can find the results of your search below.
Fulltext results:
- Question 3, Exercise 2.5
- nd{array} \right] \quad \text{by } R2-R_3 \quad \frac{1}{2} R3\\ \sim&{\text{R}} \left[ \begin{array}{c... 1 \\ 0 & 1 & 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 & \frac{1}{2} & \frac{1}{2} \end{array} \right] \quad \text{by } R1 + R2 \text{ and } \frac{1}{2} R3\\ \sim&{\text{R}} \left[ \begin{array}{c
- Question 4, Exercise 2.6
- nd{bmatrix}\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} & \vert & 1 \\ 3 & -4 & 3 & \vert & 7 \\ 4 & 2 & -5 & \vert & 10 \end{bmatrix}\quad \dfrac{1}{2}\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} & : & 1 \\ 0 & -\frac{5}{2} & \fr
- Question 6, Exercise 2.6
- 9 & -18 \\ 3 & -7 & -1 \end{bmatrix}\\ A^{-1}& = \frac{1}{|A|} \text{adj}(A) \\ &= \frac{1}{-22} \begin{bmatrix} -2 & -10 & 8 \\ -5 & 19 & -18 \\ 3 & -7 & -1 \end{bmatrix}\\ A^{-1} &= \begin{bmatrix} \frac{2}{22} & \frac{10}{22} & \frac{-8}{22} \\ \frac{5}{22} & \frac{-19}{22} & \frac{18}{22} \\ \frac{-3}{22
- Question 1, Exercise 2.5
- R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 1 & \frac{33}{26} \\ 0 & 18 & 25 \end{array}\right] \quad \frac{1}{26}R_2 \\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 1 & \frac{33}{26} \\ 0 & 0 & \frac{217}{26} \end{array}\right]\quad R_3 - 18R_2\\ \sim & \text{R} \left[\begin{ar
- Question 6, Exercise 2.3
- & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & -\frac{1}{9} & 0 & \frac{1}{9} \end{array}\right] \quad \frac{1}{9} R_3\\ &=\left[\begin{array}{ccc|ccc} 2 & 1 & 0 & 1 - 3 \left(\frac{1}{9}\right) & 0 & 3 \left(\frac{1}{9}\right) \\
- Question 1, Exercise 2.2
- t]$ 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 2, Exercise 2.5
- ]\\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & \frac{9}{5} & \frac{3}{5} \\ 3 & -5 & 6 \\ 2 & 10 & 6 \end{array} \right]\quad \frac{1}{5} R1\\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & \frac{9}{5} & \frac{3}{5} \\ 0 & -\frac{52}{5} & \frac
- Question 3, Exercise 2.6
- }{cccc} 2 & 3 & 4 & 2 \\ 0 & -2 & -3 & 3 \\ 0 & -\frac{13}{2} & -5 & -6 \end{array}\right]\quad R_2 - R_1 \text{and}\quad R_3 - \frac{3}{2}R_1\\ & \sim \text{R}\left[\begin{array}{cccc} 2 & 3 & 4 & 2 \\ 0 & -2 & -3 & 3 \\ 0 & 0 & \frac{19}{4} & -\frac{63}{4} \end{array}\right]\quad R_3 - \frac{13}{4} R_2 \end{align*} Now \begin{align*} \
- Question 7, Exercise 2.3
- rray}\right] \\ |A|& = 12 - 8 = 4\\ A^{-1} &= \dfrac{1}{4} \left[\begin{array}{ll}6 & -1 \\ -8 & 2\end{array}\right]\\ & = \left[\begin{array}{ll} \frac{6}{4} & \frac{-1}{4} \\ \frac{-8}{4} & \frac{2}{4}\end{array}\right]\\ & = \left[\begin{array}{ll} \frac{3}{2} & -\fr
- Question 7 and 8, Exercise 2.6
- 6 & -2 \\ 19 & 5 & -11 \end{bmatrix}\\ A^{-1} &= \frac{1}{62} \begin{bmatrix} -3 & 9 & 5 \\ 26 & -16 & -... & -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{
- Question 5, Exercise 2.6
- So, $A$ is non-singular. \begin{align*} x_1 &= \frac{A_1}{|A|}\\ &=\frac{ \begin{vmatrix} 8 & 1 & 2 \\ 1 & -2 & 3 \\ 10 & -7 & 4 \end{vmatrix}}{52}\\ &=\frac{8(-8 + 21) - 1(4 - 30) + 2(-7 + 20)}{52}\\ &=\frac{104 + 26 + 26}{52}\\ & = \frac{156}{52} = 3 \end{ali
- Question 5, Exercise 2.3
- y}\right] \end{align*} \begin{align*} A^{-1} &= \dfrac{1}{-9} \left[\begin{array}{ccc} -3 & -3 & 0 \\ 1 ... end{array}\right]\\ & = \left[\begin{array}{ccc} \frac{1}{3} & \frac{1}{3} & 0 \\ -\frac{1}{9} & \frac{2}{9} & -\frac{1}{3} \\ \frac{5}{9} & -\frac{1}{9} & -\frac{1}{3} \en
- Question 4, Exercise 2.2
- y}{cc} a & b \\ c & d \end{array}\right]^{-1} &= \frac{1}{ad - bc} \left[\begin{array}{cc} d & -b \\ -c ... 1 & 3 \\ 2 & 4 \end{array}\right] \\ C^{-1} &= \dfrac{1}{-2} \left[\begin{array}{cc} 4 & -3 \\ -2 & 1 \... rray}\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
- Question 2, Exercise 2.6
- 8+3)+1(-4-9)=0\\ &20+11\lambda-13=0\\ &\lambda =-\frac{7}{11}\\ \end{align*} The system becomes \begin{align*} &2 x_{1}+ \frac{7}{11}x_{2}+x_{3}=0 \cdots(iv)\\ &2 x_{1}+3 x_{2}... have\\ \begin{align*} &\begin{array}{cccc} 2x_1&+\frac{7}{11} x_{2}&+ x_{3}&=0\\ \mathop+\limits_{-}2x... ts_{-}3x_2&\mathop-\limits_{+}x_3&=0 \\ \hline &-\frac{26}{11}x_2&+2x_3 &=0\\ \end{array} \\ \implies &
- Question 4, Exercise 2.3
- bda + 16 &= 0\\ -23\lambda &= -16 \\ \lambda &= \dfrac{16}{23} \end{align*} Thus, $\lambda = \dfrac{16}{23}$. =====Question 4(ii)===== Find the value of $\lam... - 1 &= 0\\ (-14 + 2i)\lambda &= 1\\ \lambda &= \dfrac{1}{-14 + 2i}\\ \implies \lambda &= \dfrac{1}{-14 + 2i} \cdot \dfrac{-14 - 2i}{-14 - 2i}\\ & = \dfrac{-1