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- 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}\qua... dfrac{1}{2}\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} & : & 1 \\ 0 & -\frac{5}{2} & \frac{9}{2} & : & 4 \\ 4 & 2 & -5 & : & 10 \end{bmat
- 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 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 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 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
- \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} & -\frac{1}{4} \\ -2 &
- 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 1, Exercise 2.2
- \( A \) is: \[ A = \left[\begin{array}{cc} 2 & \frac{7}{2} \\ \frac{5}{2} & 4 \end{array}\right] \] **Alternative Method:** Given \( a_{ij}=\dfrac{i+3j}{2... _{21} & a_{22} \end{bmatrix} \\ &=\begin{bmatrix}\frac{1+3(1)}{2} & \frac{1+3(2)}{2} \\ \frac{2+3(1)}{2} & \frac{2+3(2)}{2} \end{bmatrix} \\ &=\begin{bmatrix}
- 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 5, Exercise 2.3
- 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} \end{array}\right] \end
- Question 1, Exercise 2.6
- x_2&+2x_3 &=0\\ \end{array} \\ \implies & x_2 = \frac{2}{5}x_3 \end{align*} Put the value of $x_2$ in (iii), we have \begin{align*} &x_1 - 4\left(\frac{2}{5}x_3\right) + 3x_3 = 0 \\ &x_1 - \frac{8}{5}x_3 + 3x_3 = 0 \\ &x_1 + \frac{7}{5}x_3 = 0 \\ &x_1 = -\frac{7}{5}x_3 \end{align*} Therefore, the
- 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 & -... -5 & 9 & 4 \\ 0 & -1 & 8 & 7 \end{bmatrix} \quad \frac{1}{2}R_3\\ &\sim \text{R} = \begin{bmatrix} 1 & 0... -1 & 8 & 7\\ 0 & 0 & 1 & 1 \end{bmatrix} \quad \frac{-1}{31}R_2\text{interchange}\quad R_3\\ &\sim \te
- 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 ... ) + 2(1) & -3\left(\dfrac{3}{2}\right) + 2\left(-\frac{1}{2}\right) \end{array}\right] \\ &= \left[\begi