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- Question 1, Exercise 2.5
- & 8 & 3 \\ -4 & 6 & 5\end{array}\right]\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 26 &... right]\quad R_2 + 6R_1 \quad R_3 + 4R_1\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 1 & ... nd{array}\right] \quad \frac{1}{26}R_2 \\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 1 & ... 26} \end{array}\right]\quad R_3 - 18R_2\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 1 &
- Question 2, Exercise 2.5
- & 6 \\ 2 & 10 & 6 \end{array} \right]\\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & \frac{9}{5} & \fr... \end{array} \right]\quad \frac{1}{5} R1\\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & \frac{9}{5} & \fr... end{array} \right]\quad R2 - 3 \cdot R1\\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & \frac{9}{5} & \fr... nd{array} \right]\quad R3 - 2 \cdot R1 \\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & \frac{9}{5} & \fr
- Question 4, Exercise 2.6
- & 7 \\ 4 & 2 & -5 & : & 10 \end{bmatrix}\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2}... ert & 10 \end{bmatrix}\quad \dfrac{1}{2}\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2}... 5 & : & 10 \end{bmatrix}\quad R_2 - 3R_1\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2}... -3 & : & 6 \end{bmatrix}\quad R_3 - 4R_1\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2}
- Question 7 and 8, Exercise 2.6
- 9 \\ 9 & -16 & 5 \\ 5 & -2 & -11 \end{bmatrix}\\ \text{adj}(A) &= \begin{bmatrix} -3 & 9 & 5 \\ 26 & -16... & 3 & 4 \\ 1 & 2 & -3 & 0 \end{bmatrix}\\ &\sim \text{R} = \begin{bmatrix} 1 & 2 & -3 & 0\\ 2 & -1 & 3 & 4 \\ 3 & 4 & 7 & 14 \end{bmatrix} \quad R_1 \text{interchange} R_3\\ &\sim \text{R} = \begin{bmatrix} 1 & 2 & -3 & 0\\ 0 & -5 & 9 & 4 \\ 0 & -2 & 16 & 14
- Question 3, Exercise 2.5
- -1 & 4 & 0 & 0 & 1 \end{array} \right]\\ \sim &{\text{R}} \left[ \begin{array}{ccc|ccc} 1 & -1 & 4 & 0 ... & -1 & -1 & 1 & 0 & 0 \end{array} \right] \quad \text{by swapping } R1 \text{ and } R3\\ \sim&{\text{R}} \left[ \begin{array}{ccc|ccc} 1 & -1 & 4 & 0 & 0 & 1 \\ 0 & 2 & 4 & 0 & 1
- Question 2, Exercise 2.6
- ** \begin{align*} &x_{1}-4 x_{2}+3 x_{3}=0 \quad \text{(i)}\\ &2 x_{1}+\lambda x_{2}+x_{3}=0 \quad \text{(ii)}\\ &x_{1}-2 x_{2}+\lambda x_{3}=0 \quad \text{(iii)} \end{align*} For the system to have a non-tri... begin{align*} x_{1} - 4x_{2} + 3x_{3} &= 0 \quad \text{(i)} \\ 2x_{1} + 2x_{2} + x_{3} &= 0 \quad \text{
- Question 3, Exercise 2.6
- 5 \\ 3 & -2 & 1 & -3 \end{array}\right]\\ & \sim \text{R}\left[\begin{array}{cccc} 2 & 3 & 4 & 2 \\ 0 & ... }{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... 1 \\ 3 & -4 & -5 & 3 \end{array}\right]\\ &\sim \text{R}\left[\begin{array}{cccc} 2 & 2 & 6 & 1 \\ 5 &
- Question 6, Exercise 2.6
- \\ -10 & 19 & -7 \\ 8 & -18 & -1 \end{bmatrix}\\ \text{adj}(A)& = \begin{bmatrix} -2 & -10 & 8 \\ -5 & 1... -7 & -1 \end{bmatrix}\\ A^{-1}& = \frac{1}{|A|} \text{adj}(A) \\ &= \frac{1}{-22} \begin{bmatrix} -2 & ... {23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix}\\ \text{Cofactor matrix} &= \begin{bmatrix} 11 & 8 & 1 \\ 4 & -8 & -4 \\ -5 & -8 & -7 \end{bmatrix}\\ \text{adj}(A) &= \begin{bmatrix} 11 & 4 & -5 \\ 8 & -8
- Question 1, Exercise 2.1
- nd{array}\right]$ ** Solution. ** \begin{align}\text{Order of A}&= 2\times 3\end{align} ===== Ques... nd{array}\right]$ ** Solution. ** \begin{align}\text{Order of B}&= 3\times 2\end{align} ===== Question... d{array}\right]$. ** Solution. ** \begin{align}\text{Order of C}&= 3\times 1\end{align} ===== Question... {array}\right]$. ** Solution. ** \begin{align}\text{Order of D}&= 1\times 4\end{align} =====Question
- Question 6, Exercise 2.3
- atrix} \end{align*} To show $A A^{-1} = I_3\quad \text{and }\quad A^{-1} A = I_3$\\ \begin{align*} A A^... \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I_3 \\ \text{Now}\quad A^{-1} A \\ A^{-1} A &= \begin{bmatri... A A^{-1} = A^{-1} A = I_3 $. ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:ex2-3-p5|< Question 5]]</btn></text> <text align="right"><btn type="success">[[math-1
- Question 1, Exercise 2.6
- lign*} So the system has non-trivial solution. \text{By}\quad(i)-2(ii), we have \begin{align*} &\begin... ** \begin{align*} &2x_1 - 3x_2 + 4x_3 = 0 \quad \text{(i)}\\ &x_1 + x_2 + x_3 = 0 \quad \text{(ii)}\\ &x_1 - 4x_2 + 3x_3 = 0 \quad \text{(iii)} \end{align*} For the system of equations, we have: \b
- Question 9, Exercise 2.2
- 23} + b_{23} & a_{33} + b_{33} \end{pmatrix} \\ \text{and}\\ (A + B)^t &= \begin{pmatrix} a_{11} + b... $\quad(A + B)^t = A^t + B^t$. ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:ex2-2-p8|< Question 8]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit02:ex2-2-p10|Question 10 >]]</btn></text
- Question 2, Exercise 2.1
- ** Solution. ** Square matrix ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:ex2-1-p1|< Question 1]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit02:ex2-1-p3|Question 3 >]]</btn></text>
- Question 3, Exercise 2.1
- lar matrix; Diagonal matrix\\ ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:ex2-1-p2|< Question 2]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit02:ex2-1-p4|Question 4 >]]</btn></text>
- Question 3, Exercise 2.2
- \end{bmatrix}\\ \end{align*} ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:ex2-2-p2|< Question 2]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit02:ex2-2-p4|Question 4 >]]</btn></text>