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- Question 2, Exercise 2.6
- ich the system of homogeneous linear equation may have non-trivial solution. Also solve the system for v... }+4 x_{3}=0\cdots(vi)\\ \end{align*} (iv)-(v), we have\\ \begin{align*} &\begin{array}{cccc} 2x_1&+\frac... \end{align*} Put the value of $x_2$ in (vi), we have \begin{align*} &3 x_{1}-2(\frac{11}{13})x_{3}+4 x... ich the system of homogeneous linear equation may have non-trivial solution. Also solve the system for v
- Question 1, Exercise 2.6
- n-trivial solution. \text{By}\quad(i)-2(ii), we have \begin{align*} &\begin{array}{cccc} 2x_1&-3 x_{2}... \end{align*} Put the value of $x_3$ in (iii), we have \begin{align*} &4 x_{1}+2x_{3}-6 x_{3}=0\\ &4 x_{... ii)} \end{align*} For the system of equations, we have: \begin{align*} A &= \left[ \begin{array}{ccc} 2 ... \end{align*} Put the value of $x_2$ in (iii), we have \begin{align*} &x_1 - 4\left(\frac{2}{5}x_3\right
- Question 3, Exercise 2.6
- \frac{63}{19}\end{align*} From the second row, we have: \begin{align*} &- 2y - 3z = 3\\ \Rightarrow &-2y... {19}\end{align*} Finally, from the first row, we have: \begin{align*} 2x + 3y + 4z &= 2 \\ 2x + 3(\frac... uad R_3 + 3R_2 \end{align*} From the last row, we have: \begin{align*} &-2z = 1 \\ &\Rightarrow \quad z
- Question 1, Exercise 2.2
- hod:** Given \( a_{ij}=\dfrac{i+3j}{2} \). So we have \begin{align*} A&=\begin{bmatrix}a_{11} & a_{12... * Given \( a_{ij}=\dfrac{i \times j}{2} \). So we have \begin{align*} A &= \begin{bmatrix}a_{11} & a_{12
- Question 13, Exercise 2.2
- end{pmatrix} \cdots (ii) \end{align*} From (i) we have \begin{align*} Y = 2X - \begin{pmatrix} 1 & 6 & -... matrix} \end{align*} Put value of $Y$ in (ii), we have \begin{align*} X + 3\left(2X - \begin{pmatrix} 1
- Question 6, Exercise 2.6
- ** Solution. ** For this system of equations; we have \begin{align*} A &= \begin{bmatrix} 5 & 3 & 1 \\ ... ** Solution. ** For this system of equations; we have \begin{align*} A& = \begin{bmatrix} 1 & 2 & -3 \\
- Question 6, Exercise 2.3
- & 0 & 1 \end{bmatrix} = I_3 \end{align*} Thus, we have shown that $ A A^{-1} = A^{-1} A = I_3 $. ===
- Question 1, Review Exercise
- eneous equations with three variables system will have unique solution if: * (a) $\operatorname{Ran