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- Question 3, Exercise 2.6
- -2 y+z=-3$\\ ** Solution. ** Given the system of equations: \begin{align*} \begin{aligned} 2x + 3y + 4z &= 2... 4 y-5 z=3$\\ ** Solution. ** Given the system of equations: \begin{align*} 5x - 2y + z &= 2 \quad \cdots (i)... + 0z = 3\), which is inconsistent, the system of equations has no solution. =====Question 3(iii)===== So... $x+3 y=5$\\ ** Solution. ** Given the system of equations: \begin{align*} 2x + z &= 2 \quad \cdots (i) \\ 2
- Question 6, Exercise 2.6
- 2 y+4 z=25$\\ ** Solution. ** For this system of equations; we have \begin{align*} A &= \begin{bmatrix} 5 & ... lign*} Therefore, the solution to the system of equations is: $$x = \frac{1}{11}, \quad y =\frac{119}{11}, ... +2 y-5 z=-3$ ** Solution. ** For this system of equations; we have \begin{align*} A& = \begin{bmatrix} 1 & ... {align*} Therefore, the solution to the system of equations is: $$ x = \frac{37}{12}, \quad y = \frac{7}{3},
- Question 7 and 8, Exercise 2.6
- ht]$; find $A^{-1}$ and hence solve the system of equations.\\ $3 x+4 y+7 z=14 ; 2 x-y+3 z=4 ; \quad x+2 y-3 ... {-11}{62} \end{bmatrix}$$ Now given the system of equations: \begin{align*} 3x + 4y + 7z &= 14 \\ 2x - y + 3z... 2&=1\\ x_3&=1 \end{align*} Now solutions of above equations are; $$ \begin{bmatrix} \dfrac{-3}{62} & \dfrac{9
- Question 1, Review Exercise
- $3$</collapse> vii. System of homogeneous linear equations has non-trivial solution if: * (a) $|A|>0$ ... $</collapse> viii. For non-homogeneous system of equations; the system is inconsistent if: * (a) $\oper... s</collapse> ix. For a system of non-homogeneous equations with three variables system will have unique solu
- Question 1, Exercise 2.6
- \quad \text{(iii)} \end{align*} For the system of equations, we have: \begin{align*} A &= \left[ \begin{array
- Question 4, Exercise 2.6
- 2\end{align*} Thus, the solution to the system of equations is: $$\boxed{x_1 = \frac{13}{3}, \quad x_2 = \fra
- Question 5, Exercise 2.6
- {align*} The solution set for the given system of equations using Cramer's rule is: $$( -\frac{1}{7}, \frac{1
- Question 9 and 10, Exercise 2.6
- n. =====Question 9===== Show that the system of equations $2 x-y+3 z=\alpha ; 3 x+y-5 z=\beta ;-5 x-5 y+21