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# Use the reduced coefficient matrix to write a general solution in terms of one or more column matrices. Also determine the dimension of the solution space and a basis for this space.\begin{aligned} 4 x_{1}-3 x_{2}+x_{4}+x_{5}-3 x_{6} &=0 \\ 2 x_{2}+4 x_{4}-x_{5}-6 x_{6} &=0 \\ 3 x_{1}-2 x_{2}+4 x_{5}-x_{6} &=0 \\ 2 x_{1}+x_{2}-3 x_{3}+4 x_{4} &=0 \end{aligned}

Solution

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The coefficient matrix is

\begin{align*} A&=\mqty[4 & -3 & 0 & 1 & 1 & -3 \\ 0 & 2 & 0 & 4 & -1 & -6 \\ 3 & -2 & 0 & 0 & 4 & -1 \\ 2 & 1 & -3 & 4 & 0 & 0 ]\\ \text{We find the reduced matrix}\\ A_R&=\mqty[1 & 0 & 0 & 0 & \frac{22}{3} & -\frac{1}{5} \\[5pt] 0 & 1 & 0 & 0 & \frac{49}{10} & \frac{1}{5} \\[5pt] 0 & 0 & 1 & 0 & \frac{11}{10} & -\frac{11}{5}\\[5pt] 0 & 0 & 0 & 1 & -\frac{27}{10} & -\frac{8}{5} ]\\ \end{align*}

$rank(A)=4 \Rightarrow$ general solution will have $m-rank(A)=6-4=2$ arbitrary constants.

Let $x_5=\alpha$ and $x_6=\beta$ to write the general solution.

$\begin{equation*} X=\alpha \mqty(-\frac{22}{3} \\[5pt] -\frac{49}{10} \\[5pt] -\frac{11}{10} \\[5pt] \frac{27}{10} \\[5pt] 1 \\[5pt] 0) + \beta \mqty(\frac{1}{5} \\[5pt] -\frac{1}{5} \\[5pt] \frac{11}{5} \\[5pt] \frac{8}{5} \\[5pt] 0 \\[5pt] 1) \end{equation*}$

The dimension of the solution space is $6-4=2$.

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