Question

Write v = (2, -5 , 3) as a linear combination of

u1=(1,−3 ,2)

u2=(2,−4 ,−1)

u3=(1,−3 , 7)

Answer 1

Let $x_1u_1+x_2u_2+x_3u_3=v\quad x\in\reals$

$\begin{bmatrix} 1&2&1 \\ -3&-4&-3\\2&-1&7 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2\\x_3 \end{bmatrix}=\begin{bmatrix} 2 \\-5\\3 \end{bmatrix}$

Considering augmented matrix $A$ for this system and applying Gauss- Jordan elimination

$A=\begin{matrix} |&1&2&1\\|&-3&-4&-3&\\|&2&-1&7 \end{matrix}\begin{vmatrix} 2\\-5\\3 \end{vmatrix}$

$-\frac{1}{3}$ $R_2-R_1\to\>R_2$

$\begin{matrix} |&1&2&1 \\ |&0&-\frac{2}{3} & 0\\ |&1&-\frac{1}{2}&\frac{7}{2} \end{matrix}\begin{vmatrix} 2\\ -\frac{1}{3}\\ \frac{3}{2} \end{vmatrix}$

$\frac{1}{2}\>R_3-R_1\>\to\>R_2$

$\begin{matrix} |&1&2&1 \\ |&0&-\frac{2}{3}&0 \\ |&0&-\frac{5}{2}&\frac{5}{2} \end{matrix}\begin{vmatrix} 2 \\ -\frac{1}{3} \\ -\frac{1}{2} \end{vmatrix}$

$-\frac{3}{2}\>R_2\to\>R_2$ $\begin{matrix} |&1&2&1 \\ |&0&1&0 \\ |&0&-\frac{5}{2}&\frac{5}{2} \end{matrix}\begin{vmatrix} 2 \\ \frac{1}{2} \\ -\frac{1}{2} \end{vmatrix}$

$-\frac{2}{5}R_3-R_2\to\>R_3$

$\begin{matrix} |&1&2&1 \\ |&0&1&0 \\ |&0&0&-1 \end{matrix}\begin{vmatrix} 2 \\ \frac{1}{2} \\ -\frac{3}{10} \end{vmatrix}$

$-1$ $R_3\to\>R_3$ $\begin{matrix} |&1&2&1 \\ |&0&1&0 \\ |&0&0&1 \end{matrix}\begin{vmatrix} 2 \\ \frac{1}{2}\\ \frac{3}{10} \end{vmatrix}$

$R_1-R_3\to\>R_1$ $\begin{matrix} |&1&2&0 \\ |&0&1&0\\ |&0&0&1 \end{matrix}\begin{vmatrix} 2 \\ \frac{1}{2} \\ \frac{3}{10} \end{vmatrix}$

$R_1-2R_2\to\>R_1$ $\begin{matrix} |&1&0&0 \\ |&0&1&0 \\ |&0&0&1 \end{matrix}\begin{vmatrix} \frac{7}{10} \\ \frac{1}{2} \\ \frac{3}{10} \end{vmatrix}$

rref $A=$ $\begin{matrix}|& 1&0&0 \\|& 0&1&0\\|&0&0&1 \end{matrix}\begin{vmatrix} \frac{7}{10}\\\frac{1}{2}\\\frac{3}{10} \end{vmatrix}$

$\implies x_1=\frac{7}{10}\quad x_2=\frac{1}{2} \quad x_3=\frac{3}{10}$

$v=\frac{7}{10}u_1+\frac{1}{2}u_2+\frac{3}{10}u_3$