Answer to Question #290370 in Linear Algebra for sagar

Question

Write v as a linear combination of u1, u2, u3, where

(a) v = (4, −9, 2), u1 = (1, 2, −1), u2 = (1, 4, 2), u3 = (1, −3, 2);

(b) v = (1, 3, 2), u1 = (1, 2, 1), u2 = (2, 6, 5), u3 = (1, 7, 8);

(c) v = (1, 4, 6), u1 = (1, 1, 2), u2 = (2, 3, 5), u3 = (3, 5, 8);

Answer 1

(a) Writing $~v$ as a linear combination of $u_1, u_2,$ and $u_3$ means finding $c_1,$ $c_2,$ and $c_3$ such that $v=c_1u_1+c_2u_2+c_3u_3.$ This is equivalent to solving a system of linear equations

$$\begin{bmatrix} 1 & 1 & 1 \\ 2 & 4 & -3 \\ -1 & 2 & 2 \end{bmatrix}\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}=\begin{bmatrix} 4 \\ -9 \\ 2 \end{bmatrix}$$

Augmented matrix

$$A=\begin{bmatrix} 1 & 1 & 1 &\ \ 4\\ 2 & 4 & -3 & \ \ -9 \\ -1 & 2 & 2 & \ \ 2 \end{bmatrix}$$

$R_2=R_2-2R_1$

$$\begin{bmatrix} 1 & 1 & 1 &\ \ 4\\ 0 & 2 & -5 & \ \ -17 \\ -1 & 2 & 2 & \ \ 2 \end{bmatrix}$$

$R_3=R_3+R_1$

$$\begin{bmatrix} 1 & 1 & 1 &\ \ 4\\ 0 & 2 & -5 & \ \ -17 \\ 0 & 3 & 3 & \ \ 6 \end{bmatrix}$$

$R_2=R_2/2$

$$\begin{bmatrix} 1 & 1 & 1 &\ \ 4\\ 0 & 1 & -5/2 & \ \ -17/2 \\ 0 & 3 & 3 & \ \ 6 \end{bmatrix}$$

$R_1=R_1-R_2$

$$\begin{bmatrix} 1 & 0 & 7/2 &\ \ 25/2\\ 0 & 1 & -5/2 & \ \ -17/2 \\ 0 & 3 & 3 & \ \ 6 \end{bmatrix}$$

$R_3=R_3-3R_2$

$$\begin{bmatrix} 1 & 0 & 7/2 &\ \ 25/2\\ 0 & 1 & -5/2 & \ \ -17/2 \\ 0 & 0 & 21/2 & \ \ 63/2 \end{bmatrix}$$

$R_3=2R_3/21$

$$\begin{bmatrix} 1 & 0 & 7/2 &\ \ 25/2\\ 0 & 1 & -5/2 & \ \ -17/2 \\ 0 & 0 & 1 & \ \ 3 \end{bmatrix}$$

$R_1=R_1-7R_3/2$

$$\begin{bmatrix} 1 & 0 & 0 &\ \ 2\\ 0 & 1 & -5/2 & \ \ -17/2 \\ 0 & 0 & 1 & \ \ 3 \end{bmatrix}$$

$R_2=R_2+5R_3/2$

$$\begin{bmatrix} 1 & 0 & 0 &\ \ 2\\ 0 & 1 & 0 & \ \ -1 \\ 0 & 0 & 1 & \ \ 3 \end{bmatrix}$$

So $c_1 = 2, c_2 = -1, c_3 = 3,$ and we can check that we can write

$$v=2u_1-u_2+3u_3$$

$$=2\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}-\begin{bmatrix} 1 \\ 4 \\ 2 \end{bmatrix}+3\begin{bmatrix} 1 \\ -3 \\ 2 \end{bmatrix}=\begin{bmatrix} 4 \\ -9 \\ 2 \end{bmatrix}$$

(b) Writing $~v$ as a linear combination of $u_1, u_2,$ and $u_3$ means finding $c_1,$ $c_2,$ and $c_3$ such that $v=c_1u_1+c_2u_2+c_3u_3.$ This is equivalent to solving a system of linear equations

$$\begin{bmatrix} 1 & 2 & 1 \\ 2 & 6& 7 \\ 1 & 5 & 8 \end{bmatrix}\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}=\begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix}$$

Augmented matrix

$$A=\begin{bmatrix} 1 & 2 & 1 &\ \ 1\\ 2 & 6 & 7 & \ \ 3 \\ 1 & 5 & 8 & \ \ 2 \end{bmatrix}$$

$R_2=R_2-2R_1$

$$\begin{bmatrix} 1 & 2 & 1 &\ \ 1\\ 0 & 2 & 5 & \ \ 1 \\ 1 & 5 & 8 & \ \ 2 \end{bmatrix}$$

$R_3=R_3-R_1$

$$\begin{bmatrix} 1 & 2 & 1 &\ \ 1\\ 0 & 2 & 5 & \ \ 1 \\ 0 & 3 & 7 & \ \ 1 \end{bmatrix}$$

$R_2=R_2/2$

$$\begin{bmatrix} 1 & 2 & 1 &\ \ 1\\ 0 & 1 & 5/2 & \ \ 1/2 \\ 0 & 3 & 7 & \ \ 1 \end{bmatrix}$$

$R_1=R_1-2R_2$

$$\begin{bmatrix} 1 & 0 & -4 &\ \ 0\\ 0 & 1 & 5/2 & \ \ 1/2 \\ 0 & 3 & 7 & \ \ 1 \end{bmatrix}$$

$R_3=R_3-3R_2$

$$\begin{bmatrix} 1 & 0 & -4 &\ \ 0\\ 0 & 1 & 5/2 & \ \ 1/2 \\ 0 & 0 & -1/2 & \ \ -1/2 \end{bmatrix}$$

$R_3=-2R_3$

$$\begin{bmatrix} 1 & 0 & -4 &\ \ 0\\ 0 & 1 & 5/2 & \ \ 1/2 \\ 0 & 0 & 1 & \ \ 1 \end{bmatrix}$$

$R_1=R_1+4R_3$

$$\begin{bmatrix} 1 & 0 & 0 &\ \ 4\\ 0 & 1 & 5/2 & \ \ 1/2 \\ 0 & 0 & 1 & \ \ 1 \end{bmatrix}$$

$R_2=R_2-5R_3/2$

$$\begin{bmatrix} 1 & 0 & 0 &\ \ 4\\ 0 & 1 & 0& \ \ -2 \\ 0 & 0 & 1 & \ \ 1 \end{bmatrix}$$

So $c_1 = 4, c_2 = -2, c_3 = 1,$ and we can check that we can write

$$v=4u_1-2u_2+u_3$$

$$=4\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}-2\begin{bmatrix} 2 \\ 6\\ 5 \end{bmatrix}+\begin{bmatrix} 1 \\ 7 \\ 8 \end{bmatrix}=\begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix}$$

(c) Writing $~v$ as a linear combination of $u_1, u_2,$ and $u_3$ means finding $c_1,$ $c_2,$ and $c_3$ such that $v=c_1u_1+c_2u_2+c_3u_3.$ This is equivalent to solving a system of linear equations

$$\begin{bmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & 8 \end{bmatrix}\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}=\begin{bmatrix} 1 \\ 4 \\ 6 \end{bmatrix}$$

Augmented matrix

$$A=\begin{bmatrix} 1 & 2 & 3 &\ \ 1\\ 1 & 3 & 5 & \ \ 4 \\ 2 & 5 & 8 & \ \ 6 \end{bmatrix}$$

$R_2=R_2-R_1$

$$\begin{bmatrix} 1 & 2 & 3 &\ \ 1\\ 0 & 1 & 2 & \ \ 3 \\ 2 & 5 & 8 & \ \ 6 \end{bmatrix}$$

$R_3=R_3-2R_1$

$$\begin{bmatrix} 1 & 2 & 3 &\ \ 1\\ 0 & 1 & 2 & \ \ 3 \\ 0 & 1 & 2 & \ \ 4 \end{bmatrix}$$

$R_1=R_1-2R_2$

$$\begin{bmatrix} 1 & 0 & -1 &\ \ -5\\ 0 & 1 & 2 & \ \ 3 \\ 0 & 1 & 2 & \ \ 4 \end{bmatrix}$$

$R_3=R_3-R_2$

$$\begin{bmatrix} 1 & 0 & -1 &\ \ -5\\ 0 & 1 & 2 & \ \ 3 \\ 0 & 0 & 0 & \ \ 1 \end{bmatrix}$$

$R_1=R_1+5R_3$

$$\begin{bmatrix} 1 & 0 & -1 &\ \ 0\\ 0 & 1 & 2 & \ \ 3 \\ 0 & 0 & 0 & \ \ 1 \end{bmatrix}$$

Then the system has no solution.

Therefore we cannot write $v$ as a linear combination of $u_1, u_2, u_3.$