Question

Given r1 = 3i − 4j + 3k, r2 = 5i + 3j − 6k, r3 = 2i + 7j + 3k and r4 = 4i + 3j + 5k.

Find the scalars a, b and c such that r4 = ar1 + br2 + cr3

Answer 1

$$3a+5b+2c=4$$

$$-4a+3b+7c=3$$

$$3a-6b+3c=5$$

$$A=\begin{pmatrix} 3 & 5 & 2 & & 4 \\ -4 & 3 & 7 & & 3 \\ 3 & -6 & 3 & & 5 \\ \end{pmatrix}$$

$R_1=R_1/3$

$$\begin{pmatrix} 1 & 5/3 & 2/3 & & 4/3 \\ -4 & 3 & 7 & & 3 \\ 3 & -6 & 3 & & 5 \\ \end{pmatrix}$$

$R_2=R_2+4R_1$

$$\begin{pmatrix} 1 & 5/3 & 2/3 & & 4/3 \\ 0 & 29/3 & 29/3 & & 25/3 \\ 3 & -6 & 3 & & 5 \\ \end{pmatrix}$$

$R_3=R_3-3R_1$

$$\begin{pmatrix} 1 & 5/3 & 2/3 & & 4/3 \\ 0 & 29/3 & 29/3 & & 25/3 \\ 0 & -11 & 1 & & 1 \\ \end{pmatrix}$$

$R_2=3R_2/29$

$$\begin{pmatrix} 1 & 5/3 & 2/3 & & 4/3 \\ 0 & 1 & 1 & & 25/29 \\ 0 & -11 & 1 & & 1 \\ \end{pmatrix}$$

$R_1=R_1-5R_2/3$

$$\begin{pmatrix} 1 & 0 & -1 & & -3/29 \\ 0 & 1 & 1 & & 25/29 \\ 0 & -11 & 1 & & 1 \\ \end{pmatrix}$$

$R_3=R_3+11R_2$

$$\begin{pmatrix} 1 & 0 & -1 & & -3/29 \\ 0 & 1 & 1 & & 25/29 \\ 0 & 0 & 12 & & 304/29 \\ \end{pmatrix}$$

$R_3=R_3/12$

$$\begin{pmatrix} 1 & 0 & -1 & & -3/29 \\ 0 & 1 & 1 & & 25/29 \\ 0 & 0 & 1 & & 76/87 \\ \end{pmatrix}$$

$R_1=R_1+R_3$

$$\begin{pmatrix} 1 & 0 & 0 & & 67/87 \\ 0 & 1 & 1 & & 25/29 \\ 0 & 0 & 1 & & 76/87 \\ \end{pmatrix}$$

$R_2=R_2-R_3$

$$\begin{pmatrix} 1 & 0 & 0 & & 67/87 \\ 0 & 1 & 0 & & -1/87 \\ 0 & 0 & 1 & & 76/87 \\ \end{pmatrix}$$

$$a=\dfrac{67}{87}, b=-\dfrac{1}{87}, c=\dfrac{76}{87}$$