Question

Find a, b, c if $a(1,3,2)+b(1,-5,6)+c(2,1,-2)=(4,10,-8)$.

Answer 1

$a(1,3,2)+b(1,-5,6)+c(2,1,-2)=(4,10,-8)$

$(a+b+2c,3a-5b+c,2a+6b-2c)=(4,10,-8)$

$a+b+2c=4$ ---- $\left(1\right)$

$3a-5b+c=10$ ----- $\left(2\right)$

$2a+6b-2c=-8$ ---- $\left(3\right)$

Adding $1$ and $3$, we get

$3a+7b=-4$ ----- $\left(4\right)$

Adding $\left(3\right)$ and $2\times \left(2\right)$, we get

$8a-4b=12$ ---- $\left(5\right)$

Multiply $\left(4\right)$ by $8$ and $\left(5\right)$ by $3$.

$24a+56b=-32$ ---- $\left(6\right)$

$24a-12b=36$ ---- $\left(7\right)$

Subtracting $\left(7\right)$ from $\left(6\right)$, we get

$68b=-68$

$b=-1$

Putting, $b=-1$ in $\left(5\right)$

$a=1$

Putting $a=1,b=-1$ in $\left(1\right)$

$c=2$

Therefore,

$a=1,b=-1,c=2$