Find two constants that when multiplied by vectors gives you another vector

Question

If we have vector $A = 5{i} -2j $ ,$B = -4i + 3 j $ and $C = 5i - 4 j$. How can I find two constants $h$ and $k$ such that $C = hA+kB$?

Thanks.

Answer 1

To see what is happening just use that $i=(1,0)$ and $j=(0,1)$

$C=(hA_1+kB_1,hA_2+kB_2)$

$A=(A_1,A_2)=(5,-2)$ and $B=(B_1,B_2)=(-4,3)$ and $C=(C_1,C_2)=(5,-4)$

Thus $(5,-4)=(5h-4k,-2h+3k)$

That gives you a system of two equations.

Answer 2

C = 5i - 4j = hA + kB,

5i - 4j = h(5i-2j) + k(-4i+3j)

5i - 4j = (5h-4k)i + (-2h+3k)j

System of two equations

5h -4k = 5

-2h +3k =4

Now you can finish :)

Answer 3

You've been given that $ C = hA + kB$ which in matrix form is:

$ \begin{bmatrix} 5 \\ -4 \end{bmatrix} = h \begin{bmatrix} 5\\ -2 \end{bmatrix} + k \begin{bmatrix} -4\\ 3 \end{bmatrix} $

$ = \begin{bmatrix} 5h\\ -2h \end{bmatrix} + \begin{bmatrix} -4k\\ 3k \end{bmatrix} = \begin{bmatrix} 5h - 4k\\ -2h + 3k \end{bmatrix} = \begin{bmatrix} 5 & -4\\ -2 & 3 \end{bmatrix} \begin{bmatrix} h\\ k \end{bmatrix} $

Let's denote this matrix by $ \begin{bmatrix} 5 & -4\\ -2 & 3 \end{bmatrix} = M $ and its inverse by $M^{-1}.$

$\implies M^{-1}\begin{bmatrix} 5\\ -4 \end{bmatrix} = \begin{bmatrix} h\\ k \end{bmatrix}$

Please tell me you know how to calculate $M^{-1}$ and can solve this now for $h$ and $k$.