Matrix Multiplication Confusioned

Question

Find a 2×2 matrix such that:

$$ \begin{pmatrix} 0&-6\\ 1&-5\\ \end{pmatrix} \begin{pmatrix} x&y\\ z&w\\ \end{pmatrix} = \begin{pmatrix} 1&0\\ 0&1\\ \end{pmatrix} $$

---

I tried to multiply above the vectors on the left and solving for each variable but it is not the right answer when I checked.

Answer 1

$$ \begin{pmatrix} 0&-6\\ 1&-5\\ \end{pmatrix} \begin{pmatrix} x&y\\ z&w\\ \end{pmatrix} = \begin{pmatrix} -6z& -6w \\ x-5z&y-5w \end{pmatrix} $$

So we need,

$-6z=1$,

$z=-(1/6)$ ,

$w=0$ ,

$x-5z=0$ ,

$x=-(5/6)$

$y=1$

Answer 2

Multiplying the left-hand side matrices gives: \begin{pmatrix} -6z&-6w\\ x-5z&y-5w\\ \end{pmatrix}

Can you go from here?

Answer 3

After multiplying we find that

$$\begin{pmatrix}-6z&-6w\\ {x-5z}&{y-5w} \end{pmatrix}=\begin{pmatrix} 1&0 \\0&1 \end{pmatrix}$$

Which then makes the matrix easy to solve for each respective term.

(Solving from easiest to hardest)

$$z= {-1\over{6}}\\w=0\\y=1\\x={5\over{6}}$$

hope this helps!

Answer 4

Alternatively:

$A\cdot A^{-1}=I$ and $A^{-1}=\begin{pmatrix} a_{11}& a_{12}\\ a_{21}& a_{22}\\ \end{pmatrix}^{-1}=\frac{1}{a_{11}a_{22}-a_{12}a_{21}}\begin{pmatrix} a_{22}& -a_{12}\\ -a_{21}&a_{11}\\ \end{pmatrix}$.

So we find the inverse of the matrix: $$\begin{pmatrix} x& y\\ z& w\\ \end{pmatrix}=\begin{pmatrix} 0&-6\\ 1&-5\\ \end{pmatrix}^{-1}=\frac{1}{6}\begin{pmatrix} -5&6\\ -1&0\\ \end{pmatrix}=\begin{pmatrix} -\frac56&1\\ -\frac16&0\\ \end{pmatrix}$$