0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What is your answer, and what were your steps? $\endgroup$ – ultrainstinct Oct 9 '17 at 3:52
  • $\begingroup$ Seems like you just multiply and solve for the variables no? $\endgroup$ – I'mAnAccountantIKnowAlotOfMath Oct 9 '17 at 3:53
2
$\begingroup$

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

So we need,

$-6z=1$,

$z=-(1/6)$ ,

$w=0$ ,

$x-5z=0$ ,

$x=-(5/6)$

$y=1$

$\endgroup$
  • $\begingroup$ Wow I totally did it the right way but got my variables confused (I did it by corresponding the variable with the number in the place of the variable on the right). $\endgroup$ – GDamer Oct 9 '17 at 4:00
1
$\begingroup$

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

Can you go from here?

$\endgroup$
0
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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: $$\begin{pmatrix} x& y\\ z& w\\ \end{pmatrix}=\begin{pmatrix} 0&-6\\ 1&-5\\ \end{pmatrix}^{-1}=\frac{1}{6}\begin{pmatrix} -5&6\\ -1&0\\ \end{pmatrix}=\begin{pmatrix} -\frac56&1\\ -\frac16&0\\ \end{pmatrix}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.