Solve for X: 2 x 2 matrices.

[learner123]

Hello everyone,

I am here asking for maths help, the question I have difficult with is question 2 , in my exam practice sheet. Here is the following:

Exam Revision Questions.jpg

Requirement:
Solve for X:

We know we need to use this method , which are two types:

Type 1: AX = B and Type 2: XA = B

But the issue is we must first do operations that will set it up to these two forms.


We know that we must remove the fraction because Matrices cant divided. I would assume we would times both sides by A to remove the fraction. But the problem is getting those two forms. With these operations.


Here is my working out: But i don't believe they're correct.

Workings out.jpg

So my problem is I cant get AX = B or XA = B. So I can solve for X, If you want anymore details please tell me and I hope posted this in the correct section as well , I apologize if I didn't.

Kindest Regards,

 

[soroban]

Hello, learner123!

Your work is excellent . . . with one small error.

\(\displaystyle \text{Given: }\;A \:=\:\begin{bmatrix}1&3\\\text{-}2&1\end{bmatrix} \qquad B \:=\:\begin{bmatrix}4&2\\\text{-}1&3\end{bmatrix} \qquad C \:=\:\begin{bmatrix}2&\text{-}1\\0&3 \end{bmatrix}\)


\(\displaystyle \text{Solve for }X\!:\quad (X + B) - \frac{C}{A} \;=\;0\)

These are your steps . . .

\(\displaystyle AB \;=\;\begin{bmatrix}1&3\\ \text{-}2&1\end{bmatrix}\,\begin{bmatrix}4&2\\\text{-}1&3\end{bmatrix} \;=\; \begin{bmatrix}1&11\\ \text{-}9&\text{-}1 \end{bmatrix}\) . Right!


\(\displaystyle C - AB \;=\;\begin{bmatrix}2&\text{-}1\\0&3\end{bmatrix} - \begin{bmatrix}1&11\\ \text{-}9&\text{-}1\end{bmatrix} \;=\;\begin{bmatrix}1&\text{-}12\\ \boxed{\text{-}9}&4\end{bmatrix}\)
. . . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle \uparrow\)
. . . . . . . . . . . . . . . . . . . . . . . . .no

\(\displaystyle A^{-1} \;=\;\tfrac{1}{7}\begin{bmatrix}1&\text{-}3 \\ 2&1\end{bmatrix} \;=\; \begin{bmatrix}\frac{1}{7} & \text{-}\frac{3}{7} \\ \\[-3mm] \frac{2}{7} & \frac{1}{7}\end{bmatrix}\) . Correct!



\(\displaystyle \text{Try it again . . . You should get: }\;X \;=\;\tfrac{1}{7}\begin{bmatrix}\text{-}26 & \text{-}24 \\ 11 & \text{-}20\end{bmatrix}\)

 

[learner123]

Thank you , that's what i got after. sorry for long reply