2

$$\begin{bmatrix}3 & -2\\-1 & -3\end{bmatrix}X + \begin{bmatrix}1 & 5\\0 & 7\end{bmatrix} = \begin{bmatrix}-5 & 8\\-1 & 3\end{bmatrix}X$$

Not sure how I am supposed to go at this ? Do I move the + to the other side first ? Can someone let me know how to tackle this, thanks.

Reza M.
  • 311
  • 1
    Yes, you could use a bit of matrix algebra to first write the equation in the form $AX=B$. (Think of how you'd solve for $x$ in the scalar equation $ax+b=cx$.) – David Mitra Feb 04 '13 at 13:07

2 Answers2

4

$$AX+B=CX\implies X=(C-A)^{-1}\cdot B\quad (\text{provided $C-A$ is invertible})$$

Did
  • 279,727
4

Taking the $x$ to LHS, we get $$\begin{pmatrix}3+5&-2-8\\-1+1&-3-3\end{pmatrix}X=-\begin{pmatrix}1&5\\0&7\end{pmatrix}$$ $$\Rightarrow \begin{pmatrix}8&-10\\0&-6\end{pmatrix}X=-\begin{pmatrix}1&5\\0&7\end{pmatrix}$$

Now note that $\begin{pmatrix}8&-10\\0&-6\end{pmatrix}$ is invertible and let A be its inverse. (To find the inverse you may look here). So premultiplying with $A$, we get $$X=-A\begin{pmatrix}1&5\\0&7\end{pmatrix}$$

Community
  • 1
Tapu
  • 3,496
  • The question I linked was by you...so you know the trick to invert a $2\times 2$ matrix. – Tapu Feb 04 '13 at 13:17