Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I posted a similar question recently but I still have problem with this problem and would appreciate any help!

$$\left[ \begin{array}{cc} 9 & -3\\ 5 & -5\end{array} \right] - X \left[ \begin{array}{cc} -9 & -2\\ 8 & 5\end{array} \right] = E$$ With $E$ i pressume they mean the identity matrix $\left[ \begin{array}{cc} 1 & 0\\ 0 & 1\end{array} \right]$.

How should I go on and solve this for the $2\times2$ matrix $X$? a full development so I can follow your solution would be very much appreciated!

Thank you kindly for you help!

share|improve this question
    
You should specify in your question whether $X$ is a scalar or a matrix. –  user1551 Mar 15 '13 at 8:55
    
Sorry for that. $X$ is a $2\times2$ matrix. –  Lukas Arvidsson Mar 15 '13 at 8:58

3 Answers 3

up vote 2 down vote accepted

I will express it as equations, as I think that is easier (at least for beginners).
With $$X=\begin{pmatrix} x_{11} & x_{12}\\ x_{21} & x_{22} \\ \end{pmatrix}$$ At first we make the multiplication \begin{align*} X \cdot \begin{pmatrix} -9 & -2\\ 8 & 3 \end{pmatrix}&= \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} \cdot \begin{pmatrix} -9 & -2 \\ 8 & 3 \end{pmatrix}\\ &= \begin{pmatrix} -9 x_{11} +8 x_{12} & -2x_{11}+ 3 x_{12}\\ -9 x_{21} + 8 x_{22} & -2 x_{21} + 3 x_{22}\\ \end{pmatrix} \end{align*} So our equation is $$\begin{pmatrix} 9 & -3 \\ 5 & -5 \end{pmatrix} - \begin{pmatrix} -9 x_{11} +8 x_{12} & -2x_{11}+ 3 x_{12}\\ -9 x_{21} + 8 x_{22} & -2 x_{21} + 3 x_{22}\\ \end{pmatrix}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$$ Now we write it as a system of equations: \begin{align*} 1&=9- (-9 x_{11} +8x_{12})\\ 0&= -3-(-2x_{11} +3x_{12}) \\ 0&=5-(-9x_{21} +8 x_{21}) \\ 1&=-5 - (-2x_{21}+3x_{22}) \\ \end{align*} If you need help solving this system tell me.

share|improve this answer
    
Thank you very much for your excellent answer! –  Lukas Arvidsson Mar 15 '13 at 9:27
    
You are welcome :) –  Dominic Michaelis Mar 15 '13 at 9:28

First of all rearrange to get X[-9 -2, 8 5] = [8 -3, 5 -6]. Then find the inverse of [-9 -2, 8 5], and multiply both sides of the equation by this inverse on the right, which will leave X = [8 -3, 5 -6][-9 -2, 8 5]^-1 as your solution.

share|improve this answer

Denote $B=\begin{pmatrix}-9&-2\\8&5\end{pmatrix}$. Then we have: $$\begin{pmatrix}9&-3\\5&-5\end{pmatrix}-E=XB$$ Observe that $\det B=-45+16\neq0$ and hence $B$ is invertible. Multiplying by $B^{-1}$ we have: $$X=\left(\begin{pmatrix}9&-3\\5&-5\end{pmatrix}-E\right)B^{-1}$$ Do you know how to find $B^{-1}$?

share|improve this answer
    
Thanks for your answer! I get the final result to be: $\left[ \begin{array}{cc} \frac{-64}{29} & \frac{-43}{29}\\ \frac{-48}{29} & \frac{-54}{29}\end{array} \right]$ Is that correct? –  Lukas Arvidsson Mar 15 '13 at 9:13
    
Appearantly, the two above is correct but not the two below... –  Lukas Arvidsson Mar 15 '13 at 9:16
    
Yes, you have a computation error somewhere. You should get $X=\frac1{29}\begin{pmatrix}-64&-43\\-73&-64\end{pmatrix}$ –  Dennis Gulko Mar 15 '13 at 9:19
    
Yes that is correct. Thank you for your help! –  Lukas Arvidsson Mar 15 '13 at 9:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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