# Matrix multiplication/addition problem

• April 11th 2011, 05:15 PM
cb220
Hi, I need to solve for a matrix X in the equation below:

$\left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]X + \left[ \begin{array}{cc} -2 & 1 \\ 4 & 2 \end{array} \right] = \left[ \begin{array}{cc} -8 & 4 \\ 5 & 3 \end{array} \right]$

What I tried to do was first minus the $\left[ \begin{array}{cc} -2 & 1 \\ 4 & 2 \end{array} \right]$ to the other side, then I took the inverse of $\left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]$ and multiplied that on both sides. I thought that taking the inverse and multiplying with it would have a division type of effect. Unfortunately I didn't get the correct answer. Would anyone mind telling me what I've done wrong?

Thanks!

• April 11th 2011, 05:17 PM
Chris L T521
Quote:

Originally Posted by cb220
Hi, I need to solve for a matrix X in the equation below:

$\left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]X + \left[ \begin{array}{cc} -2 & 1 \\ 4 & 2 \end{array} \right] = \left[ \begin{array}{cc} -8 & 4 \\ 5 & 3 \end{array} \right]$

What I tried to do was first minus the $\left[ \begin{array}{cc} -2 & 1 \\ 4 & 2 \end{array} \right]$ to the other side, then I took the inverse of $\left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]$ and multiplied that on both sides. I thought that taking the inverse and multiplying with it would have a division type of effect. Unfortunately I didn't get the correct answer. Would anyone mind telling me what I've done wrong?

Thanks!

What did you get for the inverse of $\begin{bmatrix}-7 & -6\\ -9 & 5\end{bmatrix}$ ??
• April 11th 2011, 05:19 PM
cb220
I got:

$\begin{bmatrix}-5/19 & -6/19\\ -9/19 & -7/19\end{bmatrix}$
• April 11th 2011, 05:26 PM
Chris L T521
Quote:

Originally Posted by cb220
I got:

$\begin{bmatrix}-5/19 & -6/19\\ -9/19 & -7/19\end{bmatrix}$

I don't get those values...(Worried)

I get $\begin{bmatrix}-5/89 & -6/89\\ -9/89 & 7/89\end{bmatrix}$ for the inverse matrix.
• April 11th 2011, 05:35 PM
cb220
Hmm, throwing it into a matrix calculator online seems to give these values:

-0.263157894736842 -0.315789473684210
-0.473684210526316 -0.368421052631579

Those are consistent with my values, unfortunately after trying to solve the problem with both my own values and your values, I still got the answer wrong in both cases :(.

I think I must be missing something here *frustration*!
• April 11th 2011, 06:26 PM
topsquark
Quote:

Originally Posted by cb220
Hmm, throwing it into a matrix calculator online seems to give these values:

-0.263157894736842 -0.315789473684210
-0.473684210526316 -0.368421052631579

Those are consistent with my values, unfortunately after trying to solve the problem with both my own values and your values, I still got the answer wrong in both cases :(.

I think I must be missing something here *frustration*!

Why don't you show us exactly what you have right now, step by step. Then we can pick out any mistakes that you might be making.

-Dan
• April 11th 2011, 06:38 PM
cb220
Okay, starting with:

$\left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]X + \left[ \begin{array}{cc} -2 & 1 \\ 4 & 2 \end{array} \right] = \left[ \begin{array}{cc} -8 & 4 \\ 5 & 3 \end{array} \right]$

Moving over the second matrix on the left side to the right side:

$\left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]X = \left[ \begin{array}{cc} -8 & 4 \\ 5 & 3 \end{array} \right] - \left[ \begin{array}{cc} -2 & 1 \\ 4 & 2 \end{array} \right]$

Should simplify to:

$\left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]X = \left[ \begin{array}{cc} -6 & 3 \\ 1 & 1 \end{array} \right]$

Then I took the inverse of the matrix attached to X and multiplied both sides by it, I'll use the solution that I arrived at for the inverse, though I don't know whether it is right or not at this point:

$X = \left[ \begin{array}{cc} -6 & 3 \\ 1 & 1 \end{array} \right] * \left[ \begin{array}{cc} -5/19 & -6/19 \\ -9/19 & -7/19 \end{array} \right]$

And my final answer for X came out to be:

$X = \left[ \begin{array}{cc} -3/19 & 15/19 \\ -14/19 & -13/19 \end{array} \right]$

If you want me to expand on any of what I've done let me know.

EDIT: Made a correction in my typing up of the math.
• April 11th 2011, 06:54 PM
Deveno
Quote:

Originally Posted by cb220
Hi, I need to solve for a matrix X in the equation below:

$\left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]X + \left[ \begin{array}{cc} -2 & 1 \\ 4 & 2 \end{array} \right] = \left[ \begin{array}{cc} -8 & 4 \\ 5 & 3 \end{array} \right]$

What I tried to do was first minus the $\left[ \begin{array}{cc} -2 & 1 \\ 4 & 2 \end{array} \right]$ to the other side, then I took the inverse of $\left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]$ and multiplied that on both sides. I thought that taking the inverse and multiplying with it would have a division type of effect. Unfortunately I didn't get the correct answer. Would anyone mind telling me what I've done wrong?

Thanks!

i get, as a first step:

$\left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]X = \left[ \begin{array}{cc} -6 & 3 \\ 1 & 1 \end{array} \right]$

i concur with Chris L that the inverse of the first matrix is:

$\begin{bmatrix}-5/89 & -6/89\\ -9/89 & 7/89\end{bmatrix}$

applying this to the RHS after the first step checks out in the original equation. i suggest you double-check your arithmetic, i made several errors verifyng this myself.

one more thing you need to be aware of: matrix multiplcation is NOT commutative.

if AX = B, then X = A^-1B, NOT BA^-1.

the determinant of your original matrix (the one you inverted) is (-7)(5) - (-6)(-9) = -35 - 54 = -89. by a well-known rule, to invert a 2x2 matrix, you swap the upper-left and lower-right corners, and change the sign of the other two entries (leaving them where they are), and then divide each one by the determinant.
• April 11th 2011, 07:03 PM
cb220
Ahh, I see now. I messed up on the inversion and somehow managed to even punch it into the calculator in such a way that produced the same wrong results. Also, I messed up with the order of multiplication and did BA^-1. Thanks for the help guys!