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$$ \begin{bmatrix} -5 & -9 \\ -6 & -2 \\ \end{bmatrix} X + \begin{bmatrix} -3 & 2 \\ -9 & -3 \\ \end{bmatrix} = \begin{bmatrix} -9 & -7 \\ -2 & -3 \\ \end{bmatrix} X $$

I am asked to solve for matrix X

I combine the x to the right side and get $$ \begin{bmatrix} -14 & -16 \\ -8 & -6 \\ \end{bmatrix} X $$

Then I move the other matrix to the right and get

$$ \begin{bmatrix} -14 & -16 \\ -8 & -6 \\ \end{bmatrix} X = \begin{bmatrix} 3 & -2 \\ 9 & 3 \\ \end{bmatrix} $$

I then proceed to find the inverse of the x coefficient matrix which comes out to be \begin{bmatrix} 6/44 & -16/44 \\ -8/44 & 14/44 \\ \end{bmatrix}

I know this is the correct inverse because when I multiply it by the original x coefficient matrix I get the identity matrix.

Now as far as I'm aware, all that's left is multiplying the inverse by the right side, but when I do that, my homework system tells me that I'm wrong. Where is the mistake? This is the answer that I get \begin{bmatrix} -63/22 & -15/11 \\ 51/22 & 29/22 \\ \end{bmatrix}

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  • $\begingroup$ when you "move" something to the other side it goes there with minus like $ -5x+-3=-9x$ is equivalent to $-3 = -4x $ not $ -3=-14x $ $\endgroup$ – jack Feb 18 '16 at 13:52
  • $\begingroup$ @Gyfe Is x a scalar or a matrix? $\endgroup$ – SchrodingersCat Feb 18 '16 at 14:03
  • $\begingroup$ It's a matrix, sorry for confusion $\endgroup$ – Allan Feb 18 '16 at 14:07
  • $\begingroup$ write X for a matrix to clarify this confusion $\endgroup$ – Pieter21 Feb 18 '16 at 14:19
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You have a sign error, as

$$ \left( \begin{pmatrix} -5 & -9 \\ -6 & -2 \end{pmatrix} - \begin{pmatrix} -9 & -7 \\ -2 & -3 \end{pmatrix} \right) X = \begin{pmatrix} 4 & -2 \\ -4 & 1 \end{pmatrix}X = \begin{pmatrix} 3 & -2 \\ 9 & 3 \end{pmatrix}. $$

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  • $\begingroup$ If I multiply the inverse by your right side, I still get the wrong answer $\endgroup$ – Allan Feb 18 '16 at 14:07
  • $\begingroup$ You need to multiply by the inverse on the left. That is, $x = \begin{pmatrix} 4 & -2 \\ -4 & 1 \end{pmatrix}^{-1} \begin{pmatrix} -3 & 2 \\ -9 & -3 \end{pmatrix}.$ $\endgroup$ – levap Feb 18 '16 at 14:18
  • $\begingroup$ That was the problem, I was using the wrong matrix for the inverse, however your right matrix is incorrect. $$ \begin{bmatrix} 3 & -2 \\ 9 & 3 \\ \end{bmatrix} $$ Using that matrix gives the correct answer $\endgroup$ – Allan Feb 18 '16 at 14:33
  • $\begingroup$ My right matrix is correct assuming the question you wrote is correct. If you meant to solve $$ \begin{bmatrix} -5 & -9 \\ -6 & -2 \\ \end{bmatrix} X + \begin{bmatrix} -3 & 2 \\ -9 & -3 \\ \end{bmatrix} = \begin{bmatrix} -9 & -7 \\ -2 & -3 \\ \end{bmatrix} X $$ (note the plus sign in the left side) then the right matrix should be negated. $\endgroup$ – levap Feb 18 '16 at 14:42
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    $\begingroup$ @Gyfe: That is correct... but in your original question, you asked about $$ \begin{bmatrix} -5 & -9 \\ -6 & -2 \\ \end{bmatrix} X - \begin{bmatrix} -3 & 2 \\ -9 & -3 \\ \end{bmatrix} = \begin{bmatrix} -9 & -7 \\ -2 & -3 \\ \end{bmatrix} X $$ which means that the matrix $\begin{pmatrix} -3 & 2 \\ -9 & -3 \end{pmatrix}$ should be added and not subtracted from both sides of the equation... $\endgroup$ – levap Feb 18 '16 at 14:53
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The +/- signs are wrong on various places. RHS should just be the matrix displayed, because it already has - sign.

Also with the matrix before x you made a sign error

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You have made a few mistakes in signs while reversing the sides.

You should have approached the problem as follows:

$$ \begin{bmatrix} -5 & -9 \\ -6 & -2 \\ \end{bmatrix} x - \begin{bmatrix} -3 & 2 \\ -9 & -3 \\ \end{bmatrix} = \begin{bmatrix} -9 & -7 \\ -2 & -3 \\ \end{bmatrix} x $$

$$ \begin{bmatrix} -5x & -9x \\ -6x & -2x \\ \end{bmatrix} - \begin{bmatrix} -3 & 2 \\ -9 & -3 \\ \end{bmatrix} = \begin{bmatrix} -9x & -7x \\ -2x & -3x \\ \end{bmatrix} $$

$$ \begin{bmatrix} -5x+3 & -9x-2 \\ -6x+9 & -2x+3 \\ \end{bmatrix} = \begin{bmatrix} -9x & -7x \\ -2x & -3x \\ \end{bmatrix} $$

By the equality of the matrices, we can say that the corresponding elements are the same.

But there is no $x \in \mathbb{R}$ which satisfies the above equality.

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    $\begingroup$ I was under the impression that OP was looking for a matrix $x$, not a scalar. $\endgroup$ – Roland Feb 18 '16 at 13:58
  • $\begingroup$ @Roland OP did not mention if x is a matrix or a scalar. But I think you are right. $\endgroup$ – SchrodingersCat Feb 18 '16 at 14:03
  • $\begingroup$ There is no $x \in \mathbb R$ that satisfies the OP equation; it certainly is not the case for $x=-3/4$. $\endgroup$ – Zoran Loncarevic Feb 18 '16 at 14:03
  • $\begingroup$ @ZoranLoncarevic Corrected. $\endgroup$ – SchrodingersCat Feb 18 '16 at 14:04
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You have formally $$ AX +B = CX$$ You should proceed as follows $$ \begin{array}{rll} AX + B &=& CX \\ B &=& CX - AX \\ B &=& (C- A)X \\ (C-A)^{-1} B &=& (C-A)^{-1} (C- A)X \\ (C-A)^{-1} B &=& X \\ \end{array} $$ Now $$ C-A = -\begin{bmatrix} 9 & 7 \\ 2 &3 \end{bmatrix} - (-\begin{bmatrix} 5 & 9 \\ 6 &2 \end{bmatrix}) = \begin{bmatrix} -4 & 2 \\ 4 & -1 \end{bmatrix} $$ Then $$ (C-A)^{-1} = -\begin{bmatrix} 1/4 & 1/2 \\ 1 &1 \end{bmatrix} $$
and by $ (C-A)^{-1} B=X$ the result

$$ (C-A)^{-1} B = - \begin{bmatrix} 21/4 & 1 \\ 12 &1 \end{bmatrix} = X $$

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