Basic arithmetic with matrices

Question

We have matrices. $$A=\begin{bmatrix} -2 & 0 \\ -5 & 6 \\ \end{bmatrix} B^{-1}=\begin{bmatrix} -7 & 8 \\ 2 & -8 \\ \end{bmatrix} C=\begin{bmatrix} -15 & -2 \\ -8 & -14 \\ \end{bmatrix} $$ We need to solve matrix $X$ from equation:

$$A^{-1}XB-C=0$$ $$X=AB^{-1}+C$$ $$X=\begin{bmatrix} -2 & 0 \\ -5 & 6 \\ \end{bmatrix} \begin{bmatrix} -7 & 8 \\ 2 & -8 \\ \end{bmatrix} + \begin{bmatrix} -15 & -2 \\ -8 & -14 \\ \end{bmatrix} $$ $$ X=\begin{bmatrix} -1 & -18 \\ 39 & -102 \\ \end{bmatrix} $$ This doesn't seem to be correct solutions to this equations ?

https://www.mathsisfun.com/algebra/matrix-introduction.html may help — , Sep 30 '17 at 18:12
This was just a guess based on $Ax=b$ is same as $X=A^{-1}b$ — Tuki, Sep 30 '17 at 18:13
Since order does matter when you multiply matrices for example $AB \neq BA$. How do you determine order in this. $A^{-1}XB=C$ to $X=ACB^{-1}$ ? — Tuki, Sep 30 '17 at 18:18
Also when $C$ is not initially multiplied with $A$ or $B$ how it is multiplied in this $X=ACB^{-1}$ ? — Tuki, Sep 30 '17 at 18:25

score 2 · Accepted Answer · answered Sep 30 '17 at 18:30

2

As point out by you that the order matters:

Pre-multiply and post-multiply are different.

Starting from $${A^{ - 1}}XB - C = 0$$ Add $C$ on both sides $${A^{ - 1}}XB = C$$ Now post-multiply by $B^{-1}$ on both sides $${A^{ - 1}}X{B}{B^{ - 1}} = C{B^{ - 1}}$$ $${A^{ - 1}}X{I} = C{B^{ - 1}}$$ $${A^{ - 1}}X = C{B^{ - 1}}$$ Now pre-multiply by $A$ on both sides to get the final form $${A}{A^{ - 1}}X = AC{B^{ - 1}}$$ $$X = AC{B^{ - 1}}$$

answered Sep 30 '17 at 18:30

Fahd Siddiqui

362

One thing i didn't understand is that when you multiply $B^-1$ both sides why do you get $A^{-1}XI=CB^{-1}$. specifically what does $I$ stand for ? – Tuki Sep 30 '17 at 18:50
Perhaps indetity matrix meaning $1$ in diagonal and 0 else where ? – Tuki Sep 30 '17 at 18:52
@Tuki Correct, ${I}$ is the identity matrix with ones on its leading diagnoal and zeros as other elements. – Fahd Siddiqui Oct 01 '17 at 21:18

score 1 · Answer 2 · answered Sep 30 '17 at 18:41

The equation

$X=AB^{-1}+C \tag 1$

does not follow from

$A^{-1}XB-C= 0; \tag 2$

instead, we have

$A^{-1}XB = C, \tag 3$

$XB = AC,\tag 4$

$X = ACB^{-1}; \tag 5$

if we now perform the indicated matrix arithmetic we arrive at

$X = \begin{bmatrix} -2 & 0 \\ -5 & 6 \end{bmatrix} \begin{bmatrix} -15 & -2\\-8 & -14 \end{bmatrix} \begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix}$ $= \begin{bmatrix} 30 & 4 \\ 27 & -74 \end{bmatrix}\begin{bmatrix} -7 & 8 \\ 2 & -8 \\ \end{bmatrix} = \begin{bmatrix} -202 & 208\\ -337 & 808 \\ \end{bmatrix}. \tag 6$

As a check of the above, we see that (1) implies

$A^{-1}XB = A^{-1}(AB^{-1}+C)B = A^{-1}AB^{-1}B + A^{-1}CB = I + A^{-1}CB \ne C \tag 7$

in general.

Basic arithmetic with matrices

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