# Matrices

• December 18th 2008, 05:58 AM
Joker37
Matrices
Let A=
[ 2 3 ]
[ 4 5 ],

B=
[ 5 5 ]
[-2 1 ],

C=
[ 0 -1]
[ 6 6 ]

Find X if: A−1BX= C

NOTE: A is one matrix , B is one and C is also one matrix. Sorry I don't know how to use latex. Any help will be appreciated. The 'A−1' in A−1BX= C is supposed to mean the multiplicative inverse of A.

1/15[-132 -144]
----[ 186 162 ]

Note that I have also put "----" at left of the matrix to make the bottom numbers align the top ones. They do not mean anything.

I want to see any working out. Thank you
• December 18th 2008, 06:32 AM
Mush
Quote:

Originally Posted by Joker37
Let A=
[ 2 3 ]
[ 4 5 ],

B=
[ 5 5 ]
[-2 1 ],

C=
[ 0 -1]
[ 6 6 ]

Find X if: A−1BX= C

NOTE: A is one matrix , B is one and C is also one matrix. Sorry I don't know how to use latex. Any help will be appreciated. The 'A−1' in A−1BX= C is supposed to mean the multiplicative inverse of A.

Mutliply both sides by A, giving:

$A \times A^{-1} \times B \times X = C \times A$

Which gives:

$B \times X = C \times A$

Multiply both sides by the inverse of B

$B^{-1} \times B \times X = C \times A \times B^{-1}$

giving:

$X = C \times A \times B^{-1}$

And note that the inverse of matrix B is given by $B^{-1} = \frac{Adj(B)}{Det(B)}$

PS: for a 2x2 matrix:

[a b]
[c d]

[d -b]
[-c a]

And you should know that the determinant of said matrix is:

• December 18th 2008, 06:57 AM
Joker37
Sorry, I still don't understand how to get the answer.
• December 18th 2008, 07:36 AM
Soroban
Hello, Joker37!

Quote:

Let: . $A\:=\:\begin{bmatrix}2&3\\4&5\end{bmatrix}\quad B \:=\:\begin{bmatrix}5&5\\\text{-}2&1\end{bmatrix} \quad C \:=\:\begin{bmatrix}0&\text{-}1\\6&6\end{bmatrix}$

Given: . $A^{-1}BX \:=\:C.\qquad \text{Find X.}$

Answer: . $\tfrac{1}{15}\begin{bmatrix}\text{-}132 & \text{-}144 \\ 186 & 162 \end{bmatrix}$ . . . . Why didn't they reduce?

If you know the Fast Way to find a 2-by-2 inverse, it certainly saves time.

Given: . $M \:=\:\begin{bmatrix}a & b \\ c & d \end{bmatrix}$ . the inverse is: . $M^{-1} \;=\;\frac{1}{ad-bc}\begin{bmatrix}d & \text{-}b \\ \text{-}c & a\end{bmatrix}$

In baby-talk:
. . (1) Switch the two elements on the main diagonal,
. . (2) Change the signs of the elements on the other diagonal,
. . (3) Divide by the determinant of the matrix.

We have: . $A \:=\:\begin{bmatrix}2&3\\4&5\end{bmatrix}$
. . The determinant is: . $\begin{vmatrix}2&3\\4&5\end{vmatrix} \:=\:(2)(5)-(3)(4) \:=\:-2$
. . The inverse is: . $A^{-1} \;=\;\text{-}\tfrac{1}{2}\begin{bmatrix}5&\text{-}3\\\text{-}4&2\end{bmatrix}\;=\;\begin{bmatrix}\text{-}\frac{5}{2} & \frac{3}{2} \\ 2 & \text{-}1 \end{bmatrix}$

The equation is: . $\overbrace{\begin{bmatrix}\text{-}\frac{5}{2} & \frac{3}{2}\\2&\text{-}1\end{bmatrix}}^{A^{-1}}\cdot\overbrace{\begin{bmatrix}5&5\\\text{-}2&1\end{bmatrix}}^B\cdot X \;=\;\overbrace{\begin{bmatrix}0&\text{-}1\\6&6\end{bmatrix}}^C$

. . . . . . . . . . . . . . . $\underbrace{\begin{bmatrix}\text{-}\frac{31}{2}&\text{-}11\\ 12 & 9\end{bmatrix}}_D\cdot X \;=\;\begin{bmatrix}0 & \text{-}1 \\ 6&6 \end{bmatrix}\;\;{\color{blue}[1]}$

Then we find that: . $D^{-1} \:=\:\begin{bmatrix}\text{-}\frac{6}{5} & \text{-}\frac{22}{15} \\ \\[-4mm] \frac{8}{5} & \frac{31}{15}\end{bmatrix}$

Left-multiply both side of [1] by $D^{-1}$

. . $\begin{bmatrix}\text{-}\frac{6}{5}&\text{-}\frac{22}{15} \\ \frac{8}{5}&\frac{31}{15}\end{bmatrix}\cdot\begin{ bmatrix}\text{-}\frac{31}{2} & \text{-}11 \\ 12 & 9 \end{bmatrix}\cdot X \;=\;\begin{bmatrix}\text{-}\frac{6}{5} & \text{-}\frac{22}{15} \\ \frac{8}{5} & \frac{31}{5}\end{bmatrix}\cdot \begin{bmatrix}0&\text{-}1\\6&6\end{bmatrix}$

. . . . . . . . . . . . . . . $X \;=\;\begin{bmatrix}\text{-}\frac{44}{5} & \text{-}\frac{38}{5} \\ \\[-4mm] \frac{62}{5} & \frac{54}{5} \end{bmatrix} \quad\hdots$ . . . There!

• December 18th 2008, 09:41 AM
Joker37
Quote:

Originally Posted by Soroban

Then we find that: . $D^{-1} \:=\:\begin{bmatrix}\text{-}\frac{6}{5} & \text{-}\frac{22}{15} \\ \\[-4mm] \frac{8}{5} & \frac{31}{15}\end{bmatrix}$

How did you get this? It looks strange with all the fractions.
• December 18th 2008, 01:38 PM
Soroban
Hello, Joker37!

Use the Fast Way that I explained . . . or find it yourself.