# Matrix Multiplication in Z9 and Z101

• Mar 3rd 2011, 06:07 AM
ramdrop
Matrix Multiplication in Z9 and Z101
Hey, I present to you two problems:

First one:

To find the Inverse of the matrix $A = \begin{matrix} 4 & 2\\7 & 4\end{matrix}$ in $\mathbb{Z}$9.

I have worked out the determinent which is equal to 12, but in $\mathbb{Z}$9 that is equal to 3.
I know I have to inverse the determinent in $\mathbb{Z}$9 but I am a little unsure about how to do that. But I'm stuck at the line:
$A.^-1 = \frac{1}{3}\begin{matrix} 4 & -2\\-7 & 4\end{matrix}$ in $\mathbb{Z}$9

Second one:

Given a matrix:
$M = \begin{matrix} 6 & 31 \\ 37 & 95\end{matrix}$ in $\mathbb{Z}$101. work out $M^2$ and find $M^{2*m*n}$ where m = 34 and n = 84.

I have worked out $M^2$ but I'm not sure how to proceed with the other part, I think I must use Euler's formula, but that specifies for two numbers pq that are prime, clearly 34 and 84 are not.

Regards
• Mar 3rd 2011, 06:42 AM
tonio
Quote:

Originally Posted by ramdrop
Hey, I present to you two problems:

First one:

To find the Inverse of the matrix $A = \begin{matrix} 4 & 2\\7 & 4\end{matrix}$ in $\mathbb{Z}$9.

I have worked out the determinent which is equal to 12, but in $\mathbb{Z}$9 that is equal to 3.

Nop. That matrix's determinant is $4\cdot 4-7\cdot 2=2=2\!\!\pmod 9$ ...., and thus:

$\displaystyle{A^{-1}=\frac{1}{2}\begin{pmatrix}4&-2\\-7&4\end{pmatrix}=5\,\begin{pmatrix}4&-2\\-7&4\end{pmatrix}=\begin{pmatrix}2&-1\\1&2\end{pmatrix}$

because $2^{-1}=5\!\!\pmod 9$ .

Tonio

I know I have to inverse the determinent in $\mathbb{Z}$9 but I am a little unsure about how to do that. But I'm stuck at the line:
$A.^-1 = \frac{1}{3}\begin{matrix} 4 & -2\\-7 & 4\end{matrix}$ in $\mathbb{Z}$9

Second one:

Given a matrix:
$M = \begin{matrix} 6 & 31 \\ 37 & 95\end{matrix}$ in $\mathbb{Z}$101. work out $M^2$ and find $M^{2*m*n}$ where m = 34 and n = 84.

I have worked out $M^2$ but I'm not sure how to proceed with the other part, I think I must use Euler's formula, but that specifies for two numbers pq that are prime, clearly 34 and 84 are not.

Regards

.
• Mar 3rd 2011, 06:46 AM
ramdrop
I don't know why I have written 12 here, but I have 2 on my piece of paper, many thanks!
• Mar 3rd 2011, 12:20 PM
Soroban
Hello, ramdrop!

Quote:

$\text{Find the Inverse of the matrix: }\:A \:=\:\begin{pmatrix} 4 & 2\\7 & 4\end{pmatrix}\:\text{ in }\mathbb{Z}9.$

I used the Augmented Matrix approach.

$\text{We have: }\;\left[\begin{array}{cc|cc} 4 & 2 & 1 & 0 \\ 7 & 4 & 0 & 1 \end{array}\right]\;\text{(mod 9)}$

$\begin{array}2R_1 - R_2 \\ \\ \end{array} \left[\begin{array}{cc|cc}
1 & 0 & 2 & \text{-}1 \\ 7 & 4 & 0 & 1 \end{array}{\right] \text{(mod 9)} \;\equiv \;\left[\begin{array}{cc|cc}1 & 0 & 2 & 8 \\ 7 & 4 & 1 & 2\end{array}\right]\text{(mod 9)}$

$\begin{array}{c} \\ R_2-7R_1\end{array} \left[\begin{array}{cc|cc}1 & 0 & 2 & 8 \\ 0 & 4 & \text{-}14 & \text{-}55 \end{array}\right] \text{(mod 9)} \;\equiv \;\left[\begin{array}{cc|cc}1 & 0 & 2 & 8 \\ 0 & 4 & 4 & 8 \end{array}\right]\text{(mod 9)}$

. . $\begin{array}{c} \\ \frac{1}{4}R_2 \end{array} \left[\begin{array}{cc|cc}1 & 0 & 2 & 8 \\ 0 & 1 & 1 & 2\end{array}\right] \text{(mod 9)}$

$\text{Therefore: }\:A^{-1} \;\equiv\;\begin{bmatrix}2 & 8 \\ 1 & 2 \end{bmatrix}\text{(mod 9)}$

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Check

$\begin{bmatrix}4 & 2 \\ 7 & 4 \end{bmatrix}\begin{bmatrix}2 & 8 \\ 1 & 2 \end{bmatrix} \;=\; \begin{bmatrix}10 & 36 \\ 18 & 64\end{bmatrix} \;\equiv\;\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\text{(mod 9)}$

• Mar 4th 2011, 06:51 AM
ramdrop
That works very nicely, but I'm still stuck on the second question.