MHB Solve System Using Matrix Inverse

AI Thread Summary
The discussion focuses on finding the inverse of a given 3x3 matrix using row operations. The initial matrix is transformed into the identity matrix while applying the same operations to an adjacent identity matrix. Key steps include swapping rows, performing row reductions, and scaling rows to achieve the identity form. The final result indicates that the calculated inverse matrix is correct. The process demonstrates a systematic approach to matrix inversion through elementary row operations.
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$
\begin{bmatrix}
0 &-2 &2 \\
3 & 1 &3 \\
1 &-2 &3
\end{bmatrix} \cdot\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=\begin{bmatrix}
12\\
-2\\
8
\end{bmatrix}$

follow the directions for (b) and (c)
 
There are many ways to find an inverse matrix. One I like
write the matrix and identity matrix next to each other:
[math]\begin{bmatrix}0 & -2 & 2 \\ 3 & 1 & 3 \\ 1 & -3 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}[/math].

Now use "row operations" to reduce the matrix to the identity matrix while applying the same row operation to the identity matrix.

Normally the first thing you would do is divide every number in the first row by the leftmost number in that row. But here, that is 0 so instead swap the first and third rows:
[math]\begin{bmatrix}1 & -3 & 1 \\ 3 & 1 & 3 \\ 0 & -2 & 2 \end{bmatrix}\begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{bmatrix}[/math]

Now there is already a 1 in the upper left so all we need to do to get the right first column is subtract 3 times the first row from the second row:
[math]\begin{bmatrix}1 & -3 & 1 \\ 0 & 10 & 0 \\ 0 & -2 & 2 \end{bmatrix}\begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & -1 \\ 1 & 0 & 0 \end{bmatrix}[/math]

Divide the second row by 10:
[math]\begin{bmatrix}1 & -3 & 1 \\ 0 & 1 & 0 \\ 0 & -2 & 2 \end{bmatrix}\begin{bmatrix}0 & 0 & 1 \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ 1 & 0 & 0 \end{bmatrix}[/math].

Add 3 times the second row to the first row and add 2 times the second row to the third row:
[math]\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}\begin{bmatrix}0 & \frac{3}{10} & \frac{7}{10} \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ 1 & \frac{2}{10} & \frac{8}{10} \end{bmatrix}[/math]

Divide the third row by 2:
[math]\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}0 & \frac{3}{10} & \frac{7}{10} \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ \frac{1}{2} & \frac{1}{10} & \frac{4}{10} \end{bmatrix}[/math].

Finally, subtract the third row from the first row:
[math]\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}0 & \frac{3}{10} & \frac{3}{10} \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ \frac{1}{2} & \frac{1}{10} & \frac{4}{10} \end{bmatrix}[/math]

IF I have done everything correctly, [math] \begin{bmatrix}0 & \frac{3}{10} & \frac{3}{10} \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ \frac{1}{2} & \frac{1}{10} & \frac{4}{10} \end{bmatrix}[/math] is the inverse matrix to [math]\begin{bmatrix}0 & -2 & 2 \\ 3 & 1 & 3 \\ 1 & -3 & 1 \end{bmatrix}[/math]
 
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Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
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