- #1
brinlin
- 12
- 0
Click For 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.
- #2
skeeter
- 1,103
- 1
$
\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)
\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)
- #3
HOI
- 921
- 2
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]
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]
For the sake of argument, lets assume this is, in fact, a right triangle with a rectangle inscribed in it.
Here is what looks like a very elegant solution:
Triangles A and B are similar
a/4=2/b
ab=8
But what is happening in that second line? Is there a property of similar triangles I'm forgetting?
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