- #1
brinlin
- 13
- 0
In summary, the conversation discusses one method for finding an inverse matrix by using row operations to reduce the given matrix to the identity matrix while applying the same operations to the identity matrix. The steps involved in this method include swapping rows, dividing rows, and adding or subtracting rows to obtain the identity matrix. The end result is an inverse matrix that can be used to solve for the variable matrix in a system of linear equations.
- #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:
\(\displaystyle \begin{bmatrix}0 & -2 & 2 \\ 3 & 1 & 3 \\ 1 & -3 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\).
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:
\(\displaystyle \begin{bmatrix}1 & -3 & 1 \\ 3 & 1 & 3 \\ 0 & -2 & 2 \end{bmatrix}\begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{bmatrix}\)
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:
\(\displaystyle \begin{bmatrix}1 & -3 & 1 \\ 0 & 10 & 0 \\ 0 & -2 & 2 \end{bmatrix}\begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & -1 \\ 1 & 0 & 0 \end{bmatrix}\)
Divide the second row by 10:
\(\displaystyle \begin{bmatrix}1 & -3 & 1 \\ 0 & 1 & 0 \\ 0 & -2 & 2 \end{bmatrix}\begin{bmatrix}0 & 0 & 1 \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ 1 & 0 & 0 \end{bmatrix}\).
Add 3 times the second row to the first row and add 2 times the second row to the third row:
\(\displaystyle \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}\begin{bmatrix}0 & \frac{3}{10} & \frac{7}{10} \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ 1 & \frac{2}{10} & \frac{8}{10} \end{bmatrix}\)
Divide the third row by 2:
\(\displaystyle \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}0 & \frac{3}{10} & \frac{7}{10} \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ \frac{1}{2} & \frac{1}{10} & \frac{4}{10} \end{bmatrix}\).
Finally, subtract the third row from the first row:
\(\displaystyle \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}0 & \frac{3}{10} & \frac{3}{10} \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ \frac{1}{2} & \frac{1}{10} & \frac{4}{10} \end{bmatrix}\)
IF I have done everything correctly, \(\displaystyle \begin{bmatrix}0 & \frac{3}{10} & \frac{3}{10} \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ \frac{1}{2} & \frac{1}{10} & \frac{4}{10} \end{bmatrix}\) is the inverse matrix to \(\displaystyle \begin{bmatrix}0 & -2 & 2 \\ 3 & 1 & 3 \\ 1 & -3 & 1 \end{bmatrix}\)
write the matrix and identity matrix next to each other:
\(\displaystyle \begin{bmatrix}0 & -2 & 2 \\ 3 & 1 & 3 \\ 1 & -3 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\).
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:
\(\displaystyle \begin{bmatrix}1 & -3 & 1 \\ 3 & 1 & 3 \\ 0 & -2 & 2 \end{bmatrix}\begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{bmatrix}\)
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:
\(\displaystyle \begin{bmatrix}1 & -3 & 1 \\ 0 & 10 & 0 \\ 0 & -2 & 2 \end{bmatrix}\begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & -1 \\ 1 & 0 & 0 \end{bmatrix}\)
Divide the second row by 10:
\(\displaystyle \begin{bmatrix}1 & -3 & 1 \\ 0 & 1 & 0 \\ 0 & -2 & 2 \end{bmatrix}\begin{bmatrix}0 & 0 & 1 \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ 1 & 0 & 0 \end{bmatrix}\).
Add 3 times the second row to the first row and add 2 times the second row to the third row:
\(\displaystyle \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}\begin{bmatrix}0 & \frac{3}{10} & \frac{7}{10} \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ 1 & \frac{2}{10} & \frac{8}{10} \end{bmatrix}\)
Divide the third row by 2:
\(\displaystyle \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}0 & \frac{3}{10} & \frac{7}{10} \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ \frac{1}{2} & \frac{1}{10} & \frac{4}{10} \end{bmatrix}\).
Finally, subtract the third row from the first row:
\(\displaystyle \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}0 & \frac{3}{10} & \frac{3}{10} \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ \frac{1}{2} & \frac{1}{10} & \frac{4}{10} \end{bmatrix}\)
IF I have done everything correctly, \(\displaystyle \begin{bmatrix}0 & \frac{3}{10} & \frac{3}{10} \\ 0 & \frac{1}{10} & -\frac{1}{10} \\ \frac{1}{2} & \frac{1}{10} & \frac{4}{10} \end{bmatrix}\) is the inverse matrix to \(\displaystyle \begin{bmatrix}0 & -2 & 2 \\ 3 & 1 & 3 \\ 1 & -3 & 1 \end{bmatrix}\)
FAQ: Solve System Using Matrix Inverse
What is a matrix inverse?
A matrix inverse is a mathematical operation that involves finding the reciprocal of a matrix. It is denoted by A-1 and is used to solve equations involving matrices.
How do you find the inverse of a matrix?
To find the inverse of a matrix, you need to follow a specific set of steps. First, find the determinant of the matrix. If the determinant is non-zero, then the matrix has an inverse. Next, find the adjugate of the matrix by finding the transpose of the matrix of cofactors. Finally, divide the adjugate by the determinant to get the inverse matrix.
What is the purpose of using matrix inverse to solve systems?
The purpose of using matrix inverse to solve systems is to simplify the process of solving equations involving multiple variables. It allows us to solve systems of equations in a more efficient and accurate manner.
Can all matrices be inverted?
No, not all matrices can be inverted. A matrix must have a non-zero determinant in order to have an inverse. Matrices with a determinant of zero are known as singular matrices and do not have an inverse.
Are there any limitations to using matrix inverse to solve systems?
Yes, there are some limitations to using matrix inverse to solve systems. For example, if the matrix is very large, finding the inverse can be computationally intensive and time-consuming. Additionally, if the matrix is ill-conditioned, the inverse may not be accurate and can lead to errors in the solution.