Answer :
Certainly! Let's find the inverse of the given matrices using the Gauss-Jordan method, which involves row operations to transform the given matrix into an identity matrix while simultaneously transforming an identity matrix into the inverse of the given matrix.
### a. Finding the inverse of [tex]\( \left(\begin{array}{cc} 1 & 2 \\ 2 & 5 \end{array}\right) \)[/tex]
1. Augment the given matrix with the identity matrix:
[tex]\[ \left(\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 2 & 5 & 0 & 1 \\ \end{array}\right) \][/tex]
2. Pivot on the first element (1,1):
- The pivot element is already 1, so no changes needed in the first row.
- Make the element below the pivot (2) into 0 by subtracting 2 times the first row from the second row:
[tex]\[ \left( \begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & 1 & -2 & 1 \\ \end{array}\right) \][/tex]
3. Pivot on the second element (2,2):
- Now the element at (2,2) is already 1.
- Make the element above the pivot (2) into 0 by subtracting 2 times the second row from the first row:
[tex]\[ \left( \begin{array}{cc|cc} 1 & 0 & 5 & -2 \\ 0 & 1 & -2 & 1 \\ \end{array}\right) \][/tex]
4. Read off the inverse matrix from the augmented matrix:
[tex]\[ \left(\begin{array}{cc} 5 & -2 \\ -2 & 1 \end{array}\right) \][/tex]
So, the inverse of the matrix [tex]\( \left(\begin{array}{cc} 1 & 2 \\ 2 & 5 \end{array}\right) \)[/tex] is [tex]\( \left(\begin{array}{cc} 5 & -2 \\ -2 & 1 \end{array}\right) \)[/tex].
### b. Finding the inverse of [tex]\( \left(\begin{array}{cc} -3 & -5 \\ 6 & 8 \end{array}\right) \)[/tex]
1. Augment the given matrix with the identity matrix:
[tex]\[ \left( \begin{array}{cc|cc} -3 & -5 & 1 & 0 \\ 6 & 8 & 0 & 1 \\ \end{array}\right) \][/tex]
2. Pivot on the first element (1,1):
- Multiply the first row by [tex]\(-\frac{1}{3}\)[/tex]:
[tex]\[ \left(\begin{array}{cc|cc} 1 & \frac{5}{3} & -\frac{1}{3} & 0 \\ 6 & 8 & 0 & 1 \\ \end{array}\right) \][/tex]
- Make the element below the pivot (6) into 0 by subtracting 6 times the first row from the second row:
[tex]\[ \left(\begin{array}{cc|cc} 1 & \frac{5}{3} & -\frac{1}{3} & 0 \\ 0 & -2 & 2 & 1 \\ \end{array}\right) \][/tex]
3. Pivot on the second element (2,2):
- Multiply the second row by [tex]\(-\frac{1}{2}\)[/tex]:
[tex]\[ \left( \begin{array}{cc|cc} 1 & \frac{5}{3} & -\frac{1}{3} & 0 \\ 0 & 1 & -1 & -\frac{1}{2} \\ \end{array}\right) \][/tex]
- Make the element above the pivot ([tex]\(\frac{5}{3}\)[/tex]) into 0 by subtracting [tex]\(\frac{5}{3}\)[/tex] times the second row from the first row:
[tex]\[ \left( \begin{array}{cc|cc} 1 & 0 & \frac{4}{3} & \frac{5}{6} \\ 0 & 1 & -1 & -\frac{1}{2} \\ \end{array}\right) \][/tex]
4. Read off the inverse matrix from the augmented matrix:
[tex]\[ \left( \begin{array}{cc} \frac{4}{3} & \frac{5}{6} \\ -1 & -\frac{1}{2} \end{array}\right) \][/tex]
Converting to decimals for clarity:
[tex]\[ \left( \begin{array}{cc} 1.3333333333333333 & 0.8333333333333333 \\ -1 & -0.5 \end{array}\right) \][/tex]
So the inverse of the matrix [tex]\( \left(\begin{array}{cc} -3 & -5 \\ 6 & 8 \end{array}\right) \)[/tex] is [tex]\( \left(\begin{array}{cc} 1.3333333333333333 & 0.8333333333333333 \\ -1 & -0.5 \end{array}\right) \)[/tex].
### a. Finding the inverse of [tex]\( \left(\begin{array}{cc} 1 & 2 \\ 2 & 5 \end{array}\right) \)[/tex]
1. Augment the given matrix with the identity matrix:
[tex]\[ \left(\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 2 & 5 & 0 & 1 \\ \end{array}\right) \][/tex]
2. Pivot on the first element (1,1):
- The pivot element is already 1, so no changes needed in the first row.
- Make the element below the pivot (2) into 0 by subtracting 2 times the first row from the second row:
[tex]\[ \left( \begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & 1 & -2 & 1 \\ \end{array}\right) \][/tex]
3. Pivot on the second element (2,2):
- Now the element at (2,2) is already 1.
- Make the element above the pivot (2) into 0 by subtracting 2 times the second row from the first row:
[tex]\[ \left( \begin{array}{cc|cc} 1 & 0 & 5 & -2 \\ 0 & 1 & -2 & 1 \\ \end{array}\right) \][/tex]
4. Read off the inverse matrix from the augmented matrix:
[tex]\[ \left(\begin{array}{cc} 5 & -2 \\ -2 & 1 \end{array}\right) \][/tex]
So, the inverse of the matrix [tex]\( \left(\begin{array}{cc} 1 & 2 \\ 2 & 5 \end{array}\right) \)[/tex] is [tex]\( \left(\begin{array}{cc} 5 & -2 \\ -2 & 1 \end{array}\right) \)[/tex].
### b. Finding the inverse of [tex]\( \left(\begin{array}{cc} -3 & -5 \\ 6 & 8 \end{array}\right) \)[/tex]
1. Augment the given matrix with the identity matrix:
[tex]\[ \left( \begin{array}{cc|cc} -3 & -5 & 1 & 0 \\ 6 & 8 & 0 & 1 \\ \end{array}\right) \][/tex]
2. Pivot on the first element (1,1):
- Multiply the first row by [tex]\(-\frac{1}{3}\)[/tex]:
[tex]\[ \left(\begin{array}{cc|cc} 1 & \frac{5}{3} & -\frac{1}{3} & 0 \\ 6 & 8 & 0 & 1 \\ \end{array}\right) \][/tex]
- Make the element below the pivot (6) into 0 by subtracting 6 times the first row from the second row:
[tex]\[ \left(\begin{array}{cc|cc} 1 & \frac{5}{3} & -\frac{1}{3} & 0 \\ 0 & -2 & 2 & 1 \\ \end{array}\right) \][/tex]
3. Pivot on the second element (2,2):
- Multiply the second row by [tex]\(-\frac{1}{2}\)[/tex]:
[tex]\[ \left( \begin{array}{cc|cc} 1 & \frac{5}{3} & -\frac{1}{3} & 0 \\ 0 & 1 & -1 & -\frac{1}{2} \\ \end{array}\right) \][/tex]
- Make the element above the pivot ([tex]\(\frac{5}{3}\)[/tex]) into 0 by subtracting [tex]\(\frac{5}{3}\)[/tex] times the second row from the first row:
[tex]\[ \left( \begin{array}{cc|cc} 1 & 0 & \frac{4}{3} & \frac{5}{6} \\ 0 & 1 & -1 & -\frac{1}{2} \\ \end{array}\right) \][/tex]
4. Read off the inverse matrix from the augmented matrix:
[tex]\[ \left( \begin{array}{cc} \frac{4}{3} & \frac{5}{6} \\ -1 & -\frac{1}{2} \end{array}\right) \][/tex]
Converting to decimals for clarity:
[tex]\[ \left( \begin{array}{cc} 1.3333333333333333 & 0.8333333333333333 \\ -1 & -0.5 \end{array}\right) \][/tex]
So the inverse of the matrix [tex]\( \left(\begin{array}{cc} -3 & -5 \\ 6 & 8 \end{array}\right) \)[/tex] is [tex]\( \left(\begin{array}{cc} 1.3333333333333333 & 0.8333333333333333 \\ -1 & -0.5 \end{array}\right) \)[/tex].
