Answer :
Certainly! Let's solve the given system of equations using Cramer's rule step by step.
The system of linear equations is:
[tex]\[ \begin{aligned} I_1 - 2I_2 + I_3 &= -2 \quad \text{(1)} \\ -I_1 + I_2 &= 1 \quad \text{(2)} \\ I_2 + I_3 &= 2 \quad \text{(3)} \end{aligned} \][/tex]
### Step 1: Express the system of equations in matrix form
The given equations can be written in matrix form as [tex]\( AX = B \)[/tex], where [tex]\( A \)[/tex] is the matrix of coefficients, [tex]\( X \)[/tex] is the column matrix of variables, and [tex]\( B \)[/tex] is the matrix of constants.
[tex]\[ A = \begin{pmatrix} 1 & -2 & 1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}, \quad X = \begin{pmatrix} I_1 \\ I_2 \\ I_3 \end{pmatrix}, \quad B = \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix} \][/tex]
Using Cramer's rule, we need to find the determinant of the coefficient matrix [tex]\( A \)[/tex] and the determinants of matrices [tex]\( A_1, A_2, A_3 \)[/tex] which are obtained by replacing one column of [tex]\( A \)[/tex] with the matrix [tex]\( B \)[/tex].
### Step 2: Calculate the determinant of matrix [tex]\( A \)[/tex]
[tex]\[ A = \begin{pmatrix} 1 & -2 & 1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix} \][/tex]
The determinant [tex]\( \text{det}(A) \)[/tex] of matrix [tex]\( A \)[/tex] is computed as:
[tex]\[ \text{det}(A) = 1 \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} - (-2) \begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} + 1 \begin{vmatrix} -1 & 1 \\ 0 & 1 \end{vmatrix} \][/tex]
[tex]\[ = 1 (1 \cdot 1 - 0 \cdot 1) + 2 (-1 \cdot 1 - 0 \cdot 0) + 1 (-1 \cdot 1 - 1 \cdot 0) = 1 (1) + 2 (-1) + 1 (-1) = 1 - 2 - 1 = -2 \][/tex]
So, [tex]\(\text{det}(A) = -2\)[/tex].
### Step 3: Calculate the determinants of matrices [tex]\( A_1, A_2, A_3 \)[/tex]
#### Matrix [tex]\( A_1 \)[/tex]
Replace the first column of [tex]\( A \)[/tex] with [tex]\( B \)[/tex]:
[tex]\[ A_1 = \begin{pmatrix} -2 & -2 & 1 \\ 1 & 1 & 0 \\ 2 & 1 & 1 \end{pmatrix} \][/tex]
The determinant [tex]\( \text{det}(A_1) \)[/tex] is given by:
[tex]\[ \text{det}(A_1) = -2 \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} - (-2) \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \][/tex]
[tex]\[ = -2 (1 \cdot 1 - 0 \cdot 1) + 2 (1 \cdot 1 - 0 \cdot 2) + 1 (1 \cdot 1 - 1 \cdot 2) = -2 (1) + 2 (1) + 1 (-1) = -2 + 2 - 1 = -1 \][/tex]
So, [tex]\( \text{det}(A_1) = -1 \)[/tex].
#### Matrix [tex]\( A_2 \)[/tex]
Replace the second column of [tex]\( A \)[/tex] with [tex]\( B \)[/tex]:
[tex]\[ A_2 = \begin{pmatrix} 1 & -2 & 1 \\ -1 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix} \][/tex]
The determinant [tex]\( \text{det}(A_2) \)[/tex] is:
[tex]\[ \text{det}(A_2) = 1 \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} - (-2) \begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} + 1 \begin{vmatrix} -1 & 1 \\ 0 & 2 \end{vmatrix} \][/tex]
[tex]\[ = 1 (1 \cdot 1 - 0 \cdot 2) + 2 (-1 \cdot 1 - 0 \cdot 0) + 1 (-1 \cdot 2 - 1 \cdot 0) = 1 (1) + 2 (-1) + 1 (-2) = 1 - 2 - 2 = -3 \][/tex]
So, [tex]\( \text{det}(A_2) = -3 \)[/tex].
#### Matrix [tex]\( A_3 \)[/tex]
Replace the third column of [tex]\( A \)[/tex] with [tex]\( B \)[/tex]:
[tex]\[ A_3 = \begin{pmatrix} 1 & -2 & -2 \\ -1 & 1 & 1 \\ 0 & 1 & 2 \end{pmatrix} \][/tex]
The determinant [tex]\( \text{det}(A_3) \)[/tex] is:
[tex]\[ \text{det}(A_3) = 1 \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} - (-2) \begin{vmatrix} -1 & 1 \\ 0 & 2 \end{vmatrix} + (-2) \begin{vmatrix} -1 & 1 \\ 0 & 1 \end{vmatrix} \][/tex]
[tex]\[ = 1 (1 \cdot 2 - 1 \cdot 1) + 2 (-1 \cdot 2 - 0 \cdot 1) - 2 (-1 \cdot 1 - 1 \cdot 0) = 1 (1) + 2 (-1) - 2 (1) = 1 - 2 - 2 = -1 \][/tex]
So, [tex]\( \text{det}(A_3) = -1 \)[/tex].
### Step 4: Apply Cramer's Rule to find [tex]\( I_1, I_2, I_3 \)[/tex]
Cramer's rule states that for a system [tex]\( AX = B \)[/tex],
[tex]\[ I_1 = \frac{\text{det}(A_1)}{\text{det}(A)} \quad I_2 = \frac{\text{det}(A_2)}{\text{det}(A)} \quad I_3 = \frac{\text{det}(A_3)}{\text{det}(A)} \][/tex]
Using the computed determinants:
[tex]\[ I_1 = \frac{-1}{-2} = 0.5 \quad I_2 = \frac{-3}{-2} = 1.5 \quad I_3 = \frac{-1}{-2} = 0.5 \][/tex]
### Conclusion
By applying Cramer's Rule, we find that the currents [tex]\( I_1, I_2, \)[/tex] and [tex]\( I_3 \)[/tex] are:
[tex]\[ I_1 = 0.5 \, \text{A} \][/tex]
[tex]\[ I_2 = 1.5 \, \text{A} \][/tex]
[tex]\[ I_3 = 0.5 \, \text{A} \][/tex]
These are the solutions to the given system of equations.
The system of linear equations is:
[tex]\[ \begin{aligned} I_1 - 2I_2 + I_3 &= -2 \quad \text{(1)} \\ -I_1 + I_2 &= 1 \quad \text{(2)} \\ I_2 + I_3 &= 2 \quad \text{(3)} \end{aligned} \][/tex]
### Step 1: Express the system of equations in matrix form
The given equations can be written in matrix form as [tex]\( AX = B \)[/tex], where [tex]\( A \)[/tex] is the matrix of coefficients, [tex]\( X \)[/tex] is the column matrix of variables, and [tex]\( B \)[/tex] is the matrix of constants.
[tex]\[ A = \begin{pmatrix} 1 & -2 & 1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}, \quad X = \begin{pmatrix} I_1 \\ I_2 \\ I_3 \end{pmatrix}, \quad B = \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix} \][/tex]
Using Cramer's rule, we need to find the determinant of the coefficient matrix [tex]\( A \)[/tex] and the determinants of matrices [tex]\( A_1, A_2, A_3 \)[/tex] which are obtained by replacing one column of [tex]\( A \)[/tex] with the matrix [tex]\( B \)[/tex].
### Step 2: Calculate the determinant of matrix [tex]\( A \)[/tex]
[tex]\[ A = \begin{pmatrix} 1 & -2 & 1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix} \][/tex]
The determinant [tex]\( \text{det}(A) \)[/tex] of matrix [tex]\( A \)[/tex] is computed as:
[tex]\[ \text{det}(A) = 1 \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} - (-2) \begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} + 1 \begin{vmatrix} -1 & 1 \\ 0 & 1 \end{vmatrix} \][/tex]
[tex]\[ = 1 (1 \cdot 1 - 0 \cdot 1) + 2 (-1 \cdot 1 - 0 \cdot 0) + 1 (-1 \cdot 1 - 1 \cdot 0) = 1 (1) + 2 (-1) + 1 (-1) = 1 - 2 - 1 = -2 \][/tex]
So, [tex]\(\text{det}(A) = -2\)[/tex].
### Step 3: Calculate the determinants of matrices [tex]\( A_1, A_2, A_3 \)[/tex]
#### Matrix [tex]\( A_1 \)[/tex]
Replace the first column of [tex]\( A \)[/tex] with [tex]\( B \)[/tex]:
[tex]\[ A_1 = \begin{pmatrix} -2 & -2 & 1 \\ 1 & 1 & 0 \\ 2 & 1 & 1 \end{pmatrix} \][/tex]
The determinant [tex]\( \text{det}(A_1) \)[/tex] is given by:
[tex]\[ \text{det}(A_1) = -2 \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} - (-2) \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \][/tex]
[tex]\[ = -2 (1 \cdot 1 - 0 \cdot 1) + 2 (1 \cdot 1 - 0 \cdot 2) + 1 (1 \cdot 1 - 1 \cdot 2) = -2 (1) + 2 (1) + 1 (-1) = -2 + 2 - 1 = -1 \][/tex]
So, [tex]\( \text{det}(A_1) = -1 \)[/tex].
#### Matrix [tex]\( A_2 \)[/tex]
Replace the second column of [tex]\( A \)[/tex] with [tex]\( B \)[/tex]:
[tex]\[ A_2 = \begin{pmatrix} 1 & -2 & 1 \\ -1 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix} \][/tex]
The determinant [tex]\( \text{det}(A_2) \)[/tex] is:
[tex]\[ \text{det}(A_2) = 1 \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} - (-2) \begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} + 1 \begin{vmatrix} -1 & 1 \\ 0 & 2 \end{vmatrix} \][/tex]
[tex]\[ = 1 (1 \cdot 1 - 0 \cdot 2) + 2 (-1 \cdot 1 - 0 \cdot 0) + 1 (-1 \cdot 2 - 1 \cdot 0) = 1 (1) + 2 (-1) + 1 (-2) = 1 - 2 - 2 = -3 \][/tex]
So, [tex]\( \text{det}(A_2) = -3 \)[/tex].
#### Matrix [tex]\( A_3 \)[/tex]
Replace the third column of [tex]\( A \)[/tex] with [tex]\( B \)[/tex]:
[tex]\[ A_3 = \begin{pmatrix} 1 & -2 & -2 \\ -1 & 1 & 1 \\ 0 & 1 & 2 \end{pmatrix} \][/tex]
The determinant [tex]\( \text{det}(A_3) \)[/tex] is:
[tex]\[ \text{det}(A_3) = 1 \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} - (-2) \begin{vmatrix} -1 & 1 \\ 0 & 2 \end{vmatrix} + (-2) \begin{vmatrix} -1 & 1 \\ 0 & 1 \end{vmatrix} \][/tex]
[tex]\[ = 1 (1 \cdot 2 - 1 \cdot 1) + 2 (-1 \cdot 2 - 0 \cdot 1) - 2 (-1 \cdot 1 - 1 \cdot 0) = 1 (1) + 2 (-1) - 2 (1) = 1 - 2 - 2 = -1 \][/tex]
So, [tex]\( \text{det}(A_3) = -1 \)[/tex].
### Step 4: Apply Cramer's Rule to find [tex]\( I_1, I_2, I_3 \)[/tex]
Cramer's rule states that for a system [tex]\( AX = B \)[/tex],
[tex]\[ I_1 = \frac{\text{det}(A_1)}{\text{det}(A)} \quad I_2 = \frac{\text{det}(A_2)}{\text{det}(A)} \quad I_3 = \frac{\text{det}(A_3)}{\text{det}(A)} \][/tex]
Using the computed determinants:
[tex]\[ I_1 = \frac{-1}{-2} = 0.5 \quad I_2 = \frac{-3}{-2} = 1.5 \quad I_3 = \frac{-1}{-2} = 0.5 \][/tex]
### Conclusion
By applying Cramer's Rule, we find that the currents [tex]\( I_1, I_2, \)[/tex] and [tex]\( I_3 \)[/tex] are:
[tex]\[ I_1 = 0.5 \, \text{A} \][/tex]
[tex]\[ I_2 = 1.5 \, \text{A} \][/tex]
[tex]\[ I_3 = 0.5 \, \text{A} \][/tex]
These are the solutions to the given system of equations.
