Math Problem Statement
solve by matrix inversion : 2x - y - z = 7; 3x + y - z = 7; x + y - z = 3
Solution
To solve the system of equations using matrix inversion, we follow these steps:
1. Write the system of equations in matrix form:
The given system of equations is:
2x - y - z &= 7 \quad \text{(Equation 1)} \\ 3x + y - z &= 7 \quad \text{(Equation 2)} \\ x + y - z &= 3 \quad \text{(Equation 3)} \end{aligned}$$ This can be written in the matrix form as $$AX = B$$, where: $$A = \begin{pmatrix} 2 & -1 & -1 \\ 3 & 1 & -1 \\ 1 & 1 & -1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 7 \\ 7 \\ 3 \end{pmatrix}$$ ### 2. Find the inverse of matrix $$A$$: The inverse of matrix $$A$$, denoted as $$A^{-1}$$, is calculated by using the formula: $$A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A)$$ Let's calculate the determinant of $$A$$ and then find the inverse. ### 3. Calculate the determinant of $$A$$: $$\text{det}(A) = \begin{vmatrix} 2 & -1 & -1 \\ 3 & 1 & -1 \\ 1 & 1 & -1 \end{vmatrix}$$ Expanding along the first row: $$\text{det}(A) = 2\begin{vmatrix} 1 & -1 \\ 1 & -1 \end{vmatrix} - (-1)\begin{vmatrix} 3 & -1 \\ 1 & -1 \end{vmatrix} + (-1)\begin{vmatrix} 3 & 1 \\ 1 & 1 \end{vmatrix}$$ \[ \text{det}(A) = 2[(1 \times -1) - (1 \times -1)] + (1)[(3 \times -1) - (1 \times -1)] - (1)[(3 \times 1) - (1 \times 1)] \] \[ \text{det}(A) = 2[0] + (1)[-2] - (1)[2] = -2 - 2 = -4 \] ### 4. Find the adjugate matrix of $$A$$: The adjugate matrix is the transpose of the cofactor matrix. Calculating the cofactors: \[ \text{Cof}(A) = \begin{pmatrix} (1)(-1) - (-1)(1) & -[(3)(-1) - (-1)(1)] & (3)(1) - (1)(1) \\ -[(1)(-1) - (-1)(1)] & (2)(-1) - (-1)(1) & -(2)(1) - (-1)(3) \\ [(1)(1) - (1)(-1)] & -(2)(-1) - (-1)(3) & (2)(1) - (3)(-1) \end{pmatrix} \] $$\text{Cof}(A) = \begin{pmatrix} -2 & 2 & 2 \\ 2 & -3 & -7 \\ 0 & -2 & -5 \end{pmatrix}$$ So the adjugate matrix is: $$\text{adj}(A) = \begin{pmatrix} -2 & 2 & 0 \\ 2 & -3 & -2 \\ 2 & -7 & -5 \end{pmatrix}$$ ### 5. Calculate $$A^{-1}$$: $$A^{-1} = \frac{1}{-4} \begin{pmatrix} -2 & 2 & 0 \\ 2 & -3 & -2 \\ 2 & -7 & -5 \end{pmatrix}$$ $$A^{-1} = \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{3}{4} & \frac{1}{2} \\ -\frac{1}{2} & \frac{7}{4} & \frac{5}{4} \end{pmatrix}$$ ### 6. Solve for $$X$$: Now, $$X = A^{-1}B$$: $$X = \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{3}{4} & \frac{1}{2} \\ -\frac{1}{2} & \frac{7}{4} & \frac{5}{4} \end{pmatrix} \begin{pmatrix} 7 \\ 7 \\ 3 \end{pmatrix}$$ Multiply the matrices: $$X = \begin{pmatrix} \frac{1}{2}(7) - \frac{1}{2}(7) + 0(3) \\ -\frac{1}{2}(7) + \frac{3}{4}(7) + \frac{1}{2}(3) \\ -\frac{1}{2}(7) + \frac{7}{4}(7) + \frac{5}{4}(3) \end{pmatrix}$$ $$X = \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$$ Thus, the solution is: $$\begin{aligned} x &= 0 \\ y &= 2 \\ z &= 1 \end{aligned}$$ Would you like any more details or explanations? Here are five related questions you might find useful: 1. How can matrix inversion be used to solve larger systems of equations? 2. What are the conditions for a matrix to be invertible? 3. Can matrix inversion be used for non-square matrices? 4. What are the computational challenges of matrix inversion for large matrices? 5. How does matrix inversion compare to other methods of solving linear systems, such as Gaussian elimination? **Tip:** When solving a system of equations using matrix inversion, always ensure that the determinant of the matrix is non-zero; otherwise, the matrix is not invertible.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Matrix Algebra
Systems of Equations
Formulas
Matrix inversion formula: $A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A)$
Theorems
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Suitable Grade Level
Advanced High School