Use an inverse matrix to solve each system of linear equations. (a) x1 + 2x2 + x3 = 0 x1 + 2x2 − x3 = 2 x1 − 2x2 + x3 = −4
(x1, x2, x3) =
(b) x1 + 2x2 + x3 = 1 x1 + 2x2 − x3 = 1 x1 − 2x2 + x3 = −3
(x1, x2, x3) =
To solve the systems of linear equations using the inverse matrix method, we can express the system in matrix form (AX = B), where (A) is the coefficient matrix, (X) is the vector of variables, and (B) is the constant vector. We can then find the solution by calculating (X = A^{-1}B).
The system of equations is:
[ \begin{align*} x_1 + 2x_2 + x_3 & = 0 \ x_1 + 2x_2 - x_3 & = 2 \ x_1 - 2x_2 + x_3 & = -4 \end{align*} ]
Step 1: Write in matrix form (AX = B)
[ A = \begin{bmatrix} 1 & 2 & 1 \ 1 & 2 & -1 \ 1 & -2 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix}, \quad B = \begin{bmatrix} 0 \ 2 \ -4 \end{bmatrix} ]
Step 2: Find the inverse of matrix (A)
To find (A^{-1}), we can use the formula for the inverse of a 3x3 matrix or use row reduction. Here, we will calculate the determinant and the adjugate.
[ \text{det}(A) = 1 \cdot (2 \cdot 1 - (-1) \cdot (-2)) - 2 \cdot (1 \cdot 1 - (-1) \cdot 1) + 1 \cdot (1 \cdot (-2) - 1 \cdot 1) ] [ = 1 \cdot (2 - 2) - 2 \cdot (1 + 1) + 1 \cdot (-2 - 1) = 0 - 4 - 3 = -7 ]
After calculating, we find:
[ A^{-1} = \begin{bmatrix} -1 & -1 & 0 \ -1 & 0 & 1 \ 0 & -1 & 1 \end{bmatrix} ]
Step 3: Calculate (X = A^{-1}B)
[ X = A^{-1}B = \begin{bmatrix} -1 & -1 & 0 \ -1 & 0 & 1 \ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 0 \ 2 \ -4 \end{bmatrix} ]
Calculating this gives:
[ X = \begin{bmatrix} -1 \cdot 0 + -1 \cdot 2 + 0 \cdot -4 \ -1 \cdot 0 + 0 \cdot 2 + 1 \cdot -4 \ 0 \cdot 0 + -1 \cdot 2 + 1 \cdot -4 \end{bmatrix} = \begin{bmatrix} -2 \ -4 \ -6 \end{bmatrix} ]
Thus, the solution for (a) is:
[ (x_1, x_2, x_3) = (-2, -4, -6) ]
The system of equations is:
[ \begin{align*} x_1 + 2x_2 + x_3 & = 1 \ x_1 + 2x_2 - x_3 & = 1 \ x_1 - 2x_2 + x_3 & = -3 \end{align*} ]
Step 1: Write in matrix form (AX = B)
[ A = \begin{bmatrix} 1 & 2 & 1 \ 1 & 2 & -1 \ 1 & -2 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix}, \quad B = \begin{bmatrix} 1 \ 1 \ -3 \end{bmatrix} ]
Step 2: Find the inverse of matrix (A)
The matrix (A) is the same as in part (a), so we can use the same inverse:
[ A^{-1} = \begin{bmatrix} -1 & -1 & 0 \ -1 & 0 & 1 \ 0 & -1 & 1 \end{bmatrix} ]
Step 3: Calculate (X = A^{-1}B)
[ X = A^{-1}B = \begin{bmatrix} -1 & -1 & 0 \ -1 & 0 & 1 \ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 \ 1 \ -3 \end{bmatrix} ]
Calculating this gives:
[ X = \begin{bmatrix} -1 \cdot 1 + -1 \cdot 1 + 0 \cdot -3 \ -1 \cdot 1 + 0 \cdot 1 + 1 \cdot -3 \ 0 \cdot 1 + -1 \cdot 1 + 1 \cdot -3 \end{bmatrix} = \begin{bmatrix} -2 \ -4 \ -4 \end{bmatrix} ]
Thus, the solution for (b) is:
[ (x_1, x_2, x_3) = (-2, -4, -4) ]
(a) ((x_1, x_2, x_3) = (-2, -4, -6))
(b) ((x_1, x_2, x_3) = (-2, -4, -