- We translate the system of linear equations into an augmented matrix form.
- Through Gaussian elimination, we reduce the matrix into row-echelon form to isolate variables.
- We find the value of each variable through back-substitution: \(z = -5\), \(y = -3\), and \(x = 1\).
- The solution to the system is \((x, y, z) = \boxed{(1, -3, -5)}\).
Explanation
- Analysis and Initial Setup
The objective is to solve the system of linear equations:
\[ \begin{cases} 6x + 2y - 2z = 10 \ -x - 3y + 7z = -27 \ 3x + 5y - 6z = 18 \end{cases} \]
Using a matrix-based approach, we represent this system as an augmented matrix \([A|B]\), where \(A\) is the matrix of coefficients and \(B\) is the column vector of the constants. This allows us to apply systematic row operations to isolate the variables. The augmented matrix is given by:
\[ \left[ \begin{array}{ccc|c} 6 & 2 & -2 & 10 \ -1 & -3 & 7 & -27 \ 3 & 5 & -6 & 18 \end{array} \right] \]
- Reducing the Matrix
First, we simplify the matrix to make calculations easier. Dividing the first row by \(2\), we obtain:
\[ \left[ \begin{array}{ccc|c} 3 & 1 & -1 & 5 \ -1 & -3 & 7 & -27 \ 3 & 5 & -6 & 18 \end{array} \right] \]
Next, we swap the first and second rows to have a pivot element of \(-1\) in the first position, then multiply the new first row by \(-1\) to get a leading \(1\):
\[ \left[ \begin{array}{ccc|c} 1 & 3 & -7 & 27 \ 3 & 1 & -1 & 5 \ 3 & 5 & -6 & 18 \end{array} \right] \]
We now perform row operations to eliminate the leading coefficients in the second and third rows:
- Replace \(R_2\) with \(R_2 - 3R_1\): \((3-3(1), 1-3(3), -1-3(-7) | 5-3(27)) = (0, -8, 20 | -76)\).
- Replace \(R_3\) with \(R_3 - 3R_1\): \((3-3(1), 5-3(3), -6-3(-7) | 18-3(27)) = (0, -4, 15 | -63)\).
This results in:
\[ \left[ \begin{array}{ccc|c} 1 & 3 & -7 & 27 \ 0 & -8 & 20 & -76 \ 0 & -4 & 15 & -63 \end{array} \right] \]
- Final Calculation and Solution
To further simplify, divide the second row \(R_2\) by \(-4\): \((0, 2, -5 | 19)\). Now the matrix is:
\[ \left[ \begin{array}{ccc|c} 1 & 3 & -7 & 27 \ 0 & 2 & -5 & 19 \ 0 & -4 & 15 & -63 \end{array} \right] \]
Next, replace \(R_3\) with \(R_3 + 2R_2\): \((0, -4+2(2), 15+2(-5) | -63+2(19)) = (0, 0, 5 | -25)\).
The final row-echelon form is:
\[ \left[ \begin{array}{ccc|c} 1 & 3 & -7 & 27 \ 0 & 2 & -5 & 19 \ 0 & 0 & 5 & -25 \end{array} \right] \]
From the third row, \(5z = -25\), which gives \(z = -5\). Back-substituting into the second row, \(2y - 5(-5) = 19 \implies 2y + 25 = 19 \implies 2y = -6 \implies y = -3\). Finally, substituting \(y\) and \(z\) into the first row gives \(x + 3(-3) - 7(-5) = 27 \implies x - 9 + 35 = 27 \implies x + 26 = 27 \implies x = 1\). The solution is \((x, y, z) = (1, -3, -5)\).
Examples
Systems of linear equations are fundamental in fields like logistics and engineering. For example, if a manufacturer produces three different products, each requiring specific amounts of raw materials (represented by variables \(x\), \(y\), and \(z\)), the total material usage is often constrained by a system of linear equations. Solving this system helps determine exactly how many units of each product should be manufactured to exhaust the available resources efficiently. In this case, finding the intersection point \((1, -3, -5)\) represents the precise balance point for the variables given the defined resource constraints.