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Ch. 6 - Matrices and Determinants
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 19

In Exercises 1 - 24, use Gaussian Elimination to find the complete solution to each system of equations, or show that none exists. {x+y−2z=23x−y−6z=−7\(\begin{cases}\) x + y - 2z = 2 \\ 3x - y - 6z = -7 \(\end{cases}\)

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1
Write the system of equations as an augmented matrix: \[\left[\begin{array}{ccc|c} 1 & 1 & -2 & 2 \\ 3 & -1 & -6 & -7 \end{array}\right]\]
Use the first row to eliminate the \(x\)-term in the second row. Multiply the first row by 3 and subtract it from the second row: \[R_2 \rightarrow R_2 - 3R_1\]
Perform the row operation to get the new second row: \[\left[\begin{array}{ccc|c} 1 & 1 & -2 & 2 \\ 0 & -4 & 0 & -13 \end{array}\right]\]
Solve the second equation for \(y\) by dividing the entire second row by the coefficient of \(y\): \[y = \frac{-13}{-4} = \frac{13}{4}\]
Substitute the value of \(y\) back into the first equation to solve for \(x\) in terms of \(z\): \[x + \frac{13}{4} - 2z = 2\] Then isolate \(x\) to express it as \[x = 2 - \frac{13}{4} + 2z\]

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