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Sia

Question
Math
Posted 7 months ago

Solve the linear system by Gauss-Jordan elimination and Gaussian elimination methods:
x−y+2z−w=−12x−y−2z−2w=−2−x+2y−4z+w=13x+aw=−3.
\begin{aligned}
x-y+2 z-w & =-1 \\
2 x-y-2 z-2 w & =-2 \\
-x+2 y-4 z+w & =1 \\
3 x+a w & =-3 .
\end{aligned}


Determine for which values of aa the system is consistent, and if so whether the solution is unique. Answer "inconclusive" if there is not enough information to make a decision.
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Answer from Sia
Posted 7 months ago
Solution by Steps
step 1
Write the augmented matrix for the given system of equations: (1−12−1∣−12−1−2−2∣−2−12−41∣1300a∣−3) \begin{pmatrix} 1 & -1 & 2 & -1 & | & -1 \\ 2 & -1 & -2 & -2 & | & -2 \\ -1 & 2 & -4 & 1 & | & 1 \\ 3 & 0 & 0 & a & | & -3 \end{pmatrix}
step 2
Perform row operations to get the matrix in row echelon form. Start by making the first element of the first row 1 (it already is): (1−12−1∣−12−1−2−2∣−2−12−41∣1300a∣−3) \begin{pmatrix} 1 & -1 & 2 & -1 & | & -1 \\ 2 & -1 & -2 & -2 & | & -2 \\ -1 & 2 & -4 & 1 & | & 1 \\ 3 & 0 & 0 & a & | & -3 \end{pmatrix} Subtract 2 times the first row from the second row: (1−12−1∣−101−60∣0−12−41∣1300a∣−3) \begin{pmatrix} 1 & -1 & 2 & -1 & | & -1 \\ 0 & 1 & -6 & 0 & | & 0 \\ -1 & 2 & -4 & 1 & | & 1 \\ 3 & 0 & 0 & a & | & -3 \end{pmatrix}
step 3
Add the first row to the third row: (1−12−1∣−101−60∣001−20∣0300a∣−3) \begin{pmatrix} 1 & -1 & 2 & -1 & | & -1 \\ 0 & 1 & -6 & 0 & | & 0 \\ 0 & 1 & -2 & 0 & | & 0 \\ 3 & 0 & 0 & a & | & -3 \end{pmatrix} Subtract 3 times the first row from the fourth row: (1−12−1∣−101−60∣001−20∣003−6a+3∣0) \begin{pmatrix} 1 & -1 & 2 & -1 & | & -1 \\ 0 & 1 & -6 & 0 & | & 0 \\ 0 & 1 & -2 & 0 & | & 0 \\ 0 & 3 & -6 & a+3 & | & 0 \end{pmatrix}
step 4
Subtract 3 times the second row from the fourth row: (1−12−1∣−101−60∣001−20∣00012a+3∣0) \begin{pmatrix} 1 & -1 & 2 & -1 & | & -1 \\ 0 & 1 & -6 & 0 & | & 0 \\ 0 & 1 & -2 & 0 & | & 0 \\ 0 & 0 & 12 & a+3 & | & 0 \end{pmatrix}
step 5
To make the matrix in reduced row echelon form, we need to make the third element of the third row 1. Divide the third row by 12: (1−12−1∣−101−60∣001−20∣0001a+312∣0) \begin{pmatrix} 1 & -1 & 2 & -1 & | & -1 \\ 0 & 1 & -6 & 0 & | & 0 \\ 0 & 1 & -2 & 0 & | & 0 \\ 0 & 0 & 1 & \frac{a+3}{12} & | & 0 \end{pmatrix}
step 6
Subtract 2 times the third row from the first row: (1−10−1−2(a+3)12∣−101−60∣001−20∣0001a+312∣0) \begin{pmatrix} 1 & -1 & 0 & -1 - \frac{2(a+3)}{12} & | & -1 \\ 0 & 1 & -6 & 0 & | & 0 \\ 0 & 1 & -2 & 0 & | & 0 \\ 0 & 0 & 1 & \frac{a+3}{12} & | & 0 \end{pmatrix} Subtract -6 times the third row from the second row: (1−10−1−2(a+3)12∣−10106(a+3)12∣001−20∣0001a+312∣0) \begin{pmatrix} 1 & -1 & 0 & -1 - \frac{2(a+3)}{12} & | & -1 \\ 0 & 1 & 0 & \frac{6(a+3)}{12} & | & 0 \\ 0 & 1 & -2 & 0 & | & 0 \\ 0 & 0 & 1 & \frac{a+3}{12} & | & 0 \end{pmatrix}
step 7
The system is consistent if the last row does not lead to a contradiction. For the system to be consistent, the coefficient of ww in the last row must not be zero. Therefore, a+3≠0a + 3 \neq 0, or a≠−3a \neq -3
step 8
If a≠−3a \neq -3, the system is consistent. To determine if the solution is unique, we need to check if the matrix can be reduced to the identity matrix. Since the last row has a non-zero element in the fourth column, the system has a unique solution for a≠−3a \neq -3
Answer
The system is consistent and has a unique solution for a≠−3a \neq -3.
Key Concept
Consistency and uniqueness of solutions in linear systems
Explanation
The system is consistent if there are no contradictions in the augmented matrix. The solution is unique if the matrix can be reduced to the identity matrix.

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