Question

The number of solutions of the system of equations \(2x+y-z=7\), \(x-3y+2z=1\), \(x+4y-3z=5\) is

• A. 1
• B. 0
• C. Infinite
• D. 2

Hint:

First, calculate \(\det(A)\). \(A = \begin{pmatrix} 2 & 1 & -1
1 & -3 & 2
1 & 4 & -3 \end{pmatrix}\). \(\det(A) = 2(9-8)-1(-3-2)-1(4+3) = 2+5-7 = 0\). Since \(\det(A)=0\), there are either no solutions or infinitely many. Use Gaussian elimination on the augmented matrix. This leads to a row \([0 \ 0 \ 0 \ | \ -1]\), which signifies no solution.

Answer 1

The Correct Option is B

The system of equations is: \begin{align} 2x+y-z &= 7 &(1)
x-3y+2z &= 1 &(2) \\ x+4y-3z &= 5 &(3) \end{align} We form the augmented matrix \([A|B]\): \[ \left[ \begin{array}{ccc|c} 2 & 1 & -1 & 7 \\1 & -3 & 2 & 1 \\ 1 & 4 & -3 & 5 \end{array} \right] \] Perform row operations to get to row echelon form.
\(R_1 \leftrightarrow R_2\): \[ \left[ \begin{array}{ccc|c} 1 & -3 & 2 & 1 \\ 2 & 1 & -1 & 7\\ 1 & 4 & -3 & 5 \end{array} \right] \] \(R_2 \rightarrow R_2 - 2R_1\): Row 2 becomes: \((2-2(1)) (1-2(-3)) (-1-2(2)) | (7-2(1))\) \( \Rightarrow 0 7 -5 | 5\) \(R_3 \rightarrow R_3 - R_1\): Row 3 becomes: \((1-1) (4-(-3)) (-3-2) | (5-1)\) \( \Rightarrow 0 7 -5 | 4\) The matrix is now: \[ \left[ \begin{array}{ccc|c} 1 & -3 & 2 & 1 \\ 0 & 7 & -5 & 5\\ 0 & 7 & -5 & 4 \end{array} \right] \] \(R_3 \rightarrow R_3 - R_2\): Row 3 becomes: \((0-0) (7-7) (-5-(-5)) | (4-5)\) \( \Rightarrow 0 0 0 | -1\) The matrix is now: \[ \left[ \begin{array}{ccc|c} 1 & -3 & 2 & 1 \\ 0 & 7 & -5 & 5 \\ 0 & 0 & 0 & -1 \end{array} \right] \] The last row represents the equation \(0x + 0y + 0z = -1\), which simplifies to \(0 = -1\).
This is a contradiction, indicating that the system of equations has no solution.
The number of solutions is 0.
\[ \boxed{0} \]