\[ \begin{pmatrix} 1 & 3 & -2 & | & -6 \\ 2 & 1 & 3 & | & 7 \\ 3 & -1 & 4 & | & 8 \end{pmatrix} \]
Subtract \(2 \times\) Row 1 from Row 2:
\[ \begin{pmatrix} 1 & 3 & -2 & | & -6 \\ 0 & -5 & 7 & | & 19 \\ 3 & -1 & 4 & | & 8 \end{pmatrix} \]
Subtract \(3 \times\) Row 1 from Row 3:
\[ \begin{pmatrix} 1 & 3 & -2 & | & -6 \\ 0 & -5 & 7 & | & 19 \\ 0 & -10 & 10 & | & 26 \end{pmatrix} \]
Divide Row 2 by \(-5\):
\[ \begin{pmatrix} 1 & 3 & -2 & | & -6 \\ 0 & 1 & -\frac{7}{5} & | & -\frac{19}{5} \\ 0 & -10 & 10 & | & 26 \end{pmatrix} \]
Add \(10 \times\) Row 2 to Row 3:
\[ \begin{pmatrix} 1 & 3 & -2 & | & -6 \\ 0 & 1 & -\frac{7}{5} & | & -\frac{19}{5} \\ 0 & 0 & 4 & | & 12 \end{pmatrix} \]
From Row 3:
\[ 4z = 12 \implies z = 3 \]
From Row 2:
\[ y - \frac{7}{5}z = -\frac{19}{5} \implies y = \frac{2}{5} \]
From Row 1:
\[ x + 3y - 2z = -6 \implies x = -\frac{6}{5} \]
The unique solution to the system is:
\[ x = -\frac{6}{5}, \quad y = \frac{2}{5}, \quad z = 3 \]
CloseThe augmented matrix for the system is:
\[ \begin{pmatrix} 1 & 3 & -1 & | & 5 \\ 2 & 5 & 1 & | & 10 \\ 4 & 11 & a & | & b \end{pmatrix} \]
Subtract \(2 \times\) Row 1 from Row 2:
\[ \begin{pmatrix} 1 & 3 & -1 & | & 5 \\ 0 & -1 & 3 & | & 0 \\ 4 & 11 & a & | & b \end{pmatrix} \]
Subtract \(4 \times\) Row 1 from Row 3:
\[ \begin{pmatrix} 1 & 3 & -1 & | & 5 \\ 0 & -1 & 3 & | & 0 \\ 0 & -1 & a + 4 & | & b - 20 \end{pmatrix} \]
Subtract Row 2 from Row 3:
\[ \begin{pmatrix} 1 & 3 & -1 & | & 5 \\ 0 & -1 & 3 & | & 0 \\ 0 & 0 & a + 1 & | & b - 20 \end{pmatrix} \]
For the system to have no solution, the third row must represent an inconsistency. This occurs when:
\[ a + 1 = 0 \quad \text{and} \quad b - 20 \neq 0 \]
Solving these:
\[ a = -1 \quad \text{and} \quad b \neq 20 \]
The set of values of \(a\) and \(b\) such that the three planes have no points of intersection is:
\[ a = -1 \quad \text{and} \quad b \neq 20 \]
Close\[ \begin{pmatrix} 3 & -4 & 2 & | & 5 \\ 9 & 2 & -1 & | & 7 \\ 6 & -8 & 4 & | & 10 \end{pmatrix} \]
Divide Row 1 by 3 to simplify:
\[ \begin{pmatrix} 1 & -\frac{4}{3} & \frac{2}{3} & | & \frac{5}{3} \\ 9 & 2 & -1 & | & 7 \\ 6 & -8 & 4 & | & 10 \end{pmatrix} \]
Subtract \(9 \times\) Row 1 from Row 2:
\[ \begin{pmatrix} 1 & -\frac{4}{3} & \frac{2}{3} & | & \frac{5}{3} \\ 0 & 14 & -7 & | & -8 \\ 6 & -8 & 4 & | & 10 \end{pmatrix} \]
Subtract \(6 \times\) Row 1 from Row 3:
\[ \begin{pmatrix} 1 & -\frac{4}{3} & \frac{2}{3} & | & \frac{5}{3} \\ 0 & 14 & -7 & | & -8 \\ 0 & 0 & 0 & | & 0 \end{pmatrix} \]
The third row corresponds to the equation \(0 = 0\), which indicates that the system has infinitely many solutions.
From Row 2:
\[ 14y - 7z = -8 \implies 2y - z = -\frac{8}{7} \]
Let \(z = t\), where \(t\) is a parameter. Then:
\[ y = \frac{t}{2} - \frac{4}{7} \]
From Row 1:
\[ x - \frac{4}{3}y + \frac{2}{3}z = \frac{5}{3} \]
Substitute \(y = \frac{t}{2} - \frac{4}{7}\) and \(z = t\):
\[ x = \frac{19}{21} \]
The general solution to the system is:
\[ x = \frac{19}{21}, \quad y = \frac{t}{2} - \frac{4}{7}, \quad z = t \]
where \(t\) is a parameter.
Closea) The augmented matrix for the system is:
\[ \begin{pmatrix} 1 & 2 & -1 & | & k \\ 2 & 3 & 1 & | & 5 \\ 3 & 5 & 0 & | & 7 \end{pmatrix} \]
After performing row operations, we obtain:
\[ \begin{pmatrix} 1 & 2 & -1 & | & k \\ 0 & -1 & 3 & | & 5 - 2k \\ 0 & 0 & 0 & | & 2 - k \end{pmatrix} \]
For the system to have no solution, the third row must represent an inconsistency. This occurs when:
\[ 2 - k \neq 0 \implies k \neq 2 \]
Thus, the system has no solution for:
\[ k \neq 2 \]
b) For the system to be consistent, the third row must satisfy:
\[ 2 - k = 0 \implies k = 2 \]
Substituting \( k = 2 \) into the augmented matrix and performing back substitution, we find the solution:
\[ x = 4 - 5t, \quad y = 3t - 1, \quad z = t \]
where \( t \) is a parameter.
Closea) The augmented matrix for the system is:
\[ \begin{pmatrix} 3 & -1 & 2 & | & 4 \\ 1 & 2 & -1 & | & 3 \\ 4 & 1 & 1 & | & k \end{pmatrix} \]
After performing row operations, we obtain:
\[ \begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & -7 & 5 & | & -5 \\ 0 & 0 & 0 & | & k - 7 \end{pmatrix} \]
For the system to have an infinite number of solutions, the third row must represent an identity. This occurs when:
\[ k - 7 = 0 \implies k = 7 \]
Thus, the system has an infinite number of solutions for:
\[ k = 7 \]
b) Substituting \( k = 7 \) into the augmented matrix and performing back substitution, we find the solution:
\[ x = \frac{11 - 3t}{7}, \quad y = \frac{5t + 5}{7}, \quad z = t \]
where \( t \) is a parameter.
Closea) The augmented matrix for the system is:
\[ \begin{pmatrix} 1 & 3 & 1 & | & 0 \\ 3 & 7 & 1 & | & b \\ 4 & 10 & a & | & -5 \end{pmatrix} \]
After performing row operations, we obtain:
\[ \begin{pmatrix} 1 & 3 & 1 & | & 0 \\ 0 & -2 & -2 & | & b \\ 0 & 0 & a - 2 & | & -5 - b \end{pmatrix} \]
(i) Unique solution: The system has a unique solution when \( a \neq 2 \).
(ii) No solution: The system has no solution when \( a = 2 \) and \( b \neq -5 \).
(iii) Infinitely many solutions: The system has infinitely many solutions when \( a = 2 \) and \( b = -5 \).
b) For \( a \neq 2 \), the value of \( z \) is:
\[ z = \frac{-5 - b}{a - 2} \]
c) For \( a = 2 \) and \( b = -5 \), the general solution is:
\[ x = -\frac{15}{2} + 2t, \quad y = \frac{5}{2} - t, \quad z = t \]
where \( t \) is a parameter.
Closea) The augmented matrix for the system is:
\[ \begin{bmatrix} 1 & 1 & 1 & | & 2 \\ 2 & 1 & -1 & | & 1 \\ 4 & 3 & m - 3 & | & 5 - m^2 \end{bmatrix} \]
Subtract \( 2 \times \text{Row 1} \) from Row 2:
\[ R_2 \rightarrow R_2 - 2R_1 \]
\[ \begin{bmatrix} 1 & 1 & 1 & | & 2 \\ 0 & -1 & -3 & | & -3 \\ 4 & 3 & m - 3 & | & 5 - m^2 \end{bmatrix} \]
Subtract \( 4 \times \text{Row 1} \) from Row 3:
\[ R_3 \rightarrow R_3 - 4R_1 \]
\[ \begin{bmatrix} 1 & 1 & 1 & | & 2 \\ 0 & -1 & -3 & | & -3 \\ 0 & -1 & m - 7 & | & -3 - m^2 \end{bmatrix} \]
Subtract \( \text{Row 2} \) from Row 3:
\[ R_3 \rightarrow R_3 - R_2 \]
\[ \begin{bmatrix} 1 & 1 & 1 & | & 2 \\ 0 & -1 & -3 & | & -3 \\ 0 & 0 & m - 4 & | & -m^2 \end{bmatrix} \]
The system is now in row-echelon form. The third row corresponds to the equation:
\[ (m - 4)z = -m^2. \]
- If \( m \neq 4 \), then \( z = \frac{-m^2}{m - 4} \). This gives a unique solution for \( z \), and back-substitution will yield unique solutions for \( y \) and \( x \).
- If \( m = 4 \), the third row becomes \( 0 \cdot z = -16 \), which is impossible. Hence, the system has no solution when \( m = 4 \).
The system has a unique solution when \( m \neq 4 \).
b) Plugging the value of \( z \) back into the original equation yields:
\[ x = -1 + \frac{-2m^2}{m - 4}, \quad y = 3 - \frac{-3m^2}{m - 4}, \quad z = \frac{-m^2}{m - 4}. \]
c) For \( m = 0 \):
\[ z = \frac{0}{-4} = 0, \quad y = 3, \quad x = -1. \]
Solution: \( (x, y, z) = (-1, 3, 0) \).
d) When \( m = 4 \), the third equation becomes \( 0 \cdot z = -16 \), which is impossible. Hence, the system has no solution.
Close