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Difficulty: Very hardSystems of Linear Equations
For how many real values of the constant aa does the following system of linear equations in xx, yy, and zz have no solution?
x+y−z=3x+(a−1)y+3z=5x+4y+(a+1)z=a+2\begin{aligned} x + y - z &= 3 \\ x + (a-1)y + 3z &= 5 \\ x + 4y + (a+1)z &= a + 2 \end{aligned}
  1. A
    None
  2. Exactly oneAnswer
  3. C
    Exactly two
  4. D
    Exactly three
  5. E
    Infinitely many

Answer

Exactly one
To find when the system has no solution, we first eliminate xx by subtracting the first equation from the second and third equations. This produces a two-variable system in yy and zz: (a−2)y+4z=2(a-2)y + 4z = 2 and 3y+(a+2)z=a−13y + (a+2)z = a - 1. The determinant of this system's coefficients is (a−2)(a+2)−12=a2−16(a-2)(a+2) - 12 = a^2 - 16. Setting the determinant to zero yields two critical values: a=4a = 4 and a=−4a = -4. Testing a=4a = 4 simplifies both reduced equations to y+2z=1y + 2z = 1, which means the system is consistent with infinitely many solutions. Testing a=−4a = -4 yields −3y+2z=1-3y + 2z = 1 and −3y+2z=5-3y + 2z = 5, which is impossible (1=51 = 5), making the system inconsistent. Thus, there is exactly one real value of aa (a=−4a = -4) for which the system has no solution.

Step-by-Step Solution

1
Eliminate the variable xx from the second and third equations using the first equation.
Subtracting the first equation x+y−z=3x + y - z = 3 from the second equation yields:
(a−2)y+4z=2(a-2)y + 4z = 2
Subtracting the first equation from the third equation yields:
3y+(a+2)z=a−13y + (a+2)z = a - 1
Reducing the 3×33 \times 3 system to a 2×22 \times 2 system in yy and zz simplifies the analysis of linear dependence and consistency.
2
Determine the values of aa for which the reduced 2×22 \times 2 system lacks a unique solution by setting its coefficient determinant to zero.
The determinant of the coefficient matrix is:
D=(a−2)(a+2)−(3)(4)=a2−4−12=a2−16D = (a-2)(a+2) - (3)(4) = a^2 - 4 - 12 = a^2 - 16
Setting D=0D = 0 yields a2=16a^2 = 16, which gives a=4a = 4 or a=−4a = -4.
A system of linear equations has either a unique solution (when the determinant is non-zero) or non-unique behavior—either no solution or infinitely many solutions—when the determinant is zero.
3
Test a=4a = 4 in the reduced system.
Substituting a=4a = 4 into the reduced equations gives:
2y+4z=2  ⟹  y+2z=12y + 4z = 2 \implies y + 2z = 1
3y+6z=3  ⟹  y+2z=13y + 6z = 3 \implies y + 2z = 1
Since both equations are identical, the system is consistent and has infinitely many solutions.
When equation ratios match completely including constant terms, the equations represent identical hyperplanes, yielding infinitely many solutions.
4
Test a=−4a = -4 in the reduced system.
Substituting a=−4a = -4 into the reduced equations gives:
−6y+4z=2  ⟹  −3y+2z=1-6y + 4z = 2 \implies -3y + 2z = 1
3y−2z=−5  ⟹  −3y+2z=53y - 2z = -5 \implies -3y + 2z = 5
Comparing these gives 1=51 = 5, which is a contradiction. Thus, for a=−4a = -4, the system has no solution.
When parallel equations have equal coefficient ratios but unequal constant ratios, the system is inconsistent.

Key Concept

Parametric Systems of Linear Equations and Consistency Conditions
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Topics

Systems of Linear Equations

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