Question:
For \( a \in \mathbb{R} \), consider the system of linear equations
\[
\begin{cases}
a x + a y = a + 2,
x + a y + (a - 1)z = a - 4,
a x + a y + (a - 2)z = -8,
\end{cases}
\]
in the unknowns \(x, y, z\). Then, which of the following statements is TRUE?
For \( a \in \mathbb{R} \), consider the system of linear equations
\[
\begin{cases}
a x + a y = a + 2,
x + a y + (a - 1)z = a - 4,
a x + a y + (a - 2)z = -8, \end{cases} \] in the unknowns \(x, y, z\). Then, which of the following statements is TRUE?
x + a y + (a - 1)z = a - 4,
a x + a y + (a - 2)z = -8, \end{cases} \] in the unknowns \(x, y, z\). Then, which of the following statements is TRUE?
Show Hint
The determinant of the coefficient matrix determines uniqueness. If nonzero, the system has a unique solution; if zero, check consistency for infinite or no solutions.
Updated On: Dec 6, 2025
- The given system has a unique solution for \( a = 1 \)
- The given system has infinitely many solutions for \( a = 2 \)
- The given system has a unique solution for \( a = -2 \)
- The given system has infinitely many solutions for \( a = -2 \)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Write in matrix form.
\[ \begin{bmatrix} a & a & 0
1 & a & a - 1
a & a & a - 2 \end{bmatrix} \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} a + 2
a - 4
-8 \end{bmatrix}. \]
Step 2: Find determinant of coefficient matrix.
\[ \Delta = \begin{vmatrix} a & a & 0
1 & a & a - 1
a & a & a - 2 \end{vmatrix} = a \begin{vmatrix} a & a - 1
a & a - 2 \end{vmatrix} - a \begin{vmatrix} 1 & a - 1
a & a - 2 \end{vmatrix}. \] Compute minors: \[ \begin{vmatrix} a & a - 1
a & a - 2 \end{vmatrix} = a(a - 2) - a(a - 1) = -a, \] \[ \begin{vmatrix} 1 & a - 1
a & a - 2 \end{vmatrix} = (1)(a - 2) - a(a - 1) = a - 2 - a^2 + a = -a^2 + 2a - 2. \] Thus, \[ \Delta = a(-a) - a(-a^2 + 2a - 2) = -a^2 + a^3 - 2a^2 + 2a = a^3 - 3a^2 + 2a = a(a - 1)(a - 2). \]
Step 3: Analyze cases.
\(\Delta = 0\) when \(a = 0, 1, 2\). For all other values of \(a\), the system has a unique solution. At \(a = -2\), determinant \(\neq 0\), so it has a unique solution. Final Answer: \[ \boxed{\text{The given system has a unique solution for } a = -2.} \]
\[ \begin{bmatrix} a & a & 0
1 & a & a - 1
a & a & a - 2 \end{bmatrix} \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} a + 2
a - 4
-8 \end{bmatrix}. \]
Step 2: Find determinant of coefficient matrix.
\[ \Delta = \begin{vmatrix} a & a & 0
1 & a & a - 1
a & a & a - 2 \end{vmatrix} = a \begin{vmatrix} a & a - 1
a & a - 2 \end{vmatrix} - a \begin{vmatrix} 1 & a - 1
a & a - 2 \end{vmatrix}. \] Compute minors: \[ \begin{vmatrix} a & a - 1
a & a - 2 \end{vmatrix} = a(a - 2) - a(a - 1) = -a, \] \[ \begin{vmatrix} 1 & a - 1
a & a - 2 \end{vmatrix} = (1)(a - 2) - a(a - 1) = a - 2 - a^2 + a = -a^2 + 2a - 2. \] Thus, \[ \Delta = a(-a) - a(-a^2 + 2a - 2) = -a^2 + a^3 - 2a^2 + 2a = a^3 - 3a^2 + 2a = a(a - 1)(a - 2). \]
Step 3: Analyze cases.
\(\Delta = 0\) when \(a = 0, 1, 2\). For all other values of \(a\), the system has a unique solution. At \(a = -2\), determinant \(\neq 0\), so it has a unique solution. Final Answer: \[ \boxed{\text{The given system has a unique solution for } a = -2.} \]
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