Question

The following system of linear equations \[ 7x-3y+z=0,\quad 3x-y+z=0,\quad x-y-z=0 \] has:

• A. infinitely many solutions
• B. a unique solution
• C. no solution
• D. three solutions

Hint:

For homogeneous systems, \(\det(A)=0\) \(\Rightarrow\) either infinitely many or only the trivial solution. Check rank or directly parametrize via row-reduction; any free variable \(\Rightarrow\) infinitely many solutions.

Answer 1

The Correct Option is A

Step 1: Coefficient matrix and determinant.
Coefficient matrix: $A = \begin{bmatrix} 7 & -3 & 1 \\ 3 & -1 & 1 \\ 1 & -1 & -1 \end{bmatrix}$.
Compute $\det(A)$. If $\det(A)\neq 0 \Rightarrow$ unique solution; if $\det(A)=0 \Rightarrow$ either no solution or infinitely many depending on consistency.
A quick evaluation (by expansion/row-ops) gives $\det(A)=0 \Rightarrow$ no unique solution.

Step 2: Check consistency (rank test).
Solve the homogeneous system $A\mathbf{x}=\mathbf{0}$. Row-reduce:
$7x - 3y + z = 0$,
$3x - y + z = 0$,
$x - y - z = 0$.
$\Rightarrow \; x = -z,\; y = -2z,\; z$ free.
Since a non-trivial solution exists and the equations are consistent (rank $=2 < 3$), the solution space is a line (one free parameter).

Step 3: Parametric form.
Let $z = t$. Then $(x,y,z) = (-t,-2t,t) = t(-1,-2,1),\; t \in \mathbb{R}$.
$\Rightarrow$ infinitely many solutions.

$\boxed{\text{Infinitely many solutions}}$