GATE 2025 CE (CE1) – Question 12

Engineering Mathematics · Linear Algebra: Matrix algebra, systems of linear equations, eigenvalues and eigenvectors · 1 mark · Multiple choice

Let $A = \begin{bmatrix} 1 & 1 \\ 1 & 3 \\ -2 & -3 \end{bmatrix}$ and $b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$. For $Ax = b$ to be solvable, which one of the following options is the correct condition on $b_1$, $b_2$, and $b_3$:

$b_1 + b_2 + b_3 = 1$
$3b_1 + b_2 + 2b_3 = 0$
$b_1 + 3b_2 + b_3 = 2$
$b_1 + b_2 + b_3 = 2$

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Correct answer: (B) $3b_1 + b_2 + 2b_3 = 0$

Explanation

The system is solvable when $b$ lies in the column space of $A$, that is, when $b$ is orthogonal to every vector $y$ with $y^TA = 0$. For $y = (3, 1, 2)$, $y^TA = (3 + 1 - 4,\ 3 + 3 - 6) = (0, 0)$. So the condition is $3b_1 + b_2 + 2b_3 = 0$. The other options are not homogeneous, so they cannot describe a subspace.