Matrices and Determinants – System of Linear Equations – JEE Main 23 Jan 2026 Shift 2

Solution:

For the system to have infinitely many solutions, we must have $D = D_x = D_y = D_z = 0$.

First, calculate $D$:
$$ D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & a \\ 1 & 2 & 3 \end{vmatrix} = 0 $$

$$ \Rightarrow 1(15 – 2a) – 1(6 – a) + 1(4 – 5) = 0 $$
$$ \Rightarrow 15 – 2a – 6 + a – 1 = 0 $$
$$ \Rightarrow 8 – a = 0 \Rightarrow a = 8 $$

Now, calculate $D_z$ (replacing the z-column with constants):
$$ D_z = \begin{vmatrix} 1 & 1 & 6 \\ 2 & 5 & 36 \\ 1 & 2 & b \end{vmatrix} = 0 $$

$$ \Rightarrow 1(5b – 72) – 1(2b – 36) + 6(4 – 5) = 0 $$
$$ \Rightarrow 5b – 72 – 2b + 36 – 6 = 0 $$
$$ \Rightarrow 3b – 42 = 0 \Rightarrow b = 14 $$

We verify with $D_x$ and $D_y$ for $a = 8$ and $b = 14$:
$$ D_x = \begin{vmatrix} 6 & 1 & 1 \\ 36 & 5 & 8 \\ 14 & 2 & 3 \end{vmatrix} = 0 $$

$$ D_y = \begin{vmatrix} 1 & 6 & 1 \\ 2 & 36 & 8 \\ 1 & 14 & 3 \end{vmatrix} = 0 $$

Since $D = D_x = D_y = D_z = 0$ for $a = 8$ and $b = 14$, the system has infinitely many solutions.

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