Simultaneous Equations JEE 2024: Determinant Method

Q: What does a non-trivial solution mean in simultaneous equations?

A non-trivial solution has at least one unknown that is non-zero. It does not require every unknown to be non-zero.

Q: Why is α = 5π/24 the only valid angle?

The determinant condition reduces to sin(2α − π/4) = 1/2. The given interval restricts 2α − π/4 to (−π/4, 3π/4), where only the root π/6 is allowed. This gives α = 5π/24.

Q: Does a zero determinant always mean a system has solutions?

No. A zero determinant guarantees non-trivial solutions for a square homogeneous system, but a non-homogeneous system may be inconsistent. In the article's non-homogeneous practice system, k = 5 gives a zero determinant but reduces the equations to the contradiction 0 = 1.

Which first-quadrant angle allows a non-zero solution?

The simultaneous equations JEE 2024 question has exactly one admissible angle: $α = \frac{5 π}{24}$

The determinant gives this answer only when both mixed terms are retained. The task is to find the angle for which the three supplied homogeneous equations admit a solution other than the zero vector, subject to:

0 < α < \frac{π}{2}

Non-trivial means at least one unknown is non-zero, not that all three must be non-zero.

This is a bank-labelled JEE Main 2024 question, tagged hard. Its supplied date, 2023-01-24, belongs to 2023, so the year and date conflict; no exact examination date or shift is asserted here. The bank records answer index 2 but supplies no option texts, so no option list is reconstructed here.

Why must the coefficient determinant be zero?

A square homogeneous system has a non-trivial solution exactly when its coefficient determinant is zero. A non-zero determinant makes the matrix invertible and forces all three unknowns to vanish; a zero determinant guarantees a non-zero solution.

Preserving every coefficient, write:

A = (\begin{matrix} 1 & \sqrt{2} \sin α & \sqrt{2} \cos α \\ 1 & \cos α & \sin α \\ 1 & \sin α & - \cos α \end{matrix}), A (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})

Use short names to keep the first-row expansion readable:

s = \sin α, c = \cos α, D = det A

The first-row cofactor signs are:

+, -, +

Evaluate the three minors separately. Keep the middle cofactor’s minus sign separate from the negative value of its minor.

M_{11} = | \begin{matrix} c & s \\ s & - c \end{matrix} | = - c^{2} - s^{2}

M_{12} = | \begin{matrix} 1 & s \\ 1 & - c \end{matrix} | = - c - s

M_{13} = | \begin{matrix} 1 & c \\ 1 & s \end{matrix} | = s - c

Now expand along the first row:

D = 1 M_{11} - \sqrt{2} s M_{12} + \sqrt{2} c M_{13}

D = - (c^{2} + s^{2}) - \sqrt{2} s (- c - s) + \sqrt{2} c (s - c)

The required condition is: $D = 0$

How do both mixed terms lead to the correct angle?

The two mixed products add; they do not cancel. Retaining both turns the determinant condition into a shifted sine equation with just one root in the allowed interval.

Expand every product:

D = - 1 + \sqrt{2} s^{2} + \sqrt{2} s c + \sqrt{2} s c - \sqrt{2} c^{2}

Combine like terms:

D = - 1 + \sqrt{2} (s^{2} - c^{2}) + 2 \sqrt{2} s c

The supplied solution has a transcription error: its shortened expansion drops one mixed product, $\sqrt{2} s c$

Its stated final answer is correct, but that shortened expression does not justify it. Retain both products from the first-row expansion.

Apply the double-angle identities:

s^{2} - c^{2} = - \cos 2 α, 2 s c = \sin 2 α

Setting the determinant to zero gives:

- 1 + \sqrt{2} (\sin 2 α - \cos 2 α) = 0

$\sqrt{2} (\sin 2 α - \cos 2 α) = 1$ Use the shifted-sine identity:

\sin θ - \cos θ = \sqrt{2} \sin (θ - \frac{π}{4})

Hence:

\sin (2 α - \frac{π}{4}) = \frac{1}{2}

Before choosing a root, translate the original interval. Define:

β = 2 α - \frac{π}{4}

Then:

0 < 2 α < π ⟹ - \frac{π}{4} < β < \frac{3 π}{4}

The complete sine branches are:

β = \frac{π}{6} + 2 k π or β = \frac{5 π}{6} + 2 k π, k \in Z

The second branch’s unshifted root exceeds the upper bound. Every non-zero full-turn shift also lies outside the interval, leaving only: $β = \frac{π}{6}$

Thus:

2 α = \frac{π}{6} + \frac{π}{4} = \frac{5 π}{12} ⟹ α = \frac{5 π}{24}

Writing both branches, then checking the interval, proves uniqueness. The inverse-sine value alone does not.

Does the answer actually make the determinant zero?

Substitution gives zero exactly, and the angle lies strictly inside the required first-quadrant interval. This checks the result without replacing the determinant-expansion method.

At the candidate angle:

2 α - \frac{π}{4} = \frac{10 π}{24} - \frac{6 π}{24} = \frac{π}{6}

The determinant becomes:

D = - 1 + 2 \sin (2 α - \frac{π}{4}) = - 1 + 2 \times \frac{1}{2} = 0

Check the domain:

0 < \frac{5 π}{24} < \frac{12 π}{24} = \frac{π}{2}

The answer is therefore:

α = \frac{5 π}{24}

This verifies the angle itself, not a position in an unseen option list. Both the determinant condition and the interval are satisfied.

What goes wrong if you delete the squared-term difference?

Deleting the squared-term difference produces two false candidates. The unit-circle identity concerns a sum of squares, not their difference, so it cannot justify that deletion.

The incorrect move is:

D = - 1 + \sqrt{2} (s^{2} - c^{2}) + 2 \sqrt{2} s c \overset{incorrect deletion}{\to} - 1 + 2 \sqrt{2} s c

The misconception is:

s^{2} + c^{2} = 1 ⇏ s^{2} - c^{2} = 0

Within the given interval, the difference vanishes only at: $α = \frac{π}{4}$

Setting the wrongly shortened expression to zero gives:

2 \sqrt{2} s c = 1 ⟹ \sin 2 α = \frac{1}{\sqrt{2}}

Its two candidates are:

α = \frac{π}{8} or α = \frac{3 π}{8}

Check them against the actual determinant:

D (\frac{π}{8}) = - 1 + 2 \sin 0 = - 1

D (\frac{3 π}{8}) = - 1 + 2 \sin \frac{π}{2} = 1

Neither determinant is zero, so both candidates fail. Passing the angle interval cannot rescue a root obtained from a false equation.

The option texts were not supplied. These are demonstrated wrong candidates, not verified listed distractors.

How does the determinant test apply to two related systems?

For a homogeneous square system, a zero determinant guarantees non-trivial solutions. For a non-homogeneous system, it does not guarantee consistency. These two original practice questions, not additional JEE PYQs, isolate that difference.

Which parameter gives the homogeneous system a non-trivial solution?

The required parameter is 5. Find it for:

x + y + z & = 0 x + 2 y + 3 z & = 0 x + 3 y + k z & = 0

Subtract the first coefficient row from the second and third:

R_{2} \leftarrow R_{2} - R_{1}, R_{3} \leftarrow R_{3} - R_{1}

These operations preserve the determinant. Expanding down the first column gives:

det A = | \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 2 & k - 1 \end{matrix} | = (k - 1) - 4 = k - 5

Thus:

k = 5

At this value, back-substitution gives the full solution family:

(x, y, z) = (t, - 2 t, t), t \in R

A non-trivial solution requires: $t \neq 0$

What changes when the third right-hand side becomes 1?

The system has a unique solution unless the parameter is 5; at 5, it has no solution. Classify:

x + y + z & = 0 x + 2 y + 3 z & = 0 x + 3 y + k z & = 1

The coefficient determinant is unchanged. The first two equations give:

y = - 2 z, x = z

Substitution into the third gives: $(k - 5) z = 1$

Therefore:

k \neq 5 : (x, y, z) = \frac{(1, - 2, 1)}{k - 5}

At the exceptional value, the third left-hand side equals twice the second minus the first:

x + 3 y + 5 z = 2 (x + 2 y + 3 z) - (x + y + z)

Its right-hand side should therefore be zero, but is 1. The system is inconsistent.

Check the right-hand sides before using the zero-determinant rule: it guarantees non-trivial solutions for a homogeneous square system, but does not establish consistency for a non-homogeneous system.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Keep going with Oscillations and Waves JEE 2019: Four-String Tensions.

Frequently asked questions

What is the answer to the simultaneous equations JEE 2024 question?

For the system discussed, the unique admissible angle is α = 5π/24. It lies in 0 < α < π/2 and makes the coefficient determinant zero.

What does a non-trivial solution mean in simultaneous equations?