If the system of equations x + (√2 sin α)y + (√2 cos α)z = 0, x + (cos α)y + (sin α)z = 0, x + (sin

Q: Wait, why do we set the determinant to zero? Can't we just solve the equations?

That is a great question! In a homogeneous system, the trivial solution $x=0, y=0, z=0$ always exists. If we try to solve it using standard substitution or elimination, we might just end up with $0=0$, which doesn't tell us anything about the non-trivial solutions. By setting the determinant to zero, we are mathematically enforcing the condition that the equations are linearly dependent, which is the only way to have infinitely many solutions (non-trivial ones).

Q: Why do we need to check the domain of the angle?

Trigonometric functions are periodic, meaning there are infinitely many angles that satisfy $\cos \theta = -\frac{1}{2}$. If we don't restrict our search to the given domain $\alpha \in (0, \frac{\pi}{2})$, we might find a solution that is mathematically correct but physically or contextually invalid for the problem. Always map the domain of the variable to the domain of the argument of the trigonometric function!

Q: Is there a trick to expanding the 3x3 determinant faster?

While you can use Sarrus' rule or expansion by minors, the key is to look for patterns. Notice how the rows are structured. Sometimes, adding or subtracting rows (row operations) can create zeros, making the expansion much simpler. In this problem, direct expansion is manageable, but always keep an eye out for row operations that could simplify the matrix before you start calculating.

Question

If the system of equations $x + (\sqrt{2} \sin \alpha)y + (\sqrt{2} \cos \alpha)z = 0$, $x + (\cos \alpha)y + (\sin \alpha)z = 0$, $x + (\sin \alpha)y - (\cos \alpha)z = 0$ has a non-trivial solution, then $\alpha \in (0, \frac{\pi}{2})$ is equal to :

Options:

(A) $\frac{11\pi}{24}$

(B) $\frac{5\pi}{24}$

(C) $\frac{7\pi}{24}$

(D) $\frac{3\pi}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex system of linear equations. You see three equations, all perfectly balanced with zeros on the right-hand side. This is a homogeneous system, and it holds a secret.

It is not just a collection of numbers; it is a geometric statement. For a non-trivial solution to exist, the vectors defined by the coefficients must be linearly dependent. This means the volume they span must be zero.

Mathematically, this is the moment we invoke the determinant: $|A| = 0$. This is our gateway to the solution.

### Constructing the Matrix

The first step is to translate the problem into the language of matrices. We extract the coefficients of $x$, $y$, and $z$ to form our matrix $A$:

$$
\begin{vmatrix}
1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \
1 & \cos \alpha & \sin \alpha \
1 & \sin \alpha & -\cos \alpha
\end{vmatrix} = 0
$$

Take a deep breath. It looks intimidating, but it is just a structure waiting to be dismantled. We expand this determinant along the first row.

The expansion gives us:

$$
1(-\cos^2 \alpha - \sin^2 \alpha) - \sqrt{2} \sin \alpha (-\cos \alpha - \sin \alpha) + \sqrt{2} \cos \alpha (\sin \alpha - \cos \alpha) = 0
$$

### The Beauty of Simplification

Now, watch the magic happen. The first term, $-(\cos^2 \alpha + \sin^2 \alpha)$, is simply $-1$. Our equation becomes:

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

By distributing the terms, we get:

$$
-1 + \sqrt{2} \sin \alpha \cos \alpha + \sqrt{2} \sin^2 \alpha + \sqrt{2} \sin \alpha \cos \alpha - \sqrt{2} \cos^2 \alpha = 0
$$

Grouping the terms, we find:

$$
-1 + 2\sqrt{2} \sin \alpha \cos \alpha - \sqrt{2}(\cos^2 \alpha - \sin^2 \alpha) = 0
$$

This is where the double-angle identities shine. We recognize $2 \sin \alpha \cos \alpha$ as $\sin 2\alpha$ and $\cos^2 \alpha - \sin^2 \alpha$ as $\cos 2\alpha$. The equation collapses into:

$$
-1 + \sqrt{2} \sin 2\alpha - \sqrt{2} \cos 2\alpha = 0
$$

### The Final Transformation

We are almost there. Rearranging gives us $\sqrt{2} \sin 2\alpha - \sqrt{2} \cos 2\alpha = 1$. Dividing by $\sqrt{2}$ and rearranging further, we get:

$$
\cos 2\alpha - \sin 2\alpha = -\frac{1}{\sqrt{2}}
$$

To combine these into a single cosine term, we divide by $\sqrt{2}$ again:

$$
\frac{1}{\sqrt{2}} \cos 2\alpha - \frac{1}{\sqrt{2}} \sin 2\alpha = -\frac{1}{2}
$$

Using the compound angle formula $\cos(A+B) = \cos A \cos B - \sin A \sin B$, we identify $A = 2\alpha$ and $B = \frac{\pi}{4}$. Thus:

$$
\cos\left(2\alpha + \frac{\pi}{4}ight) = -\frac{1}{2}
$$

### The Domain Trap

Finally, we must respect the domain. Given $\alpha \in (0, \frac{\pi}{2})$, we find that $2\alpha + \frac{\pi}{4} \in (\frac{\pi}{4}, \frac{5\pi}{4})$.

In this interval, $\cos 	heta = -\frac{1}{2}$ occurs at $	heta = \frac{2\pi}{3}$. Setting $2\alpha + \frac{\pi}{4} = \frac{2\pi}{3}$, we solve for $\alpha$:

$$
2\alpha = \frac{2\pi}{3} - \frac{\pi}{4} = \frac{5\pi}{12}
$$

This leads us to the final result:

**$\alpha = \frac{5\pi}{24}$**

Animated Solution for Mathematics - Matrices and Determinants: If the system of equations $x + (2 sin α) y + (2 cos α) z = 0$ , $x + (cos α) y + (sin α) z = 0$ , $x + (sin α) y - (cos α) z = 0$ has a non-trivial solution, then $α \in (0, \frac{π}{2})$ is equal to :

Select Answer:

Visualized Solution

Condition for Non-Trivial Solution

Constructing the Determinant

Expanding the Determinant

Simplifying the First Term

Distributing the Coefficients

Applying Double Angle Identities

Rearranging the Equation

Normalizing for Compound Angle

Forming the Cosine Compound Angle

Analyzing the Domain of $α$

Solving for the Argument

Calculating the Final Value of $α$

Conclusion and Takeaways

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

Constructing the Matrix

The Beauty of Simplification

The Final Transformation

The Domain Trap

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