The values of α, for which det([[1, 3/2, α + 3/2], [1, 1/3, α + 1/3], [2α + 3, 3α + 1, 0]]) = 0, lie

Q: Wait, why do we expand along the third row? Can't we just use row operations?

You absolutely could use row operations to create more zeros, but in this specific problem, the third row already has a zero, which makes direct expansion along that row extremely efficient. In JEE, time is your most precious resource; always choose the path that minimizes algebraic steps.

Q: Do I really need to calculate the exact decimal values of the roots to find the interval?

Not necessarily! You can estimate the value of $\sqrt{7}$ (which is between $2$ and $3$, specifically $\approx 2.645$) and use that to bound your roots. However, for absolute certainty, calculating the approximate values is a foolproof way to verify which interval your roots fall into.

Q: What if I made a mistake in the sign of the minors? How do I avoid that?

The sign convention for cofactors is $(-1)^{i+j}$. For the third row, the signs are $(+), (-), (+)$. So, the expansion is $a_{31}(+) - a_{32}(-) + a_{33}(+)$. Always write out the signs explicitly before you start calculating to avoid those frustrating, tiny errors.

Question

The values of $\alpha$, for which $\begin{vmatrix} 1 & \frac{3}{2} & \alpha + \frac{3}{2} \ 1 & \frac{1}{3} & \alpha + \frac{1}{3} \ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0$, lie in the interval

Options:

(A) $(-2, 1)$

(B) $(-3, 0)$

(C) $(-\frac{3}{2}, \frac{3}{2})$

(D) $(0, 3)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex, intimidating wall of numbers—a $3 	imes 3$ determinant. It looks like a fortress, but every fortress has a weak point. In this problem, that weak point is the third row.

We are not just crunching numbers; we are peeling back the layers of a mathematical structure to find the hidden values of $\alpha$.

### The Strategic Choice

When you first see the determinant equation:

$$
\begin{vmatrix} 1 & \frac{3}{2} & \alpha + \frac{3}{2} \ 1 & \frac{1}{3} & \alpha + \frac{1}{3} \ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0
$$

Your instinct might be to panic. Don't. Look at that zero in the bottom right corner; it is a gift.

By choosing to expand along the third row ($R_3$), we effectively turn a complex $3 	imes 3$ problem into a much simpler $2 	imes 2$ problem. The expansion formula is $a_{31}C_{31} + a_{32}C_{32} + a_{33}C_{33} = 0$.

Because $a_{33} = 0$, the entire third term vanishes. We are left with:

$$
(2\alpha + 3) \begin{vmatrix} \frac{3}{2} & \alpha + \frac{3}{2} \ \frac{1}{3} & \alpha + \frac{1}{3} \end{vmatrix} - (3\alpha + 1) \begin{vmatrix} 1 & \alpha + \frac{3}{2} \ 1 & \alpha + \frac{1}{3} \end{vmatrix} = 0
$$

### The Algebraic Dance

Now, let's tackle these minors. For the first minor, we cross-multiply: $\frac{3}{2}(\alpha + \frac{1}{3}) - \frac{1}{3}(\alpha + \frac{3}{2})$.

As you expand this, watch the magic happen. The constant terms $\frac{1}{2}$ and $\frac{1}{2}$ appear with opposite signs and cancel out completely. We are left with $\frac{3\alpha}{2} - \frac{\alpha}{3}$, which simplifies to $\frac{7\alpha}{6}$.

For the second minor, the process is even more satisfying: $1(\alpha + \frac{1}{3}) - 1(\alpha + \frac{3}{2})$. The $\alpha$ terms subtract to zero, leaving us with $\frac{1}{3} - \frac{3}{2} = -\frac{7}{6}$.

Substituting these back into our expansion, we get:

$$
(2\alpha + 3)\left(\frac{7\alpha}{6}ight) - (3\alpha + 1)\left(-\frac{7}{6}ight) = 0
$$

### The Quadratic Climax

Notice that $\frac{7}{6}$ is a common factor in both terms. We can factor it out:

$$
\frac{7}{6} [ \alpha(2\alpha + 3) + (3\alpha + 1) ] = 0
$$

Since $\frac{7}{6} 
eq 0$, the expression inside the bracket must be zero. Expanding this gives us $2\alpha^2 + 3\alpha + 3\alpha + 1 = 0$, which simplifies to the standard quadratic:

$$
2\alpha^2 + 6\alpha + 1 = 0
$$

Using the quadratic formula, we find the roots:

$$
\alpha = \frac{-6 \pm \sqrt{36 - 8}}{4} = \frac{-6 \pm \sqrt{28}}{4} = \frac{-3 \pm \sqrt{7}}{2}
$$

**The final values for $\alpha$ are $\frac{-3 + \sqrt{7}}{2}$ and $\frac{-3 - \sqrt{7}}{2}$.** Both of these values sit comfortably within the interval $(-3, 0)$.

Animated Solution for Mathematics - Matrices and Determinants: The values of $α$ , for which $11 2 α + 3 \frac{3}{2} \frac{1}{3} 3 α + 1 α + \frac{3}{2} α + \frac{1}{3} 0 = 0$ , lie in the interval

Select Answer:

Visualized Solution

The Determinant Equation

Expansion Strategy

Expanding Along Row $3$

Evaluating the First Minor

Evaluating the Second Minor

Substituting Minors Back

Factoring and Simplifying

Forming the Quadratic Equation

Solving the Quadratic Equation

Simplifying the Roots

Approximating the Values

Determining the Interval

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Strategic Choice

The Algebraic Dance

The Quadratic Climax

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The values of α, for which ​112α+3​23​31​3α+1​α+23​α+31​0​​=0, lie in the interval