Q.If $x, y, z \in \mathbb{R}$, then the value of determinant $\begin{vmatrix} (2^x + 2^{-x})^2 & (2^x - 2^{-x})^2 & 1 \\ (3^x + 3^{-x})^2 & (3^x - 3^{-x})^2 & 1 \\ (4^x + 4^{-x})^2 & (4^x - 4^{-x})^2 & 1 \end{vmatrix}$ is equal to ________ .

Concept: Determinant Equality Equation — The determinant is zero because the first two columns are linearly dependent.

Step 1: Expand each entry.

For any $a > 0$,

$(a^x + a^{-x})^2 = a^{2x} + a^{-2x} + 2$,

$(a^x - a^{-x})^2 = a^{2x} + a^{-2x} - 2$.

Step 2: Subtract the second column from the first column.

Column 1 − Column 2 gives:

$(a^{2x} + a^{-2x} + 2) - (a^{2x} + a^{-2x} - 2) = 4$ for each row. …

We are asked to evaluate

The key observation: each row has the same pattern — the first entry is $(a + a^{-1})^2$, the second is $(a - a^{-1})^2$, and the third is $1$, where $a$ is $2^x$, $3^x$, or $4^x$. If we can show that these three numbers are linearly dependent (i.e., one is a linear combination of the other two with constant coefficients), then the rows become dependent and the determinant is zero.

Let’s check:

Subtract them:

So for any $a \neq 0$,

That is, the first entry equals the second entry plus 4. The third entry is just 1. So each row is of the form

Now we can see the linear dependence:

Row$_i$ = (second entry$_i$) $\times (1,1,0) + 4 \times (1,0,0) + 1 \times (0,0,1)$. But more directly, subtract the second column from the first column in the determinant — that operation does not change the determinant’s value.

Let’s do it step by step.

1. Apply column operation: Replace $C_1$ by $C_1 - C_2$.

2. Simplify the new first column: As we computed, each difference is exactly $4$. So $\Delta = \begin{vmatrix} 4 & (2^x-2^{-x})^2 & 1 \\ 4 & (3^x-3^{-x})^2 & 1 \\ 4 & (4^x-4^{-x})^2 & 1 \end{vmatrix}.$ …

Method: Spotting Column (or Row) Dependence Instead of Expanding Directly

When a determinant's columns are built from algebraic expressions that share a hidden identity — most commonly $(u+v)^2-(u-v)^2=4uv$ — check for that identity FIRST. If it turns two columns into (a multiple of) the same column, the determinant is zero immediately, with no further computation needed.

Steps

Step 1: Look for a shared algebraic identity across the columns/rows

Before expanding, examine what each column represents. If one column is $(\text{expr}+\text{expr}^{-1})^2$ and a neighbouring column is $(\text{expr}-\text{expr}^{-1})^2$, recall the identity

and check whether $uv$ is the same constant in every row (here, $u=a,\ v=1/a \Rightarrow uv=1$ regardless of what $a$ is), which is exactly what makes the difference between the two columns a fixed number in every row.

Step 2: Apply a column (or row) operation to make that dependence explicit …

Common Mistakes

Mistake 1: Expanding the determinant directly instead of spotting the identity

Why it's wrong: brute-force cofactor expansion of expressions like $(2^x+2^{-x})^2$ is algebraically heavy and highly error-prone (multiple exponent terms to track across three rows), when the entire determinant collapses to $0$ from a simple structural observation. Correct approach: before expanding, check whether any pair of columns differs by a row-independent constant — that's the signal to stop expanding and use column operations instead.

Mistake 2: Assuming the difference between the two squared columns still depends on the row's variable …