Q.Using the properties of determinants, evaluate: $\begin{vmatrix} x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4 \end{vmatrix}$

Concept: Determinant Evaluation Using Identities (Row/Column Operations)

Step 1: Apply $C_1 \to C_1 + C_2 + C_3$. Each entry in $C_1$ becomes $(x+4) + x + x = 3x+4$.

Step 2: Factor $(3x+4)$ from $C_1$:

Step 3: Apply $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$: …

Why This Approach Works

When a determinant has a repeating pattern — here every diagonal entry is $x+4$ and every off-diagonal is $x$ — the structure is symmetric. The most efficient way to evaluate it is not to expand directly (which is messy) but to use row operations that exploit the symmetry. Adding all rows to the first row makes every entry in that row identical, which lets us factor out a common term. After that, subtracting rows creates zeros below the diagonal, turning the determinant into an upper triangular form — and the value is just the product of the diagonal entries.

Let’s walk through it.

1. Add rows 2 and 3 to row 1. The determinant is unchanged when we add a multiple of one row to another. Here, we do $R_1 \to R_1 + R_2 + R_3$. The first row becomes:

So the determinant is now:

2. Factor out the common factor $(3x+4)$ from the first row. Since every entry in row 1 is the same, we pull it out:

3. Create zeros in the first column below the pivot. Perform $R_2 \to R_2 - xR_1$ and $R_3 \to R_3 - xR_1$. For row 2:

For row 3:

The determinant becomes: …

Method: Equal-Row-Sum Determinants — Fold Then Factor

Many CBSE determinant problems give a matrix where one value $a$ sits on the diagonal and a different value $b$ fills every off-diagonal entry. Because every row then sums to the same total ($a$ plus $(n-1)$ copies of $b$), the determinant collapses using the same row-sum trick used for any equal-row-sum determinant.

Steps

Step 1: Confirm the row-sum pattern

Add across each row. For a diagonal-$a$, off-diagonal-$b$ matrix of size $n$, every row sums to $a+(n-1)b$. Recognising this pattern up front tells you column-folding will work before you do any algebra.

Step 2: Fold all columns into one

Apply $C_1 \to C_1+C_2+\cdots+C_n$. Every entry of the new first column equals the common row sum, so it factors out:

Step 3: Subtract the first row from every other row

$R_i \to R_i - R_1$ for $i \ge 2$ clears the rest of the first column to zero and turns the remaining block into a simple diagonal (each surviving diagonal entry becomes $a-b$). …

Common Mistakes

Mistake 1: Forgetting to pull the common factor fully out

After $C_1\to C_1+C_2+C_3$ every entry in the first column is $(3x+4)$. Some students leave this factor multiplied into the column instead of taking it outside the determinant, and then accidentally count it twice (once left inside, once written as a multiplier) later on.

Mistake 2: Row-subtraction arithmetic slips …