Q.Evaluate $\begin{vmatrix} x & x+1 \\ x-1 & x \end{vmatrix}$.

The key idea is to evaluate a $2 \times 2$ determinant directly using the formula $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$.

Step 1: Identify the entries: $a = x$, $b = x+1$, $c = x-1$, $d = x$.

Step 2: Apply the formula: …

The determinant of a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $ad - bc$. That’s the only formula you need here — no row operations or factorisations required. The trick is to expand carefully and watch the signs.

1. Write down the determinant formula

   For $\begin{vmatrix} x & x+1 \\ x-1 & x \end{vmatrix}$, we have $a = x$, $b = x+1$, $c = x-1$, $d = x$.

   So the value is $a d - b c = x \cdot x - (x+1)(x-1)$.

2. Simplify the first term

   $x \cdot x = x^2$.

3. Simplify the second term

   $(x+1)(x-1)$ is a difference of squares: $(x+1)(x-1) = x^2 - 1$.

4. Subtract

   $x^2 - (x^2 - 1) = x^2 - x^2 + 1 = 1$. …

Method: Evaluating a 2×2 Determinant with Algebraic Entries

This method evaluates a $2\times2$ determinant whose entries are algebraic expressions rather than plain numbers, using the same $ad-bc$ formula but with careful expansion.

Steps

Step 1: Write down the formula with the algebraic entries substituted

Substitute the given expressions for $a,b,c,d$ exactly as printed, keeping each in its own bracket.

Step 2: Expand each product separately

Expand $ad$ and $bc$ as separate algebraic products before subtracting — don't try to combine the subtraction and the expansion in one step, since that's where sign errors creep in.

Step 3: Recognize special product patterns

Watch for a difference-of-squares pattern like $(x+1)(x-1)=x^2-1$ — spotting it saves a full expansion.

Step 4: Subtract and simplify …

Common Mistakes

Mistake 1: Failing to bracket $(x+1)(x-1)$ before subtracting, leading to a sign error

Why it's wrong: writing the determinant as $x\cdot x - x+1\cdot x-1$ without parentheses around the product $(x+1)(x-1)$ makes it easy to subtract only the first term of the product, not the whole product — turning $x^2-(x^2-1)=1$ into an incorrect expression. Correct approach: always keep the product $bc$ in its own bracket, i.e. write $ad-(bc)$, before expanding anything.

Mistake 2: Not recognizing the difference-of-squares pattern in $(x+1)(x-1)$ …