Q.Find the value of the following:

Concept: Determinant Evaluation Using Identities — for a $2 \times 2$ matrix $\begin{vmatrix} p & q \\ r & s \end{vmatrix}$, the value is $ps - qr$.

Step 1: For (i), identify $p = 2$, $q = -4$, $r = 0$, $s = 3$.

Step 2: Compute $ps - qr = (2)(3) - (-4)(0) = 6 - 0 = 6$.

Step 3: For (ii), identify $p = a$, $q = c$, $r = b$, $s = d$.

Step 4: Compute $ps - qr = a \cdot d - c \cdot b = ad - bc$.

The determinant of a $2\times 2$ matrix is one of the simplest and most fundamental ideas in linear algebra. It tells you, geometrically, the signed area of the parallelogram formed by the two row vectors (or column vectors). But for computation, we just need a clean formula.

The pattern is: multiply the main diagonal (top-left to bottom-right) and subtract the product of the other diagonal (top-right to bottom-left). That's all there is to it.

Let's apply it to each part.

1. For part (i), the matrix is $\begin{pmatrix} 2 & -4 \\ 0 & 3 \end{pmatrix}$.

Here $p = 2$, $q = -4$, $r = 0$, $s = 3$.

Using the formula:

2. For part (ii), the matrix is $\begin{pmatrix} a & c \\ b & d \end{pmatrix}$.

Here $p = a$, $q = c$, $r = b$, $s = d$.

Using the formula:

Method: Evaluating a $2\times2$ Determinant With Literal or Numeric Entries

Applies to any $2\times2$ determinant, whether entries are plain numbers or single letters.

Steps

Step 1: Read off the entries exactly as printed, position by position

Label the determinant $\begin{vmatrix} p & q \\ r & s \end{vmatrix}$ using the actual symbols in their actual positions — don't assume a matrix always uses $a,b,c,d$ in that order.

Step 2: Apply the formula $ps-qr$

Multiply the main-diagonal entries, then subtract the product of the off-diagonal entries.

Step 3: Simplify carefully

For numeric entries, compute directly; for literal entries, leave the answer as the algebraic expression $ps-qr$ in terms of whatever letters were actually given.

Common Mistakes

Mistake 1: Writing the literal determinant $\begin{vmatrix}a&c\\b&d\end{vmatrix}$ as $ab-cd$ by matching positions to the familiar template $\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc$

Why it's wrong: the letters here are arranged differently — top-right is $c$ and bottom-left is $b$ — so blindly reusing the $a,b,c,d$ template without relabeling gives the wrong pairing of entries. Correct approach: always relabel the four positions as $p,q,r,s$ (top-left, top-right, bottom-left, bottom-right) first, then apply $ps-qr$, regardless of what letters happen to sit in those positions.

Mistake 2: Treating a zero entry as making the whole term "disappear" rather than confirming the product is genuinely zero

Why it's wrong: when one entry is $0$ (as in part (i)), it's easy to skip verifying that the corresponding product really is zero and accidentally drop a nonzero cross-term instead. Correct approach: compute both diagonal products explicitly, even when one entry is $0$, before subtracting.