Question

If \[ A= \begin{bmatrix} 1&2\\ 3&4 \end{bmatrix}, \] then \(\det(A)\) is equal to:

• A. \(-2\)
• B. \(-1\)
• C. \(1\)
• D. \(2\)

Hint:

For a \(2\times2\) matrix, \[ \det \begin{bmatrix} a&b\\ c&d \end{bmatrix} = ad-bc. \] Always multiply the main diagonal first and then subtract the product of the other diagonal.

Answer 1

The Correct Option is A

Step 1: Look at the matrix.
We are given $A=\begin{bmatrix}1&2\\3&4\end{bmatrix}$. We need its determinant.

Step 2: Recall the rule for a $2\times2$ determinant.
For $\begin{bmatrix}a&b\\c&d\end{bmatrix}$, the determinant is $ad-bc$. You multiply the main diagonal and subtract the product of the other diagonal.

Step 3: Read off the entries.
Here $a=1$, $b=2$, $c=3$, $d=4$.

Step 4: Multiply the main diagonal.
The main diagonal product is $a\cdot d=1\times4=4$.

Step 5: Multiply the other diagonal.
The other diagonal product is $b\cdot c=2\times3=6$.

Step 6: Subtract.
\[ \det(A)=ad-bc=4-6=-2. \] So the determinant is negative two. \[ \boxed{-2} \]