Question

The adjoint (adjugate) of the matrix \(\begin{bmatrix}1&2 \\ 3&4\end{bmatrix}\) is

• A. \(\begin{bmatrix}4&-2
  [2pt]-3&1\end{bmatrix}\)
• B. \(\begin{bmatrix}4&-3
  [2pt]-2&1\end{bmatrix}\)
• C. \(\begin{bmatrix}4&3
  [2pt]2&1\end{bmatrix}\)
• D. \(\begin{bmatrix}4&2
  [2pt]3&1\end{bmatrix}\)

Hint:

Memorize the \(2\times2\) adjoint: swap the diagonal entries, change signs of the off–diagonals: \(\operatorname{adj}\begin{bmatrix}a&b \\ c&d\end{bmatrix}=\begin{bmatrix}d&-b \\ -c&a\end{bmatrix}\).

Answer 1

The Correct Option is A

Step 1: Recall the 2\(\times\)2 formula. For \(A=\begin{bmatrix}a&b \\ c&d\end{bmatrix}\), \[ \operatorname{adj}(A)=\begin{bmatrix}d&-b \\ -c&a\end{bmatrix}. \] This comes from taking cofactors and then transposing the cofactor matrix (for a \(2\times2\), that reduces to swapping \(a\leftrightarrow d\) and negating the off–diagonals). \\ [4pt]

Step 2: Apply it to \(A=\begin{bmatrix}1&2 \\ 3&4\end{bmatrix}\). Here \(a=1,\ b=2,\ c=3,\ d=4\). Therefore \[ \operatorname{adj}(A)=\begin{bmatrix}4&-2 \\ [2pt]-3&1\end{bmatrix}. \] This matches option (A).