Which of the following are correct?(A) If A= […] , then A^-1= […] (B)…

The correct answer is Option (3) → (A) and (D) only

For (A), let \( A=\begin{bmatrix}7 & 5\\4 & 3\end{bmatrix} \):

    \( \displaystyle |A|=21-20=1 \)

    \( \displaystyle A^{-1}=\frac{1}{|A|}\,\mathrm{adj}\,A \)

    \( \displaystyle =\frac{1}{1}\begin{bmatrix}3 & -5\\-4 & 7\end{bmatrix} \)

For (B), the determinant of \( \begin{bmatrix}3 & 5\\6 & 10\end{bmatrix} \) is

    \( \displaystyle 30-30=0 \)

For (D), \( A=A^T \) means

    \( \displaystyle \begin{bmatrix}6 & x\\y & 0\end{bmatrix}=\begin{bmatrix}6 & y\\x & 0\end{bmatrix} \)

So the inverse in (A) is \( \begin{bmatrix}3 & -5\\-4 & 7\end{bmatrix} \), the matrix in (B) has determinant 0, the matrix in (C) has \( a_{12}=-5 \) but \( a_{21}=5 \), and (D) gives x = y.

Checking the statements

(A) If \( A=\begin{bmatrix}7 & 5\\4 & 3\end{bmatrix} \), then \( A^{-1}=\begin{bmatrix}3 & -5\\-4 & 7\end{bmatrix} \) — Correct. |A| = 1, so \( A^{-1} \) equals adj A.

(B) If \( A=\begin{bmatrix}3 & 5\\6 & 10\end{bmatrix} \), then \( A \) is a non-singular matrix — Incorrect. |A| = 30 − 30 = 0, so A is singular (its second row is twice the first).

(C) \( A=\begin{bmatrix}0 & -5\\5 & 3\end{bmatrix} \) is a symmetric matrix — Incorrect. \( a_{12}=-5 \) but \( a_{21}=5 \), so \( A\ne A^T \).

(D) If \( A=\begin{bmatrix}6 & x\\y & 0\end{bmatrix} \) and \( A=A^T \), then \( x=y \) — Correct. Comparing the (1, 2) entries of A and \( A^T \) gives x = y.

Shortcut: For (A), multiply instead of inverting:

    \( \displaystyle \begin{bmatrix}7 & 5\\4 & 3\end{bmatrix}\begin{bmatrix}3 & -5\\-4 & 7\end{bmatrix} \)

    \( \displaystyle =\begin{bmatrix}21-20 & -35+35\\12-12 & -20+21\end{bmatrix} \)

    \( \displaystyle =I \)