Match List - I with List - II.

Choose the correct answer from the options given below :

(A) - (IV), (B) - (I), (C) - (II), (D) - (III)

(A) $\begin{bmatrix}0 & -5 & 9\\5 & 0 & -3\\-9 & 3 & 0\end{bmatrix}$

for diagonal = $\begin{bmatrix}a & 0 & 0\\0 & b & 0\\0 & 0 & c\end{bmatrix}$

not a diagonal matrix.

for symmetric = A^T ≠ T

$A+\begin{bmatrix}0 & -5 & 9\\5 & 0 & -3\\-9 & 3 & 0\end{bmatrix}⇒A^T=\begin{bmatrix}0 & 5 & -9\\-5 & 0 &3\\9 & -3 & 0\end{bmatrix}$

$A^T=\begin{bmatrix}0 & 5 & -9\\-5 & 0 &3\\9 & -3 & 0\end{bmatrix}⇒-\begin{bmatrix}0 & -5 & 9\\5 & 0 &-3\\-9 & 3 & 0\end{bmatrix}$

$A^T=A⇒\begin{bmatrix}0 & -5 & 9\\5 & 0 & -3\\-9 & 3 & 0\end{bmatrix}$

So its a skew symmetric matrix.

(B) $\begin{bmatrix}5 & 0 & 0\\0 & 5 & 0\\0 & 0 & 5\end{bmatrix}$

its a scalar matrix as its all elements in all principal diagonal are equal to some nor-zero constant.

(C) $\begin{bmatrix}5 & 0 & 0\\0 & -5 & 0\\0 & 0 & 7\end{bmatrix}$

its a diagonal matrix because all its non-zero diagonal elements are zero.

(D) $\begin{bmatrix}3 & -2 & 1\\-2 & -5 & 6\\1 & 6 & 0\end{bmatrix}=A$

$A^T=\begin{bmatrix}3 & -2 & 1\\-2 & -5 & 6\\1 & 6 & 0\end{bmatrix}$

$∴A^T=A$

So it is symmetric matrix.

(A) - (IV), (B) - (I), (C) - (II), (D) - (III)

option 3 is correct.