Answer

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**Hint:**

We recall that definition of symmetric matrix and the condition for invertible matrices. We take $ M=\left[ \begin{matrix}

a & b \\

c & d \\

\end{matrix} \right] $ and find the condition for symmetric matrix as $ b=c $ and the condition for invertible is $ ad-bc\ne 0 $ .We check each option and see if we obtain the conditions $ b=c,ad-bc\ne 0 $ to find their truth values.\[\]

**Complete step by step answer:**

We know that a symmetric matrix is a square matrix whose transpose is the matrix itself. Let us denote the four integral entries of the given matrix $ M $ as $ a,b,c,d $ as

\[M=\left[ \begin{matrix}

a & b \\

c & d \\

\end{matrix} \right]\]

We take the transpose of the above matrix and have

\[{{M}^{T}}=\left[ \begin{matrix}

a & c \\

b & d \\

\end{matrix} \right]\]

Since we are give in the question $ M $ is symmetric we have;

\[\begin{align}

& M={{M}^{T}} \\

& \Rightarrow \left[ \begin{matrix}

a & b \\

c & d \\

\end{matrix} \right]=\left[ \begin{matrix}

a & c \\

b & d \\

\end{matrix} \right] \\

& \Rightarrow b=c \\

\end{align}\]

So we can write the matrix as

\[M=\left[ \begin{matrix}

a & b \\

b & d \\

\end{matrix} \right]\]

We know that we call matrix invertible when the determinant of the matrix is not zero. So we have;

\[\begin{align}

& \det \left( M \right)\ne 0 \\

& \Rightarrow \det \left( \left| \begin{matrix}

a & b \\

b & d \\

\end{matrix} \right| \right)\ne 0 \\

& \Rightarrow ad-{{b}^{2}}\ne 0 \\

\end{align}\]

Let us check option A. The first column of $ M $ is $ \left[ \begin{matrix}

a \\

c \\

\end{matrix} \right] $ and the transpose of the second row of $ M $ is $ {{\left[ \begin{matrix}

b & d \\

\end{matrix} \right]}^{T}}=\left[ \begin{matrix}

b \\

d \\

\end{matrix} \right] $ . We are given that

\[\begin{align}

& \left[ \begin{matrix}

a \\

b \\

\end{matrix} \right]=\left[ \begin{matrix}

b \\

d \\

\end{matrix} \right] \\

& \Rightarrow a=b=d \\

& \Rightarrow ad-{{b}^{2}}=0 \\

\end{align}\]

So option A is incorrect. Let us check option B. The second row of $ M $ is $ \left[ \begin{matrix}

b & d \\

\end{matrix} \right] $ and the transpose of the first column of $ M $ is $ {{\left[ \begin{matrix}

a \\

b \\

\end{matrix} \right]}^{T}}=\left[ \begin{matrix}

a & b \\

\end{matrix} \right] $ . We are given

\[\begin{align}

& \left[ \begin{matrix}

b & d \\

\end{matrix} \right]=\left[ \begin{matrix}

a & b \\

\end{matrix} \right] \\

& \Rightarrow b=a,d=b \\

& \Rightarrow a=b=d \\

& \Rightarrow ad-{{b}^{2}}=0 \\

\end{align}\]

So option B is incorrect. Let us check option C. Since $ M $ is diagonal matrix with non-zero entries in the main diagonal we have $ b=c=0 $ . So we have

\[\det \left( M \right)=ad-{{b}^{2}}=ad-0=ad\ne 0\left( \because a,d\in Z \right)\]

So option C is correct. Let us check option D. The product of entries in the main diagonal of $ M $ is $ a\times d=ad $ . We are given $ ad $ is not perfect square and hence $ ad\ne {{b}^{2}} $ for any integer $ b $ . So we have

\[\det \left( M \right)=ad-{{b}^{2}}\ne 0\]

So option D is correct. The correct options are C and D. \[\]

**Note:**

We note that we only have equality between matrices $ A=B $ when they have the same order and corresponding elements in the positions are equal which means $ {{a}_{ij}}={{b}_{ij}} $ . We use invertible matrices to solve a system of equations. We find the transpose by writing rows of an original matrix as columns or writing columns of the original matrix as rows. $ A $ is skew-symmetric matrix if $ A=-{{A}^{T}} $ .

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