 Let M be a 2 x 2 symmetric matrix with integer entries. Then M is invertible if A. The first column of M is the transpose of the second row of MB. The second row of M is the transpose of the first column of M C. M is a diagonal matrix with non-zero entries in the main diagonal D. The product of entries in the main diagonal of M is not the square of an integer Verified
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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.

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 $A=-{{A}^{T}}$ .