Question

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 M\[\]
B. 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\[\]

Answer 1

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}} $ .