Question

Let M be a $2\times 2$ 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

Answer: (C), (D)

The correct options are
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

We know that,
A square matrix is invertible if |A| ≠ 0.
Let
$A=(\begin{array}{cc}a& b\\ b& c\end{array})$
Check through the options:
(a) Given,
$\left(\frac{a}{b}\right)=\left(\frac{b}{c}\right)\Rightarrow a=b=c=k\left(\frac{a}{b}\right)=\left(\frac{b}{c}\right)\Rightarrow a=b=c=k\Rightarrow A=(\begin{array}{cc}k& k\\ k& k\end{array})=0\Rightarrow \left|A\right|=0$
So, A is non-invertible matrix.
(b) Given,
(b c) = (a b)
$\Rightarrow$ a = b = c = k
Again, |A| = 0
So, A is non-invertible matrix.
(c) Given,
$A=(\begin{array}{cc}a& b\\ b& c\end{array})$
$\Rightarrow$ |A| = ac
$\Rightarrow$ |A| ≠ 0
So, A is invertible matrix.
(d) Given,
$A=(\begin{array}{cc}a& b\\ b& c\end{array})$
$\Rightarrow$ |A| = ac - |A| = ac - $b^2\ne$ 0
(Since ac is not equal to square of an integer.)
So, A is invertible in this case.
Out of 4 options ,options (c) and (d) will lead to invertible matrices.