Checking whether a matrix is invertible

Question

Consider the matrix $$A(x)=\left (\begin{matrix}1+x^2&7&11\\3x&2x&4\\8x&17&13\end{matrix}\right)$$ How do I show without calculating the determinant explicitly, whether $A(x)$ is invertible for all $x\in \Bbb{R}$?

Answer 1

It is easy to observe (without computing coefficients: $cx^3$ may be obtained only on the main diagonal) that the determinant is a polynomial of degree 3, hence it has at least one real root. So the answer is negative.

Answer 2

Solve the system $$ \left (\begin{matrix}1+x^2&7&11\\3x&2x&4\\8x&17&13\end{matrix}\right)\pmatrix{a\\b\\c}=\pmatrix{0\\0\\0} $$ If it has solutions other than $a=b=c=0$ then the matrix is singular.

Answer 3

The determinant of $A$ is 3rd degree polynomial in $x$ so there is atleast 1 value of x for which the matrix isn't invertible.

Answer 4

Note: $$A^{-1}=\frac{\operatorname{Adj}A}{\det(A)}.$$ So, whenever the determinant of a matrix is zero, it is not invertible.

Hence: $$\begin{vmatrix}1+x^2&7&11\\3x&2x&4\\8x&17&13\end{vmatrix}=26x^3-244x^2+538x-68=0 \Rightarrow x_{1,2,3}=0.13;3.23;6.02.$$