Proving a 3x3 matrix is non-singular for all values

Key concept: This problem uses determinant expansion and completing the square to prove that a parameterized matrix has no zero determinant for any real value of the variable.

Step by step

What the problem is asking

A matrix is non-singular when its determinant is not zero. Because the entries of the matrix depend on the variable $x$, the goal is to compute $\det A$ and show that it is always positive for every real $x$.

The key idea is not just finding a formula, but rewriting that formula into a form that makes its sign obvious. That is why completing the square is useful here.

Step-by-step determinant expansion

For

$A=\begin{pmatrix}2 & -2 & 4\\3 & x & -2\\-1 & 3 & x\end{pmatrix},$

expand along the first row:

$\det A=2\begin{vmatrix}x & -2\\3 & x\end{vmatrix}+2\begin{vmatrix}3 & -2\\-1 & x\end{vmatrix}+4\begin{vmatrix}3 & x\\-1 & 3\end{vmatrix}.$

Now evaluate each $2\times 2$ determinant:

$=2(x^2+6)+2(3x-2)+4(9+x).$

Simplify:

$=2x^2+6x+4x+12-4+36$

$=2x^2+10x+44.$

Why completing the square proves non-singularity

Rewrite the quadratic:

$2x^2+10x+44=2\left(x^2+5x+22\right)=2\left(x+\frac{5}{2}\right)^2+\frac{63}{2}.$

Because a square is always nonnegative,

$\left(x+\frac{5}{2}\right)^2\ge 0,$

so

$\det A \ge \frac{63}{2} > 0.$

That means the determinant is never zero, so the matrix is non-singular for all real $x$.

Useful sign test

A quadratic of the form $ax^2+bx+c$ is always positive if its completed-square form has a positive constant term and the coefficient of the square is positive. Here, both conditions hold, so there are no real roots and no value of $x$ that makes the matrix singular.

Pitfall alert

A frequent mistake is to stop after computing $\det A=2x^2+10x+44$ and conclude it is positive just because the leading coefficient is positive. That is not enough for a quadratic; it could still cross zero. You must either complete the square or compute the discriminant. Another error is using the wrong cofactor signs during expansion, which changes the final quadratic completely. It is also important to say clearly that non-singular means determinant nonzero, not merely nonnegative. The completed-square step is what gives the proof its strength.

Try different conditions

If one entry changes, for example replacing the $-2$ in the second row by a parameter $k$, the determinant may no longer be always positive. In that variant, you would recompute $\det A$ as a quadratic in $x$ with coefficients depending on $k$, then check whether its discriminant is negative or whether the completed-square form has a positive minimum. For a different variant where the matrix is symmetric, you could sometimes use eigenvalue arguments instead of direct expansion, but the conclusion still depends on proving the determinant cannot be zero for any real parameter value.

Further reading

cofactor expansion, completed square, non-singular matrix