Find values for x such that A is not invertible.

Question

For $A$ not to be invertible, i.e. $\det(A) = 0$

$A =$ $$ \begin{bmatrix} 3 & 1 & 7-x \\ 3 & 2-x & 4 \\ 4 & 2 & 8 \\ \end{bmatrix} $$

After some row operations (step 1: R1 - R2 => R1 then R2 - R3 => R2 and lastly R1+R2 => R1) I got:

$A =$ $$ \begin{bmatrix} -1 & -1 & 7-x \\ -1 & -x & -4 \\ 4 & 2 & 8 \\ \end{bmatrix} $$ I computed the determinant.
Then checked when it equals $0$ and for values of $x$ I got $\dfrac {1}{4} (19\pm \sqrt{305})$.

Could you please confirm with me whether I am on the right track? Thanks.

Answer 1

I obtain $$detA=-4 x^2 + 6 x + 2=0\implies x=\frac34\pm \frac{\sqrt {17}}4$$

To find the values for which $det A=0$, you could simplify the matrix as follow

$$\begin{bmatrix} 3 & 1 & 7-x \\ 3 & 2-x & 4 \\ 4 & 2 & 8 \\ \end{bmatrix} \to\begin{bmatrix} 3 & 1 & 1-x \\ 3 & 2-x & -2 \\ 2 & 1 & 0 \\ \end{bmatrix} $$

Answer 2

We have $$A = \begin{bmatrix} 3 & 1 & 7-x \\ 3 & 2-x & 4 \\ 4 & 2 & 8 \\ \end{bmatrix} $$ Step 1: R1-R2 -> R1 $$\text{det}A= \left\vert \begin{array}{ccc} 0 & -1+x & 3-x \\ 3 & 2-x & 4 \\ 4 & 2 & 8 \\ \end{array} \right\vert $$ Step 2: R2-R3 -> R2 $$ \text{det}A= \left\vert\begin{array}{ccc} 0 & -1+x & 3-x \\ -1 & -x & -4 \\ 4 & 2 & 8 \\ \end{array}\right\vert $$ Step 3: R3 + $4$ R2 -> R3 $$\text{det}A= \left\vert \begin{array}{ccc} 0 & -1+x & 3-x \\ -1 & -x & -4 \\ 0 & 2-4x & -8 \\ \end{array}\right\vert $$ Then, using minors of the first column, we obtain $$ \text{det}A = (-1) (-1)^{2+1} \left\vert \begin{array}{cc} -1+x &3-x\\ 2-4x &-8 \end{array} \right\vert\\ = 8 - 8x - 6 +2x+12x-4x^2\\ = -4x^2+6x +2\\ =-2 (2x^2 -3x -1)$$ Then, $2x^2 - 3x-1 = 2\left(x-\dfrac{3+\sqrt{17}}{4}\right)\left(x-\dfrac{3-\sqrt{17}}{4}\right)$, if I haven't made a mistake.