Question

solve the equation for x, using the method for evaluating determinants. \\(\

$$\begin{vmatrix} 6 & -2 & -1 \\\\ -6 & x & 0 \\\\ 4 & 5 & 1 \\end{vmatrix}$$

= -12\\)

Explanation:

Step1: Expand the determinant

To expand a 3x3 determinant

$$\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}$$

, we use the formula $a(ei - fh)-b(di - fg)+c(dh - eg)$.
For the given determinant

$$\begin{vmatrix}6&-2&-1\\-6&x&0\\4&5&1\end{vmatrix}$$

, we have:
$6(x\times1 - 0\times5)-(-2)(-6\times1 - 0\times4)+(-1)(-6\times5 - x\times4)$

Step2: Simplify each term

First term: $6(x - 0)=6x$
Second term: $-(-2)(-6 - 0)=-(-2)(-6)=-12$ (because $-2$ times $-6$ is $12$, and then we have a negative sign in front)
Third term: $-1(-30 - 4x)=30 + 4x$ (distributing the $-1$)
Now combine all terms: $6x-12 + 30 + 4x$
Combine like terms: $(6x + 4x)+(-12 + 30)=10x + 18$

Step3: Solve for x

We know the determinant equals -12, so:
$10x + 18=-12$
Subtract 18 from both sides: $10x=-12 - 18=-30$
Divide both sides by 10: $x=\frac{-30}{10}=-3$

Answer:

$x = -3$