How to find the values of $a$ that make $\det A= 0$? $$A=\begin{pmatrix}a &amp; 6 &amp; 7\\
a &amp;-8 &amp; 9\\
3 &amp;-9 &amp; a\end{pmatrix}$$
  
  Find the values of $a$ so that $\det A=0$


When I do it I get $a = 5.54$ or $a = -4.25$  but when I substitute those values back to find determinant I don't get $\det A=0$
Just solve the following equation.
$$a\cdot(-8)\cdot a+3\cdot6\cdot9+7\cdot a\cdot(-9)-3\cdot(-8)\cdot7-a\cdot(-9)\cdot9-a\cdot a\cdot6=0.$$
I got $$7a^2-9a-165=0,$$ which gives $$\left\{\frac{9+\sqrt{4701}}{14},\frac{9-\sqrt{4701}}{14}\right\}.$$
Now, if you'substitute these values, you'll get that the determinat is equal to $0$.

You got that $\det A\neq0$ because you took an approximated values of $a$.
For a matrix $$\begin{pmatrix}a&amp;b&amp;c\\d&amp;e&amp;f\\g&amp;h&amp;i\end{pmatrix}$$ the determinant is equal to $$a(ei-fh)-b(di-fg)+c(dh-eg)$$ 

For your example, $$A=\begin{pmatrix}a&amp;6&amp;7\\a&amp;-8&amp;9\\3&amp;-9&amp;a\end{pmatrix}$$

We put this into the formula and get

\begin{align}\det A &amp;= a(-8a-(-9\times 9))-6(a\times a-(9\times 3))+7(-9a-(-8\times 3))\\
&amp;=a(-8a+81)-6(a^2-27)+7(-9a+24)\\
&amp;=-8a^2+81a-6a^2+162-63a+168\\
&amp;=-14a^2+18a+330\end{align}

Now we need to solve this for $\det A=0$, so we say \begin{align}-14a^2+18a+330&amp;=0\\
14a^2-18a-330&amp;=0\end{align}

This is a quadratic equation, so we can solve this in the normal way to get \begin{align}a_1&amp;=\frac1{14}(9-\sqrt{4701})\\
a_2&amp;=\frac 1{14}(9+\sqrt{4701})\end{align}

If we round these to 2 decimal places, as you have done, then we get \begin{align}a_1&amp;\approx-4.25\\
a_2&amp;\approx5.54\end{align}

which are the values that you had found. 

When checking the answers, you need to use the exact result, i.e. the ones with the square roots in them, otherwise you will encounter rounding errors which is presumably where your confusion is coming from.