Animated Solution for Mathematics - Matrices and Determinants: Let $f(x)=\left|\begin{array}{ccc}a& -1& 0\\ ax& a& -1\\ a{x}^2& ax& a\end{array}\right|,a\in R$. Then the sum of which the squares of all the values of $a$ for $2f^{\prime }(10)-f^{\prime }(5)+100=0$ is :

We are given a function $f(x)$ defined as a determinant:

The first step is to transform this matrix into a simple polynomial. Determinants are functions in disguise, and we can simplify our work by expanding along the first row to take advantage of the zero in the top-right corner.

Expanding along the first row, we get:

Evaluating these $2\times 2$ minors requires precision. The first minor is $(a^2-(-ax))=(a^2+ax)$.

The second minor is $(ax(a)-(-1)(a{x}^2))=(a^{2}x+a{x}^2)$.

Putting it all together, we get:

Multiplying this out, we find $f(x)=a^3+a^{2}x+a^{2}x+a{x}^2$, which simplifies to $f(x)=a{x}^2+2a^{2}x+a^3$.

If you look closely, you will see that this is a perfect square. Factoring out an $a$, we get:

Now that we have our function $f(x)=a(x+a{)}^2$, the calculus part becomes a breeze. We need the derivative $f^{\prime }(x)$.

Using the power rule and the chain rule, we differentiate with respect to $x$:

The problem asks us to evaluate this derivative at $x=10$ and $x=5$. Substituting these values, we get:

We are given the condition $2f^{\prime }(10)-f^{\prime }(5)+100=0$. Let's plug in our expressions:

Expanding this, we get $40a+4a^2-10a-2a^2+100=0$. Combining like terms, we arrive at:

Dividing by 2, we obtain the quadratic equation:

Factoring this, we find $(a+10)(a+5)=0$, which gives us the roots $a=-10$ and $a=-5$.

The question asks for the sum of the squares of these values: