MaxWong

If you simplify the determinant, the equation is just $x^3 - 3x + 2 = 0$. You can solve that by factorizing the left-hand side.

Note that a_n is just an arithmetic sequence. You can use the sum of A.S. formula to rewrite the equation a_1 + a_2 + ... + a_k = 330.

After simplifying the determinant in $\det(tI - A) = 0$, you will get $-40 + 38 t - 11 t^2 + t^3 = 0$. Then you can solve the cubic equation to find the values of t.

Hint: $-40 + 38 t - 11 t^2 + t^3 = (t - 2)(t - 4)(t - 5)$.

Let (n + 1)/n be the first fraction.

$\dfrac{n + 1}n \cdot \dfrac{n + 2}{n + 1} \cdots \dfrac{n+20}{n + 19} = 3$

Note that the left-hand side is just a telescoping product, so $\dfrac{n + 20}n = 3$.

Please solve the equation for n and then find the value of (n + 1)/n on your own.

Directly use the following property of determinants:

Suppose A is a square matrix. Then $\det A = \det(A^{\top})$, where $A^\top$ is the transpose of A.

For a homogeneous system $A \vec x = \vec 0$ to have non-trivial solutions, we must have $\det A = 0$.

Then $\det \begin{pmatrix}1&k&-1\\k&-1&-1\\7&3&-2k\end{pmatrix}= 0$. Please simplify the determinant and solve the resulting cubic equation.

Yes, that is the correct answer.

Use the following property of determinants directly:

Suppose A is a n by n matrix, and c is a real number. Then $\det (cA) = c^n \det(A)$.

What if x = y = 0?

$(f(0))^2 = f(0) + f(0)\\ $

Let x = f(0), then x^2 - 2x = 0. This implies f(0) = 0 or f(0) = 2.

Case 1: f(0) = 2

Substituting x = 0 gives:

$f(0) f(y) = f(y) + f(0)\\ f(y) = 2$

Note that this holds for any real number y. That means the constant function $f(x) = 2$ is a solution to this functional equation.

Case 2: f(0) = 0

Substituting x = 0 gives:

$f(0) f(y) = f(y) + f(0)\\ f(y) = 0$

Note that this holds for any real number y. That means the constant function $f(x) = 0$ is a solution to this functional equation.

The only solutions to this functional equation are $f(x) = 0$ or $f(x) = 2$.

Note that the third row is just the first row plus the second row.

Doing row operation yields $\det \begin{pmatrix} F_{1000} & F_{1001} & F_{1002} \\ F_{1001} & F_{1002} & F_{1003} \\ F_{1002} & F_{1003} & F_{1004} \end{pmatrix}=\det \begin{pmatrix} F_{1000} & F_{1001} & F_{1002} \\ F_{1001} & F_{1002} & F_{1003} \\ 0&0&0 \end{pmatrix}$.

The answer is now obvious if you are familiar with the properties of determinant.