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.