Question

Let \( M \) and \( N \) be any two \( 4 \times 4 \) matrices with integer entries satisfying

\[ MN = \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} = 2 \]

Then the maximum value of \( \det(M) + \det(N) \) is ...............

Hint:

When multiplying matrices, the determinant of the product is the product of the determinants. Use this property to simplify the problem.

Answer 1

Correct Answer: 17

Step 1: Understand the problem.
We are given that \( M \) and \( N \) are \( 4 \times 4 \) matrices such that their product equals the matrix

\[ MN = \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}. \]

We are asked to find the maximum value of \( \det(M) + \det(N) \).
Step 2: Use properties of determinants.
From the properties of determinants, we know that:

\[ \det(MN) = \det(M) \cdot \det(N). \]

We can calculate \( \det(MN) \). Since the given matrix is a square matrix, we compute:

\[ \det(MN) = \det\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} = 1. \]

Thus, we have:

\[ \det(M) \cdot \det(N) = 1. \]

This means that the product of the determinants of \( M \) and \( N \) is 1.
Step 3: Maximize \( \det(M) + \det(N) \).
Since \( \det(M) \cdot \det(N) = 1 \), the two determinants must be multiplicative inverses of each other. Let \( \det(M) = x \), so \( \det(N) = \frac{1}{x} \). We want to maximize:

\[ x + \frac{1}{x}. \]

To find the maximum, we differentiate \( x + \frac{1}{x} \) and set it equal to zero:

\[ \frac{d}{dx}\left(x + \frac{1}{x}\right) = 1 - \frac{1}{x^2}. \]

Setting this equal to zero:

\[ 1 - \frac{1}{x^2} = 0 \quad \Rightarrow \quad x^2 = 1 \quad \Rightarrow \quad x = \pm 1. \]

Thus, the maximum value of \( x + \frac{1}{x} \) occurs when \( x = 1 \), and we get:

\[ 1 + \frac{1}{1} = 2. \]

Step 4: Conclusion.
Therefore, the maximum value of \( \det(M) + \det(N) \) is:

\[ \boxed{2}. \]