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Home » Next, we set the derivative equal to zero to find the critical points: - Upplift

Next, we set the derivative equal to zero to find the critical points: - Upplift

📅 February 28, 2026 👤 scraface

Feb 28, 2026

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📌 Case 2:** \( a + d = 0 \)

📌 Let \( d = -a \). Substituting into equations 1 and 4:

📌 \( bc + a^2 = 1 \)

📌 Both equations are identical. So, \( a^2 + bc = 1 \) holds. Thus, the conditions are \( d = -a \) and \( a^2 + bc = 1 \).

📌 The valid matrices \( T \) are:

📌 T = egin{pmatrix} a & b \ c & -a \end{pmatrix} \quad ext{with} \quad a^2 + bc = 1

📌 oxed{egin{cases} T = egin{pmatrix} a & b \ c & -a \end{pmatrix}, & a^2 + bc = 1 \end{cases}}

📌 Question:** A mathematician working in applied mathematics at a financial institution models compound interest with a discrete-time model. Suppose an investment grows according to the recurrence \( A_{n+1} = r A_n \) with \( A_0 = P_0 \). If \( A_5 = 3P_0 \), find the annual growth rate \( r \) and the smallest integer \( n \) such that \( A_n > 10P_0 \).

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Substitute \( d = 20 \) and \( v = 8 \) into the equation:
= 20a + b
5a + b &= 3 \quad \text{(Equation 1)} \\
20a + b &= 8 \quad \text{(Equation 2)}
15a = 5
a = \frac{5}{15} = \frac{1}{3}
Substitute \( a = \frac{1}{3} \) back into Equation 1:
5\left(\frac{1}{3}\right) + b = 3
\frac{5}{3} + b = 3
b = 3 - \frac{5}{3} = \frac{9}{3} - \frac{5}{3} = \frac{4}{3}

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