Question

If \( A = \begin{bmatrix} 2 & n \\ 1 & 4 \end{bmatrix} \) such that \( A^3 = 27 \begin{bmatrix} 4 & q \\ p & r \end{bmatrix} \), then \( p + q + r \) equals ________

Hint:

For matrix algebra, be very careful with the multiplication process (row by column). An alternative to direct multiplication for higher powers is the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic equation. This can sometimes simplify calculations.

Answer 1

Correct Answer: 109

Step 1: Understanding the Concept:
This problem involves matrix multiplication and solving a system of equations derived from the equality of two matrices.

Step 2: Key Formula or Approach: 
1. Calculate \(A^2\) and then \(A^3\) by performing matrix multiplication. 
2. Equate the resulting matrix for \(A^3\) with the given expression \( 27 \begin{bmatrix} 4 & q \\ p & r \end{bmatrix} \). 
3. Solve for the variables \(n, p, q,\) and \(r\) by comparing the corresponding elements of the matrices. 
4. Calculate the final sum \(p + q + r\).

Step 3: Detailed Explanation: 
First, calculate \(A^2\): 

\[ A^2 = A \cdot A = \begin{bmatrix} 2 & n \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 2 & n \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} (2)(2)+(n)(1) & (2)(n)+(n)(4) \\ (1)(2)+(4)(1) & (1)(n)+(4)(4) \end{bmatrix} = \begin{bmatrix} 4+n & 6n \\ 6 & n+16 \end{bmatrix} \] Next, calculate \(A^3\): 

\[ A^3 = A^2 \cdot A = \begin{bmatrix} 4+n & 6n \\ 6 & n+16 \end{bmatrix} \begin{bmatrix} 2 & n \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} (4+n)(2)+(6n)(1) & (4+n)(n)+(6n)(4) \\ (6)(2)+(n+16)(1) & (6)(n)+(n+16)(4) \end{bmatrix} \] \[ A^3 = \begin{bmatrix} 8+2n+6n & 4n+n^2+24n \\ 12+n+16 & 6n+4n+64 \end{bmatrix} = \begin{bmatrix} 8+8n & n^2+28n \\ n+28 & 10n+64 \end{bmatrix} \] Given that \[ A^3 = 27 \begin{bmatrix} 4 & q \\ p & r \end{bmatrix} = \begin{bmatrix} 108 & 27q \\ 27p & 27r \end{bmatrix} \] By equating corresponding elements:

1. \( 8+8n = 108 \implies 8n = 100 \implies n = \frac{25}{2} \)
2. \( n+28 = 27p \)
3. \( n^2+28n = 27q \)
4. \( 10n+64 = 27r \)

Substitute \(n = \frac{25}{2}\): 

For \(p\): 
\[ 27p = \frac{25}{2} + 28 = \frac{81}{2} \implies p = \frac{81}{54} = \frac{3}{2} \] For \(q\): 
\[ 27q = \frac{25}{2} \times \frac{81}{2} = \frac{2025}{4} \implies q = \frac{2025}{108} = \frac{75}{4} \] For \(r\): 
\[ 27r = 10\left(\frac{25}{2}\right) + 64 = 125 + 64 = 189 \implies r = \frac{189}{27} = 7 \] Finally, \[ p + q + r = \frac{3}{2} + \frac{75}{4} + 7 = \frac{109}{4} \]

Step 4: Final Answer: 
The value of \(p + q + r\) is: \[ \boxed{\frac{109}{4}} \]