Question

If \( P = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix} \) is the modal matrix of \( A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{pmatrix} \) then the sum of all the elements of \( P^{-1}AP \) is

• A. 4
• B. 5
• C. 6
• D. 3

Hint:

For triangular matrices, eigenvalues are the diagonal entries. Use properties of diagonalization to simplify such problems.

Answer 1

The Correct Option is C

To find the sum of all elements of \( P^{-1}AP \), we need to determine \( P^{-1} \) and then compute \( P^{-1}AP \).

Step 1: Compute \( P^{-1} \)

Given \( P = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix} \), an upper triangular matrix. The inverse of an upper triangular matrix is also upper triangular, and we calculate it by back substitution:

\( P^{-1} = \begin{pmatrix} 1 & -1 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix} \)

Step 2: Compute \( P^{-1}AP \)

Given \( A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{pmatrix} \), compute \( P^{-1}AP \):

1. Compute \( AP \):

\( AP = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 2 & 6 \\ 0 & 0 & 3 \end{pmatrix} \)

2. Compute \( P^{-1}AP \):

\( P^{-1}AP = \begin{pmatrix} 1 & -1 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 0 & 2 & 6 \\ 0 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \)

Step 3: Sum of all elements of \( P^{-1}AP \)

The matrix \( P^{-1}AP = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \) is diagonal, and the sum of its elements is the trace:

Trace = \( 1 + 2 + 3 = 6 \).

Thus, the sum of all elements of \( P^{-1}AP \) is 6.