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

Since \( P \) is the modal matrix (matrix of eigenvectors), then \( P^{-1}AP = D \), where \( D \) is a diagonal matrix with eigenvalues of \( A \). From the structure of \( A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{pmatrix}\), it is an upper triangular matrix. The eigenvalues of an upper triangular matrix are on the diagonal: \[ \lambda_1 = 1,\quad \lambda_2 = 2,\quad \lambda_3 = 3 \] So \( P^{-1}AP = D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}\) and the sum of all elements in \( D \) is: \[ 1 + 2 + 3 = 6 \]