If \(A=\left[\begin{array}{cc} 1 & 3 \\ 0 & -1 \end{array}\right]\), then trace of 5A⁵ + 4A⁴ + 3A³ + 2A² + A + I² is :

Concept:

The trace of a matrix expression is the sum of the diagonal elements of the resulting matrix. The trace of a sum of matrices is the sum of their traces. Also, the trace of any power of a matrix can be computed using the cyclic properties of matrix multiplication if eigenvalues are known or the matrix is triangularizable.

Calculation:

Given:

\( A = \begin{bmatrix} 1 & 3 \\ 0 & -1 \end{bmatrix} \)

We are to find:

\( \text{Trace}(5A^5 + 4A^4 + 3A^3 + 2A^2 + A + I^2) \)

Since A is upper triangular, powers of A remain upper triangular, and the trace of Aⁿ is simply the sum of powers of diagonal elements.

Diagonal elements of A: 1 and -1

So, for each power:

• \( \text{Trace}(A^n) = 1^n + (-1)^n \)

Now compute each term:

• \( \text{Trace}(A^1) = 1 + (-1) = 0 \)
• \( \text{Trace}(A^2) = 1 + 1 = 2 \)
• \( \text{Trace}(A^3) = 1 - 1 = 0 \)
• \( \text{Trace}(A^4) = 1 + 1 = 2 \)
• \( \text{Trace}(A^5) = 1 - 1 = 0 \)
• \( \text{Trace}(I^2) = \text{Trace}(I) = 2 \)

Now compute total:

\( 5 \cdot 0 + 4 \cdot 2 + 3 \cdot 0 + 2 \cdot 2 + 0 + 2 = 0 + 8 + 0 + 4 + 0 + 2 = 14 \)