For the matrix

\(A = \left[ {\begin{array}{*{20}{c}} 1&4\\ 2&3 \end{array}} \right]\)

the expression A⁵ – 4A⁴ – 7A³ + 11A² – A – 10I is equivalent to

Concept:

Cayley-Hamilton theorem: According to the Cayley-Hamilton theorem, every matrix 'A' satisfies its own characteristic equation.

Characteristic equation: If A is any square matrix of order n, we can form the matrix [A – λI], where I is the nth order unit matrix. The determinant of this matrix equated to zero i.e. |A – λI| = 0 is called the characteristic equation of A.

Calculation:

Characteristic equation: |A – λI| = 0

\(A - \lambda I = \left[ {\begin{array}{*{20}{c}} 1&4\\ 2&3 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} \lambda &0\\ 0&\lambda \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {1 - \lambda }&4\\ 2&{3 - \lambda } \end{array}} \right]\)

\(\left| {A - \lambda I} \right| = \left| {\begin{array}{*{20}{c}} {1 - \lambda }&4\\ 2&{3 - \lambda } \end{array}} \right| = 0\)

⇒ (1 – λ) (3 – λ) – 8 = 0

⇒ λ² – 4 λ – 5 = 0

From Cayley-Hamilton theorem,

A² – 4 A – 5I = 0

⇒ A² = 4A + 5I

By multiplying with A, we get

⇒ A³ = 4A² + 5 A = 4 (4A + 5I) + 5A = 21 A + 20I

By multiplying with A, we get

⇒ A⁴ = 21 A² + 20 A = 104 A + 105I

By multiplying with A, we get

⇒ A⁵ = 104 A² + 105 A = 521 A + 520I

Now, consider the given expression: A⁵ – 4A⁴ – 7A³ + 11A² – A – 10I

= 521A + 520I – 4 (104 A + 105I) – 7(21A + 20I) + 11 (4A + 5I) – A – 10I

= A + 5I