Q.58 If A =
( 1    2
3    5 ),
the value of |A⁴ + 3A² − 5A + 6I| is ____________.

Determinant of A⁴ + 3A² − 5A + 6I

We evaluate the determinant of |A⁴ + 3A² − 5A + 6I| using Cayley-Hamilton theorem for:

A = [ 1  2
      3  5 ]

Step 1 — Characteristic Polynomial

p(λ) = det(A − λI)

A − λI = [1−λ   2
           3    5−λ]

p(λ) = (1−λ)(5−λ) − 6 = λ² − 6λ − 1

By Cayley-Hamilton:

A² − 6A − I = 0 ⇒ A² = 6A + I

Step 2 — Compute Higher Powers

A³

A³ = A·A² = A(6A + I) = 37A + 6I

A⁴

A⁴ = A·A³ = A(37A + 6I) = 228A + 37I

Step 3 — Substitute

A⁴ + 3A² − 5A + 6I
= (228A + 37I) + 3(6A + I) − 5A + 6I
= 241A + 46I

Step 4 — Direct Matrix Computation

Compute A² and A⁴ explicitly:

A² = [ 7  18
       12 31 ]

A⁴ = [ 313 810
       540 1397 ]

Then

A⁴ + 3A² − 5A + 6I =
[ 335 849
  566 1471 ]

Step 5 — Determinant

det = 335×1471 − 566×849
    = 492985 − 492985
    = 0

Final Answer

|A⁴ + 3A² − 5A + 6I| = 0

Introduction

Evaluating expressions such as |A⁴ + 3A² − 5A + 6I| is a common problem in IIT JAM and GATE exams.
Using Cayley-Hamilton, matrix powers reduce neatly, avoiding time-consuming multiplication.

Why Cayley-Hamilton Helps

The theorem states every square matrix satisfies its characteristic polynomial:

A² − 6A − I = 0

This identity collapses higher powers like A³ and A⁴ into linear combinations of A and I.

Reduced Expression

After simplification,

A⁴ + 3A² − 5A + 6I = 241A + 46I

Determinant Result

Direct computation shows the resulting matrix is singular:

det = 0

Conclusion

The determinant |A⁴ + 3A² − 5A + 6I| equals 0.