Let A be a 2 × 2 real matrix with detA = 1 and trace A = 3. What is the value of trace A²?

Calculation: 

Let a and b are two eigen values of A .

a + b = 3   . . . . .  . 1

ab = 1  . . . . .  . . 2

(a - b )² = (a +b)² - 4ab 

(a-b ) ² = 5 

a - b = \(\sqrt{5}\)   . . . .  3 

now from equation 1 and 3 , we get 

a = \(\frac{3 + \sqrt{5}}{2}\)  and  b = \(\frac{3 - \sqrt{5}}{2}\) 

trace ( A² ) = a² + b² = 7 

Hence option (4) is correct

Alternate Method

Method -I Let λ₁ and λ₂ be two eigen values of A. 

Then det A = λ₁ λ₂ = 1

& Trace A = λ₁ + λ₂ = 3

And, Eigen values of A² are λ₁², λ₂².

⇒ Trace (A²) = λ12 + λ22

Now, (λ₁ + λ₂)² = λ12 + λ22 + 2λ₁λ₂

⇒ (3)² = λ₁² + λ₂² + 2

⇒ λ12 + λ22 = 9 - 2 = 7

Hence option (4) is correct

Method - II

Ch_A (x) = x² - Trace (A) x + Det (A) = 0 

⇒ x² - 3x + 1 = 0 ⇒ \(x=\frac{3 \pm \sqrt{9-4}}{2}\)

\(\Rightarrow λ_1=\frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}=\) λ₂ which are eigenvalues of A .

then eigenvalues of A² are,

\(λ_1{ }^2=\left(\frac{3+\sqrt{5}}{2}\right)^2 =\frac{9+5+6 \sqrt{5}}{4}=\frac{14+6 \sqrt{5}}{4}\)

\(+\lambda_2^2=\left(\frac{3-\sqrt{5}}{2}\right)^2=\frac{9+5-6 \sqrt{5}}{4}=\frac{14-6 \sqrt{5}}{4}\)

Then trace \(\left(A^2\right)=\lambda_1^2+\lambda_2^2=\frac{14+6 \sqrt{5}}{4}+\frac{14-6 \sqrt{5}}{4}=\frac{28}{4}\) = 7

Hence option (4) is correct