Question

Let \(A = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}\) and (mI + nA)² = A where m, n are positive real numbers and I is the identify matrix. What is (m + n) equal to?

Answer 1

Correct Answer - Option 4 : \(\frac{3}{2}\)

Formula used:

Multiplication matrix: 

\(\rm \begin{bmatrix} a & b\\ c & d \end{bmatrix} \times \begin{bmatrix} w & x\\ y & z \end{bmatrix} = \begin{bmatrix} aw + by & ax + bz\\ cw + dy & cx + dz \end{bmatrix}\)

Identity matrix: 

An identity matrix is a given square matrix of any order which contains on its main diagonal elements with a value of one, while the rest of the matrix elements are equal to zero.

I = \(\rm \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \)

Calculation:

\(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

⇒  \(I^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

\(A = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}\)

⇒  A² = \( \begin{bmatrix} -4 & 0 \\ 0 & -4 \end{bmatrix}\)

According to question,

(mI + nA)2 = A

⇒ m²I² + n²A² + 2mnA = A   

⇒  m2I2 + n2A2 = A(1 - 2mn)

⇒  m² \(\rm \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) + n² \( \rm \begin{bmatrix} -4 & 0 \\ 0 & -4 \end{bmatrix}\) = (1 - 2mn)\(\rm \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}\) \(\rm \begin{bmatrix} m^{2} & 0\\ 0 & m^{2} \end{bmatrix} + \begin{bmatrix} -4n^{2} & 0\\ 0 & -4n^{2} \end{bmatrix} = \begin{bmatrix} 0 & 2(1-2mn)\\ -2(1-2mn) & 0 \end{bmatrix} \)

On equating on both sides 

⇒ m² - 4n² = 0

⇒ m = 2n     ----(1)

⇒ 2(1 - 2mn) = 0

⇒ 1 - 2mn = 0

⇒  mn = \(\rm \frac{1}{2}\)

From equation (1)

(2n)n = \(\rm \frac{1}{2}\)

⇒ n = \(\rm ± \frac{1}{2}\)

Taking positive sign 

When n =  \(\rm \frac{1}{2}\) then, m = 1

∴ m + n = 1 + \(\rm \frac{1}{2}\) = \(\rm \frac{3}{2}\)