Question

Consider the following matrix.

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

If the eigenvalues of A are 4 and 8, then

1. x = 4, y = 10
2. x = 5, y = 8
3. x = -3, y = 9
4. x = -4, y = 10

Answer 1

Correct Answer - Option 4 : x = -4, y = 10

Concept:

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.

The roots of the characteristic equation are called Eigenvalues or latent roots or characteristic roots of matrix A.

Properties of Eigenvalues:

1. If λ is an Eigenvalue of a matrix A, then λn will be an Eigenvalue of a matrix An.
2. If λ is an Eigenvalue of a matrix A, then kλ will be an Eigenvalue of a matrix kA where k is a scalar.
3. Sum of Eigenvalues is equal to the trace of that matrix.
4. The product of Eigenvalues of a matrix A is equal to the determinant of that matrix A.
5. If λ is an Eigenvalue of matrix A, then λ2 will be an Eigenvalue of matrix A2.
6. If λ1 is an Eigenvalue of matrix A, then (λ1 + 1) will be an Eigenvalue of the matrix (A + I).
7. Eigenvalues of a matrix and its transpose are the same because the transpose matrix will also have the same characteristic equation.

Calculation:

Given:

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

Sum of eigen values = Trace (A) = 2 + y

Product of eigen values = |A| = 2y – 3x

∴ 4 + 8 = 2 + y     … i)

4 × 8 = 2y – 3x      … ii)

∴ 2 + y = 12     … iii)

 2y – 3x = 32     … iv)

∴ Solving i) and ii) we get x = -4 and y = 10