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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, then1. x = 4, y = 102. x = 5, y = 83. x = -3, y = 94. x = -4, y = 10 |
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Answer» 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:
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 |
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