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 If \(A = \left( {\begin{array}{*{20}{c}} 2&3&4\\ 0&4&2\\ 0&0&3 \end{array}} \right)\), the Eigenvalues of adj A and [(adjA)2 – 2(adj A) + I] are:
1. 2, 4, 3 and 49, 121, 25
2. 8, 12, 6 and 49, 121, 25
3. 4, 16, 9 and 16, 256, 81
4. 8, 12, 6 and -49, 100, 12

1 Answer

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Best answer
Correct Answer - Option 2 : 8, 12, 6 and 49, 121, 25

Concept:

Upper triangular matrix

When all the elements below the diagonal in the matrix are zero

  • From the properties of Eigenvalue, we know that the Eigenvalue of an upper triangular matrix is the diagonal element of that matrix.
  • This property is also valid for diagonal matrix, scalar matrix, unit matrix, and a lower triangular matrix.

The eigenvalue of Adj A matrix(λ'is given by,

λ\(\frac{{\left| A \right|}}{λ }\;\)

The determinant of the upper triangular matrix = λ1 × λ2 × λ3

Augmented matrix is given by, \(\left| {{\bf{A}} - {\bf{\lambda I}}} \right| = 0\), where I is identity matrix.

Calculation:

Given:

\(A = \left( {\begin{array}{*{20}{c}} 2&3&4\\ 0&4&2\\ 0&0&3 \end{array}} \right)\)

Now, we know that

The given matrix is an upper triangular matrix

Eigenvalues of the matrix are

, λ1 = 2, λ2 = 4 and λ3 = 3

And determinant of matrix A = 

λ1 × λ2 × λ3 = 2 × 4 × 3 = 24

If λ is an Eigenvalue of a matrix,

then the Eigenvalues of Adj A = \(\frac{{\left| A \right|}}{λ_1 }\;\)\(\frac{{\left| A \right|}}{λ_2}\;\)\(\frac{{\left| A \right|}}{λ_3 }\;\)\(\frac{24}{2 }, \frac{24}{4 }, \frac{24}{3 } = 12, 6, 8\)

If λ is the Eigenvalue of matrix A, then Augmented matrix  is \(\left| {{\bf{A}} - {\bf{\lambda I}}} \right| = 0\) and eigenvalues follows this equation.

i.e. from given equation A 2 - 2A + I ⇒\({λ ^2} - 2λ + 1 ⇒ {\left( {λ - 1} \right)^2}\)

Shortcut:

Put the Eigenvalues 12, 6, and 8 in the above equation, then eigenvalues of A2 - 2A + I are

(12 - 1)2, (6 - 1)2, (8 - 1)2 = 121, 25, 49.

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