Question #201572

Show that the matrix [ 4 6 6 1 −1 3 2 −5 −2 ] is not diagonalizable


1
Expert's answer
2021-06-02T11:27:50-0400
(466113252)\begin{pmatrix} 4& 6 & 6 \\ 1 & -1 & 3 \\ 2 & -5 & -2 \\ \end{pmatrix}

Start from forming a new matrix by subtracting λ\lambda  from the diagonal entries of the given matrix:


(4λ6611λ3252λ)\begin{pmatrix} 4-\lambda & 6 & 6 \\ 1 & -1-\lambda & 3 \\ 2 & -5 & -2-\lambda \\ \end{pmatrix}

4λ6611λ3252λ\begin{vmatrix} 4-\lambda & 6 & 6 \\ 1 & -1-\lambda & 3 \\ 2 & -5 & -2-\lambda \\ \end{vmatrix}

=(4λ)1λ352λ61322λ=(4-\lambda)\begin{vmatrix} -1-\lambda & 3 \\ -5 & -2-\lambda \end{vmatrix}-6\begin{vmatrix} 1 & 3 \\ 2 & -2-\lambda \end{vmatrix}

+611λ25+6\begin{vmatrix} 1 & -1-\lambda \\ 2 & -5 \end{vmatrix}


=(4λ)(1+λ)(2+λ()+15(4λ)=(4-\lambda)(1+\lambda)(2+\lambda()+15(4-\lambda)

+6(2+λ)+3630+12(1+λ)+6(2+\lambda)+36-30+12(1+\lambda)

=8+12λ+4λ22λ3λ2λ3+6015λ=8+12\lambda+4\lambda^2-2\lambda-3\lambda^2-\lambda^3+60-15\lambda

+12+6λ+6+12+12λ+12+6\lambda+6+12+12\lambda

=λ3+λ2+13λ+98=-\lambda^3+\lambda^2+13\lambda+98

Solve the equation 


λ3+λ2+13λ+98=0-\lambda^3+\lambda^2+13\lambda+98=0

Only approximate roots can be found.

The roots are 

λ15.9513233156912\lambda_1\approx5.9513233156912


λ22.4756616578456+3.21527996418223i\lambda_2\approx−2.4756616578456+3.21527996418223i


λ32.47566165784563.21527996418223i\lambda_3\approx−2.4756616578456-3.21527996418223i


Since there are complex roots, the matrix is not diagonalizable.




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