consider-a-non-singular-2-2-square-matrix-T-If-trace-T-4-and-trace-T-2-5-what-is-determinant-of-the-matrix-T-

Question Number 117973 by bemath last updated on 14/Oct/20

consider a non - singular 2 \times 2

square matrix T . If trace (T) = 4

and trace (T^{2}) = 5 what is determinant

of the matrix T ?

Answered by bobhans last updated on 14/Oct/20

letting p and q denote the eigenvalues of T

we have tr (T) = p + q = 4 \dots (i)

then because p^{2} and q^{2} are the eigenvalues

of T^{2} it follows that tr (T^{2}) = p^{2} + q^{2} = 5 \dots (ii)

solving eq (i) & (ii)

{2 p}^{2} - 8 p + 11 = 0 \Rightarrow p = \frac{4 \pm i \sqrt{6}}{2} \land q = \frac{4 \mp i \sqrt{6}}{2}

D eterminan of T is the product of its

eigenvalues, we get ∣ T ∣ = (\frac{4 + i \sqrt{6}}{2}) (\frac{4 - i \sqrt{6}}{2}) = \frac{11}{2}

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