Question

Consider a non-singular 2 x 2 square matrix A. If trance (A) = 4 and trace $(A{)}^2$ = 5, the determinant of the matrix A is (upto 1 decimal place).

Answer 1

Let ${\lambda }_1$ and ${\lambda }_2$ Eigen value of A then ${\lambda }_1+{\lambda }_2$ = 4 and ${\lambda }_1^2$ and ${\lambda }_2^2$ will be eigen value of $A^2$ the ${\lambda }_1^2$+ ${\lambda }_2^2$ = 5.

$({\lambda }_1+{\lambda }_2{)}^2={\lambda }_1^2+{\lambda }_2^2+2{\lambda }_1{\lambda }_2$

$(4{)}^2$ = 5 + $2{\lambda }_1{\lambda }_2$

So, |A| = Product of values

= ${\lambda }_1{\lambda }_2$ = $\frac{11}{2}$ = 5.5