GATE2018-2-19

Question

If $A=\begin{bmatrix}1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 1 \end{bmatrix}$ then $\text{det}(A^{-1})$ is _______ (correct to two decimal palces).

Answer 1

A = Upper Triangular Matrix, Hence Eigen Values = Leading Diagonal Elements = 1,4,1

Eigen Values of A^-1 = Inverse of corresponding Eigen Values of A = 1/1 , 1/4 , 1 = 1 , 0.25 , 1

Determinant of A^-1 = Product of Eigen Values of A^-1 = 1*0.25*1 = 0.25

Hence 0.25 is the answer :)

http://linear.ups.edu/html/section-PEE.html