1 votes 1 votes If $I$ is the unit matrix of order $n$ , where $k!=0$ is a constant then $adj \ kI$ is Linear Algebra engineering-mathematics linear-algebra + – Anjan asked Jan 11, 2018 Anjan 430 views answer comment Share Follow See all 9 Comments See all 9 9 Comments reply Show 6 previous comments Shubhanshu commented Jan 11, 2018 reply Follow Share $Adj (kI) = (kI)^{-1} |kI|$ $Adj (kI) = \frac{1}{k} * I^{-1} * k^n |I|$ Note :- $I^{-1} = I \text{ and } |I| = 1$ Plugging these values we get $Adj (kI) = \frac{1}{k}*I * k^{n}$ $Adj (kI) = Ik^{n-1}$ 1 votes 1 votes srivivek95 commented Jan 11, 2018 reply Follow Share Adj(kI)=(kI)−1|kI| Shouldn't it be Adj(kI)=(kI)−1|kI| I ? Though the answer will remain same. @ Shubhanshu 0 votes 0 votes Shubhanshu commented Jan 11, 2018 reply Follow Share @srivivek95 I took simple matrix inverse operation, which is:- $A^{-1} = \frac{Adj(A)}{|A|}$ and then Cross multiplication. 0 votes 0 votes Please log in or register to add a comment.