1 votes 1 votes Given A = $\begin{bmatrix} 1 &2 &2 \\ 2&1 &-2 \\ a&2 &b \end{bmatrix}$ And AAAT = 3I I is a 3x3 identity matrix. (a,b) can be: A. (2,-1) B. (-2,1) C. (-2,-1) D. (2,1) How to solve it fast? Samujjal Das asked Jan 24, 2017 Samujjal Das 482 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Gate Mission 1 commented Jan 24, 2017 reply Follow Share AAAt = 3I Since we know |A|=|At| So above equation is |A|3 = 3 [as determinant of identity matrix is 1] Now solving futher we end up at 2a+b = 3..and only option a satisfies. 1 votes 1 votes Sushant Gokhale commented Jan 25, 2017 reply Follow Share Hw did u get LHS = 2a+b ? I am getting |A| = -3b -6a + 12 1 votes 1 votes Gate Mission 1 commented Jan 25, 2017 i edited by Gate Mission 1 Jan 25, 2017 reply Follow Share ................. Yes initially i did a blunder i equated |A| = 3 but equation is |A|3 = 3 Eq becomes (2a+b-4)3 = -1/9 None option is satisfying !! 0 votes 0 votes Sushant Gokhale commented Jan 25, 2017 reply Follow Share If A-1 exists means |A| $\neq$ 0 i.e. 2a+b $\neq$ 4 So, all options satisfying :P 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes Simple,first find determinant of matrix which is -3b-6a+12 then select the value of a and b which will give you value 3 only option a satify the requirement .so ans is a. ANKITA MALL answered Jan 12, 2018 ANKITA MALL comment Share Follow See all 0 reply Please log in or register to add a comment.