Let A= bmatrix2&1&2\\6&2&11\\3&3&2 bmatrix and P= bmatrix1&2&0\\5&0&2\\7&1&5 bmatrix. The sum of the prime factors of |P-1AP-2I | is equal to
(4) 26
P-1AP-2I=P-1AP-2P-1P=P-1(A-2I)P
|XYZ|=|X||Y||Z| and |P-1||P|=1 required =|A-2I|
|A-2I|= vmatrix0&1&2\\6&0&11\\3&3&0 vmatrix=0-1(0-33)+2(18-0)=69
69=3 23 sum =3+23=26
Answer: (4)
Let for any three distinct consecutive terms a, b, c of an A.P, the lines ax+by+c=0 be concurrent at the point P and Q( , ) be a point such that the system of equations x+y+z=6, 2x+5y+ z= and x+2y+3z=4, has infinitely many solutions. Then (PQ)2 is equal to .
Let A= bmatrix2&0&1\\1&1&0\\1&0&1 bmatrix, B= [B1, B2, B3 ], where B1, B2, B3 are column matrices, and AB1= bmatrix1\\0\\0 bmatrix, AB2= bmatrix2\\3\\0 bmatrix, AB3= bmatrix3\\2\\1 bmatrix If =|B| and is the sum of all the diagonal elements of B, then 3+ 3 is equal to .
The values of , for which vmatrix1 & 32 & + 32\ 1 & 13 & + 13\ 2 +3 & 3 +1 & 0 vmatrix=0, lie in the interval