Determinants

JEE Main 2025 (Online) 23rd January Morning ShiftMCQMedium📈 Topic in 42% papers+4−1

Ques. 1JEE Main 2025, 23 January (Morning Shift)

If A, B, and (adj (A-1 )+adj (B-1 ) ) are non-singular matrices of same order, then the inverse of A (adj (A-1 )+adj (B-1 ) )-1B, is equal to

AB-1|A|+ BA-1|B|
adj (B-1 )+adj (A-1 )
AB-1+A-1B
1|AB| (adj(B)+adj(A) )

Answer

(4) 1|AB| (adj(B)+adj(A) )

Solution

(XYZ)-1=Z-1Y-1X-1 and adj (M-1 )= M|M|

X-1=B-1 ( A|A|+ B|B| )A-1= B-1|A|+ A-1|B|

Use B-1= adj(B)|B|, A-1= adj(A)|A|:

X-1= 1|AB| (adj(B)+adj(A) )

Answer: (4)

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Ques. 2JEE Main 2025, 23 January (Evening Shift)

The system of equations x+y+z=6, x+2y+5z=9, x+5y+ z= , has no solution if

=17, =18
=17, 18
=15, 17
17, 18

Ques. 3JEE Main 2025, 22 January (Morning Shift)

Let A be a square matrix of order 3 such that (A)=-2 and (3 adj (-6 adj(3A) ) )=2m+n 3mn, m>n. Then 4m+2n is equal to .

Ques. 4JEE Main 2025, 22 January (Evening Shift)

If the system of linear equations : x+y+2z=6 2x+3y+az=a+1 -x-3y+bz=2b where a,b , has infinitely many solutions, then 7a+3b is equal to :

Ques. 5JEE Main 2025, 22 January (Evening Shift)

For a 3 3 matrix M, let trace (M) denote the sum of all the diagonal elements of M. Let A be a 3 3 matrix such that |A|= 12 and trace (A)=3. If B=adj(adj(2A)), then the value of |B|+ trace (B) equals :

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