JEE Advanced Level Test: Matrices and Determinants - JEE MCQ

A is a square matrix and B is non-singular matrix, so B is invertible and a square matrix.
 
det(B⁻¹ AB)
 
=det(B⁻¹ BA)
 
=det(I_nA)  (where I_n is an identity matrix of order n )
 
=det(A)

Correct Answer :- d

Explanation : x+2y+3z=4

x+py+2z=3

x+4y+z=3

x = Δx/Δ, y = Δy/Δ, z = Δz/Δ

Δ = 0, Δx = Δy = Δz = 0

As we know that Δy = 0

A = {(1,4,3) (1,3,2) (1,3,μ)} = 0

3μ + 9 + 8 - 9 - 6 - 4μ = 0

=> μ = 2

As Δz = 0

= {(1,2,4) (1,p,3) (1,4,3)} = 0

=> 3p + 16 - 6 - 4p - 12 -6 = 0

=> p = 4

For non trival solution, D = 0
{(α + a, α, α)(α, α + b, α)(α, α, α + c)} = 0
R1→ R1 - R2,  R2 → R2 - R3
{(a, -b ,0)(0, b ,-c)(α, α, α + c)}
abα + abc + acα + bcα = 0
⇒ abc = −α(ab + bc + ca)
⇒ 1/α = −(1/c+ 1 / a + 1 / b)
α⁻¹ = −(a⁻¹+b⁻¹+c⁻¹)

Since, B is the inverse of matrix A.
So, 10A⁻¹ = {(4,2,2) (-5,0,α) (1,-2,3)}
⇒ 10A^-1 . A = {(4,2,2) (-5,0,α) (1,-2,3)} . {(1,-1,1) (2,1,-3) (1,1,1)}
⇒ 10I = {(10,0,0) (-5+α, 5+α, -5+α) (0,0,10)}
⇒ {(10,0,0) (0,10,0) (0,0,10)} = {(10,0,0) (-5+α, 5+α, -5+α) (0,0,10)}
−5+α=0
⇒ α=5

Two matrices A and B are multiplied to get AB if the number of columns of matrix A is equals to the number of rows of matrix B. 
The product of two matrices A and B is defined if the number of columns of matrix A is equal to number of rows of matrix B. Let A = [a_ij] be an mxn matrix & B = [b_jk] be an nxp matrix. Then the product of A and B is order of mxp. 
Hence, the answer is no of columns of A is equal to rows of B.

ANSWER :- b

Solution :- sin(A+B+C) = cos(A+B+C)/2 = 0

Hence, sinA/2 sinB/2 sinC/2 ≤ 1/8

f(x) ={(a^-x, e^x/na, x^2) (a^-3x, e^3x/na, x⁴) (a^-5x, e^5x/na, 1)}
f(-x) = - {(a^-x, e^x/na, x²) (a^-3x, e^3x/na, x⁴) (a^-5x, e^5x/na, 1)}
Adding (1) and (2) we get
f(x) + f(-x) = 0

a/sinA = b/sinB = c/sinC = 2R
A = 2RsinA,   b = 2RsinB,   c = 2RsinC
{(1, 4sinB/b, cosA) (2a, 8sinA, 1) (3a, 12sinA, cosB)}
= {(1, 4sinB/(2RsinB), cosA) (4RsinA, BsinA, 1) (6RsinA, 12sinA, cosB)}
= 1(8sinAcosB- 12sinA) - 2/R(4RsinAcosB - 6RsinA) + cosA(48Rsin^2A - 48Rsin^2A)
= 8sinAcosB - 12sinA -8sinAcosB + 12sinA
= 0