If B is a non–singular matrix and A is a square matrix, then det (B–1 AB) is equal to
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(InA) (where In is an identity matrix of order n )
=det(A)
If the system of equations x + 2y + 3z = 4, x + λy + 2z = 3, x + 4y + μz = 3 has an infinite anumber of solutions then
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
If a, b, c are non zeros, then the system of equations (α + a) x + αy + αz = 0
αx + (α + b)y + αz = 0
αx + αy + (α + c) z = 0
has a non–trivial solution if
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)
Let and
. If B is the inverse of matrix A, then α is
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
The value of a for which system of equations,
a3x + (a + 1)3 y + (a + 2)3z = 0,
ax + (a + 1) y + (a + 2) z = 0,
x + y + z = 0, has a non–zero solution is
Two matrices A and B are multiplied to get AB, if
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 = [aij] be an mxn matrix & B = [bjk] 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.
Let , then A-1 exists if
Let where 0 ≤ θ < 2π, then
then which one of the following holds for all n ≥ 1, by the principle of mathematical induction ?
is the unit matrix of order 2 and a, b are arbitrary constants, then
is equal to
Identity the correct statement(s)
If a, b, c > 0 & x, y, z ∈ R then the determinant
IF x, y , z ∈ R Δ = = -16 then value of x is
The determinant is
If A, B, C are angles of a triangle ABC, then
is less than or equal to
ANSWER :- b
Solution :- sin(A+B+C) = cos(A+B+C)/2 = 0
Hence, sinA/2 sinB/2 sinC/2 ≤ 1/8
Let f(x) = then the maximum value of f(x) is
Value of the D = is
If f(x) = , then
f(x) ={(a-x, ex/na, x^2) (a-3x, e3x/na, x4) (a-5x, e5x/na, 1)}
f(-x) = - {(a-x, ex/na, x2) (a-3x, e3x/na, x4) (a-5x, e^5x/na, 1)}
Adding (1) and (2) we get
f(x) + f(-x) = 0
D = is (where a, b, c are the sides opposite to angles A, B, C respectively in a triangle)
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
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