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  • Complete JEE Main/Advanced Course and Test Series
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Solved Examples on Determinants

Illustration 1: Let M and N be two 3 x 3 non-singular skew-symmetric matrices such that MN = NM. If PT denotes the transpose of P then M2N2 (MTN)-1(MN-1)T is equal to (2011)

1. M2                                                                2. –N2

3. –M2                                                              4. MN

Solution: It is given that M and N are two 3 x 3 non-singular skew-symmetric matrices, so MT = -M and NT = -N

Also, MN = NM    …….. (1)

Now, we need to compute M2N2 (MTN)-1 (MN-1)T

Using the property of inverse, we have

M2N2 (N)-1 (MT)-1 (N-1)T (M)T

⇒ M2N (NN-1) (-M)-1 (NT)-1 (-M)     (using the results MT = -M and NT = -N)

= M2NI (-M)-1 (-N)-1 (-M)

= -M2NM-1N-1M

= -M (MN) M-1N-1M

= -M (NM) M-1N-1M

= -MN (MM-1) N-1M

= -M (NN-1) M

= -M2

 

Illustration 2: Let μ and α be real. Find the set of all values of μ for which the system of linear equations

μx + (sin α)y + (cos α) z = 0

x + (cos α) y + (sin α) z = 0

and -x + (sin α) y - (cos α) z = 0

has a non-trivial solution. For μ =1, find all values of α. (1993)

Solution: Given system is μx + (sin α)y + (cos α) z = 0

x + (cos α) y + (sin α) z = 0

also, -x + (sin α)y - (cos α) z = 0 has non-trivial solution.

Therefore, determinant i.e. Δ = 0


This gives μ (-cos2α - sin2α) – sin α (-cos α + sin α) + cos α (cos α + sin α) = 0.

⇒ -μ + sin α cos α + sin α cos α - sin2α + cos2α = 0.

⇒ μ = cos 2α + sin 2α (since – √a2+b2 ≤ a sin θ + b cos θ ≤ √a2+b2)

Hence, we get –√2 ≤ μ ≤ √2    …….. (1)

Again, when μ = 1, cos 2α + sin 2α = 1

⇒ 1/√2 cos 2α + 1/√2 sin 2α = 1/√2 

⇒ cos (2α – π/4) = cos (π/4)

Therefore, 2α – π/4 = 2nπ ± π/4

⇒ 2α = 2nπ - π/4 + π/4

⇒ 2α = 2nπ + π/4 + π/4

Hence, α = nπ or α = nπ + π/4

 

Illustration 3: Suppose f(x) is a function satisfying the following conditions:

(a) f(0) = 2, f(1) = 1

(b) f has a minimum value at x =5/2 and

(c) for all x,



Hence, f’(x) = 2ax + b

On integrating, we get f(x) = ax2 + bx + c, where c is an arbitrary constant.

Since f has maximum value at x = 5/2, so f’(5/2) = 0.

⇒ 5a+b = 0               ….. (1)

Also f(0) = 2

⇒ c =2

And f(1) = 1

⇒ a + b + c = 1      ……. (2)

On solving equations (1) and (2) for a, b we get

a =1/4 and b = -5/4

Thus, f(x) = 1/4 x2 – 5/4 x +2.

Hence, the required function is f(x) = 1/4 x2 – 5/4 x +2 and the values of a and b are 1/4 and -5/4.

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