Example 1:

        If f(θ) = Then find the value of f(∏/6)

Solution:

= 3/4((1/4)0 + √3/4(-√3/4)-1/2(-(3/8)-(1/8))) (Expanding along first row)

       => f(∏/6) = 9/16 + 3/16 + 4/16 = 1                                   (Ans.)

Example 2:

        Show that the determinant

        is divisible by x².

Solution:

At x = 0, Δ = 0 with three columns identical. Therefore (x - 0) is a repeated root of the equation Δ = 0 => x² is a factor of determinant.

=> Δ is divisible by x².                                                   (Proved)

Example 3:

Determine the value of 'a' for which the following system of equation has a non trivial solution.

a³ x + (a+1)³y = (a+2)³z = 0

ax + (a+1)y + (a+2)z = 0

x + y + z = 0

Solution:

        For this system of equations to have a non trivial solution.

                => (3a² + 3a + 1)(2) + (6a² + 12a + 8)(-1) = 0

                => 6a² + 6a + 2 - 6a² - 12a - 8 = 0

                => -6a - 6 = 0

                => a = - 1                                                          (Proved)

Example 4:

        If m & p are positive (m > p) and

        Δ (m, p) =

        then prove that Δ(m, p) = ^m+2C₃/^p+2C₃  Δ(m - 1, P - 1)

Solution:

        We know that,

                ^mC_p = m!/(m-p)!p!  = m/p . ^m-1C_p-1

                => Δ(m, p) =

                => (m(m+1)(m+2))/(p(p+1)(p+2)) Δ (m-1, p-1)

               => ^m+2C₃/^p+2C₃ Δ (m-1, p-1)                              (Proved)

Example 5:

        Without expanding at any stage show that

        = x A + B

        Where A and B are determinants of order 3 not involving x

Solution:

              =  xA+B                                                                (proved)

               Example 6:

        Show that the following determinant is independent of x.

Solution:

        First two determinants are vanishing. In the third determinant

        C₁ → C₁ - cos x. C₃ and C₂ → + sin x C₃

        Since dΔ/dx = 0 so Δ is independent of x                              (Proved)

Example 7:

 Example 8:

If a, b, c ≠ 0 and  = 0, then find the value of 1/a + 1/b + 1/c

Solution:

                Let Δ =

                Δ = abc (taking a, b, c common from C₁

                = abc (C₁ → C₁ + C₂ + C₃)

                = (abc)(1+1/a+1/b+1/c)

                = (abc) (1+1/a+1/b+1/c).1

                But given that Δ = 0 .·. abc (1+1/a+1/b+1/c)

                But a, b, c ≠ 0 therefore 1/a + 1/b + 1/c = -1            (Ans.)

Example 9:

        If a and x are real numbers and n is a positive integer then show that

         = 0, for any x.

Solution:

        Two columns are identical, so Δ = 0 for any x.                 (Proved)

Example 10:

Let α be a repeated root of the quadratic equation f(x) = 0 with the leading coefficient a as unity and A(x), B(x) and C(x) are polynomials of degree 3, 4 and 5 respectively show that

is divisible by f(x)

Solution:

        Since α is the repeated root of the quadratic equation f(x) = 0

        Therefore f(x) can be written as : f(x) = (x - a)²

        Let, f(x) =

        Since two rows are coming identical for x = a, (x - a) is a root of f(x).

        f'(a) = = 0

Since in f'(x) two rows are coming identical for x = α, therefore (x-α) is a factor of f'(x). Thus (x-α)² is a factor of f(x), which completes the proof.