Home » Using the answer in (b)(i), solve the system of equations

# Using the answer in (b)(i), solve the system of equations

(a) Differentiate (frac{x^{2} + 1}{(x + 1)^{2}}) with respect to x.

(b)(i) Evaluate (begin{vmatrix} 1 & 2 & -1 \ 2 & 3 & -1 \ -1 & 1 & 3 end{vmatrix}).

(ii) Using the answer in (b)(i), solve the system of equations.

(x + 2y – z = 4)

(2x + 3y – z = 2)

(-x + y + 3z = -1).

##### Explanation

(a) Let (y = frac{x^{2} + 1}{(x + 1)^{2}})

Let (u = x^{2} + 1 ; v = (x + 1)^{2})

(frac{mathrm d u}{mathrm d x} = 2x ; frac{mathrm d v}{mathrm d x} = 2(x + 1))

Using the quotient rule,

(frac{mathrm d y}{mathrm d x} = frac{v frac{mathrm d u}{mathrm d x} – u frac{mathrm d v}{mathrm d x}}{v^{2}})

= (frac{(x + 1)^{2} (2x) – (x^{2} + 1)(2(x + 1))}{((x + 1)^{2})^{2}})

= (frac{(x + 1)[(x + 1)(2x) – (x^{2} + 1)(2)}{(x + 1)^{4}})

= (frac{2x^{2} + 2x – 2x^{2} – 2}{(x + 1)^{3}})

= (frac{2x – 2}{(x + 1)^{3}})

(b)(i) (begin{vmatrix} 1 & 2 & -1 \ 2 & 3 & -1 \ -1 & 1 & 3 end{vmatrix})

= (1(9 + 1) – 2(6 – 1) – 1(2 + 3))

= (10 – 10 – 5 = -5)

(ii) (x + 2y – z = 4)

(2x + 3y – z = 2)

(-x + y + 3z = -1)

In matrix form, this is

(begin{pmatrix} 1 & 2 & -1 \ 2 & 3 & -1 \ -1 & 1 & 3 end{pmatrix} begin{pmatrix} x \ y \ z end{pmatrix} = begin{pmatrix} 4 \ 2 \ -1 end{pmatrix})

Cofactors :

(A_{11} = +(3 times 3 + 1) = 10)

(A_{12} = -(2 times 3 – 1) = -5)

(A_{13} = +(2 times 1 + 3) = 6)

(A_{21} = -(2 times 3 + 1) = -7)

(A_{22} = +(1 times 3 – 1) = 2)

(A_{23} = -(1 times 1 + 1 times 2) = -3)

(A_{31} = +(2 times -1 + 3) = 1)

(A_{32} = -(-1 + 2) = -1)

(A_{33} = +(1 times 3 – 2 times 2) = -1)

(C = begin{pmatrix} 10 & -5 & 5 \ -7 & 2 & -3 \ 1 & -1 & -1 end{pmatrix})

adj A = (C^{T} = begin{pmatrix} 10 & -7 & 1 \ -5 & 2 & -1 \ 5 & -3 & -1 end{pmatrix})

= (frac{1}{-5} begin{pmatrix} 10 & -7 & 1 \ -5 & 2 & -1 \ 5 & -3 & -1 end{pmatrix})

= (begin{pmatrix} -2 & frac{7}{5} & frac{-1}{5} \ 1 & frac{-2}{5} & frac{1}{5} \ -1 & frac{3}{5} & frac{1}{5} end{pmatrix})

(therefore x = A^{-1} . b )

= (begin{pmatrix} -2 & frac{7}{5} & frac{-1}{5} \ 1 & frac{-2}{5} & frac{1}{5} \ -1 & frac{3}{5} & frac{1}{5} end{pmatrix} begin{pmatrix} 4 \ 2 \ -1 end{pmatrix})

= (begin{pmatrix} – 8 + frac{14}{5} + frac{1}{5} \ 4 – frac{4}{5} + frac{1}{5} \ -4 + frac{6}{5} – frac{1}{5} end{pmatrix})

= (begin{pmatrix} -5 \ 3 \ -3 end{pmatrix})

x = -5 ; y = 3 ; z = -3.