(a) Differentiate \(\frac{x^{2} + 1}{(x + 1)^{2}}\) with respect to x. (b)(i) Evaluate \(\begin{vmatrix} 1
(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.