(a) Use the trapezium rule with five ordinates to evaluate \(\int_{0} ^{1} \frac{3}{1 + x^{2}} \mathrm {d} x\), correct to four significant figures.
(b) If \(A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\), find the image of the point (1, 2) under the linear transformation \(A^{2} + A + 2I\), where I is the \(2 \times 2\) unit matrix.
(b) If \(A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\), find the image of the point (1, 2) under the linear transformation \(A^{2} + A + 2I\), where I is the \(2 \times 2\) unit matrix.
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Correct Answer: Option n
Explanation:
(a) \(\int _{0} ^{1} \frac{3}{1 + x^{2}} \mathrm {d} x\)
h = 0.25 ;
\(y_{1} = 3, y_{2} = 2.82 , y_{3} = 2.40 , y_{4} = 1.92 , y_{5} = 1.5\)
\(y_{1} + y_{5} = 3 + 1.5 = 4.5\)
\(y_{2} + y_{3} + y_{4} = 2.82 + 2.40 + 1.92 = 7.14\)
\(2(y_{2} + y_{3} + y_{4}) = 14.28\)
= \(3[\frac{1}{2} h ( y_{1} + y_{5} + 2(y_{2} + y_{3} + y_{4})]\)
= \(3[\frac{1}{2} (0.25) (4.5 + 14.28)]\)
= \(3(0.125)(18.78) = 2.3475\)
\(\approxeq 2.348\) (4 sig. figures).
(b) \(A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\)
\(A^{2} + A + 2I = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} + 2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
= \(\begin{pmatrix} 4 + 3 & 2 + 2 \\ 6 + 6 & 3 + 4 \end{pmatrix} + \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\)
= \(\begin{pmatrix} 7 & 4 \\ 12 & 7 \end{pmatrix} + \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\)
= \(\begin{pmatrix} 11 & 5 \\ 15 & 11\end{pmatrix}\)
\(T : \begin{pmatrix} 1 \\ 2 \end{pmatrix} \to \begin{pmatrix} 11 & 5 \\ 15 & 11 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix}\)
=\(\begin{pmatrix} 11 + 10 \\ 15 + 22 \end{pmatrix} = \begin{pmatrix} 21 \\ 37 \end{pmatrix}\)
The image of (1, 2) = (21, 37).
(a) \(\int _{0} ^{1} \frac{3}{1 + x^{2}} \mathrm {d} x\)
| \(x\) | 0 | 0.25 | 0.50 | 0.75 | 1.0 |
| \(x^{2}\) | 0 | 0.06 | 0.25 | 0.56 | 1.0 |
| \(1 + x^{2}\) | 1 | 1.06 | 1.25 | 1.56 | 2.0 |
| \(\frac{3}{1 + x^{2}}\) | 3 | 2.82 | 2.40 | 1.92 | 1.5 |
h = 0.25 ;
\(y_{1} = 3, y_{2} = 2.82 , y_{3} = 2.40 , y_{4} = 1.92 , y_{5} = 1.5\)
\(y_{1} + y_{5} = 3 + 1.5 = 4.5\)
\(y_{2} + y_{3} + y_{4} = 2.82 + 2.40 + 1.92 = 7.14\)
\(2(y_{2} + y_{3} + y_{4}) = 14.28\)
= \(3[\frac{1}{2} h ( y_{1} + y_{5} + 2(y_{2} + y_{3} + y_{4})]\)
= \(3[\frac{1}{2} (0.25) (4.5 + 14.28)]\)
= \(3(0.125)(18.78) = 2.3475\)
\(\approxeq 2.348\) (4 sig. figures).
(b) \(A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\)
\(A^{2} + A + 2I = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} + 2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
= \(\begin{pmatrix} 4 + 3 & 2 + 2 \\ 6 + 6 & 3 + 4 \end{pmatrix} + \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\)
= \(\begin{pmatrix} 7 & 4 \\ 12 & 7 \end{pmatrix} + \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\)
= \(\begin{pmatrix} 11 & 5 \\ 15 & 11\end{pmatrix}\)
\(T : \begin{pmatrix} 1 \\ 2 \end{pmatrix} \to \begin{pmatrix} 11 & 5 \\ 15 & 11 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix}\)
=\(\begin{pmatrix} 11 + 10 \\ 15 + 22 \end{pmatrix} = \begin{pmatrix} 21 \\ 37 \end{pmatrix}\)
The image of (1, 2) = (21, 37).