Further Mathematics Questions and Answers

\(\begin{pmatrix} 1 & 2 \\ 5 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 3 \end{pmatrix}\)

= \(\begin{pmatrix} 1\times 0 + 2\times 1 & 1\times 1 + 2\times3 \\ 5\times0 + 1\times1 &  5\times1 + 1\times 3 \end{pmatrix}\)

= \(\begin{pmatrix} 2 & 7 \\ 1 & 8 \end{pmatrix}\)

\(\begin{vmatrix} m-2 & m+1 \\ m+4 & m-2 \end{vmatrix} = -27\)

\((m^{2} - 4m + 4) - (m^{2} + 5m + 4) = -27\)

\(-9m = -27 \implies m = 3\)

\(T_{n} = a + (n - 1)d\) (for a linear or arithmetic progression)

Given: \(T_{n} = 71, a = -6, d = -2\frac{1}{2} - (-6) = 3\frac{1}{2}\)

\(\implies 71 = -6 + (n - 1)\times 3\frac{1}{2}\)

\(71 = -6 + 3\frac{1}{2}n - 3\frac{1}{2} = -9\frac{1}{2} + 3\frac{1}{2}n\)

\(71 + 9\frac{1}{2} = 3\frac{1}{2}n  \implies  n = \frac{80\frac{1}{2}}{3\frac{1}{2}}\)

\(= 23\)

\(T_{n} = ar^{n-1}\) (for a geometric progression)

\(T_{3} = ar^{3-1} = ar^{2} = \frac{8}{3}\)

\(T_{6} = ar^{6-1} = ar^{5} = \frac{64}{81}\)

Dividing \(T_{6}\) by \(T_{3}\),

\(\frac{ar^{5}}{ar^{2}} = \frac{\frac{64}{81}}{\frac{8}{3}} \implies r^{3} = \frac{8}{27}\)

\(\therefore r = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}\)

Listing out in order, the terms of this expansion have coefficients: \(^{6}C_{6}, ^{6}C_{5}, ^{6}C_{4}, ^{6}C_{3}, ^{6}C_{2}, ^{6}C_{1}, ^{6}C_{0}\)

The 4th term = \(^{6}C_{3}(3x)^{3}(-y)^{3}  = 20 \times 27x^{3} \times -y^{3}\)

= \(-540x^{3}y^{3}\).