JAMB Mathematics Past Questions & Answers - Page 329

Let A = \(\begin{pmatrix} 5 & 3 \\ 6 & 4 \end{pmatrix}\)

Then |A| = \(\begin{pmatrix} 5 & 3 \\ 6 & 4 \end{pmatrix}\) = 20 - 18 = 2

Hence A-1 = \(\frac{1}{|A|}\begin{pmatrix} 4 & -3 \\ -6 & 5 \end{pmatrix}\)

= \(\frac{1}{2}\begin{pmatrix} 4 & -3 \\ -6 & 5 \end{pmatrix}\)

= \(\begin{pmatrix} 4 \times 1/2 & -3 \times 1/2 \\ -6 \times 1/2 & 5 \times 1/2 \end{pmatrix}\)

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

2P + Q = 2\(\begin{pmatrix} 5 & 3 \\ 2 & 1 \end{pmatrix}\) + \(\begin{pmatrix} 4 & 2 \\ 3 & 5 \end{pmatrix}\)

= \(\begin{pmatrix} 10 & 6 \\ 4 & 2 \end{pmatrix}\) + \(\begin{pmatrix} 4 & 2 \\ 3 & 5 \end{pmatrix}\)

= \(\begin{pmatrix} 14 & 8 \\ 7 & 7 \end{pmatrix}\)

x * y = x + 2y (given)

3 * 4 = 3 + 2(4) = 11

Hence, 2 * (3 * 4) = 2 * 11

= 2 + 2(11)

= 2 + 22

= 24

Using T_n = \(\frac{3n + 1}{n + 1}\),

T₁ = \(\frac{3(1) + 1}{(1) + 1} = \frac{4}{2}\)

T₂ = \(\frac{3(2) + 1}{(2) + 1} = \frac{7}{3}\)

T₃ = \(\frac{3(3) + 1}{(3) + 1} = \frac{10}{4}\)

|x - 2| < 3 implies -(x - 2) < 3 .... or .... +(x - 2) < 3 -x + 2 < 3 .... or .... x - 2 < 3 -x < 3 - 2 .... or .... x < 3 + 2 x > -1 .... or .... x < 5 combining the two inequalities results, we get; -1 < x < 5