(a) Using completing the square method, solve, correct to 2 decimal places, the equation \(3y^{2} - 5y + 2 = 0\).
(b) Given that \(M = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}, N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}\) and \(MN = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}\), find the matrix N.
Show Answer Show Explanation Explanation (a) \(3y^{2} - 5y + 2 = 0\) \(\frac{3y^{2}}{3} - \frac{5y}{3} = \frac{-2}{3}\) \(y^{2} - \frac{5}{3}y = \frac{-2}{3}\) \(\frac{1}{2} of \frac{-5}{3} = \frac{-5}{6}\) \(y^{2} - \frac{5}{3}y + (-\frac{5}{6})^{2} = \frac{-2}{3} + (\frac{-5}{6})^{2}\) \((y - (\frac{5}{6}))^{2} = \frac{1}{36}\) Taking square root of both sides, \(\sqrt{(y - (\frac{5}{6}))^{2}} = \sqrt{\frac{1}{36}}\) \(y - \frac{5}{6} =\pm {\frac{1}{6}}\) \(y = \frac{5}{6} \pm \frac{1}{6}\) \(y = \frac{5}{6} + \frac{1}{6} = 1.00\) or \(y = \frac{5}{6} - \frac{1}{6} = \frac{2}{3} = 0.67\). (b) \(M = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}, N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}\) \(MN = \begin{pmatrix} m + 2n & x + 2y \\ 4m + 2n & 4x + 2y \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}\) \(m + 2n = 2 ... (1)\) \(4m + 2n = 3 ... (2)\) Multiplying (1) by 4, we have \(4m + 8n = 8 ... (3)\) (3) - (1) : \(8n - 2n = 8 - 2 = 6\) \(6n = 6 \implies n = 1\) \(m + 2n = m + 2(1) = m + 2 = 2\) \(m = 2 - 2 = 0\) \(x + 2y = 1 ... (1)\) \(4x + y = 4 ... (2)\) Multiply (1) by 4, \(4x + 8y = 4 ... (3)\) (3) - (2) : \(8y - y = 4 - 4 = 0\) \(\implies y = 0\) \(x + 2y = x + 2(0) = x + 0 = 1\) \(x = 1 - 0 = 1\) \(\therefore N = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).