(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...
Question 1 Report
(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.
Answer Details
(a) Solve \(3y^2 - 5y + 2 = 0\) by completing the square. Divide through by 3:
\[y^2 - \frac{5}{3}y + \frac{2}{3} = 0 \Rightarrow y^2 - \frac{5}{3}y = -\frac{2}{3}.\]
Half the coefficient of \(y\) is \(\dfrac{5}{6}\); add \(\left(\dfrac{5}{6}\right)^2 = \dfrac{25}{36}\) to both sides:
\[\left(y - \frac{5}{6}\right)^2 = -\frac{2}{3} + \frac{25}{36} = \frac{-24 + 25}{36} = \frac{1}{36}.\]
\[y - \frac{5}{6} = \pm\frac{1}{6} \Rightarrow y = \frac{5}{6} + \frac{1}{6} = 1 \quad\text{or}\quad y = \frac{5}{6} - \frac{1}{6} = \frac{2}{3}.\]
Hence \(y = 1.00\) or \(y = 0.67\) (2 d.p.).
(b) Let \(N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}\). Then
\[MN = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}\begin{pmatrix} m & x \\ n & y \end{pmatrix} = \begin{pmatrix} m + 2n & x + 2y \\ 4m + 3n & 4x + 3y \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}.\]
First column: \(m + 2n = 2\) and \(4m + 3n = 3\). From the first, \(m = 2 - 2n\); substituting: \(4(2 - 2n) + 3n = 3 \Rightarrow 8 - 5n = 3 \Rightarrow n = 1,\; m = 0\).
Second column: \(x + 2y = 1\) and \(4x + 3y = 4\). From the first, \(x = 1 - 2y\); substituting: \(4(1 - 2y) + 3y = 4 \Rightarrow 4 - 5y = 4 \Rightarrow y = 0,\; x = 1\).
\[N = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.\]
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