1. **Problem:** Given matrix $$M = \begin{pmatrix} 2 & 5 \\ 7 & 15 \end{pmatrix}$$ i) Show that $$M$$ is non-singular. ii) Write down the inverse of $$M$$. iii) Write down the 2x2 matrix equal to the product $$M \times M^n$$ (where $$M^n$$ is the inverse of $$M$$). iv) Pre-multiply both sides of the matrix equation $$M \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -3 \\ 17 \end{pmatrix}$$ by $$M^{-1}$$ and solve for $$x$$ and $$y$$. --- **Step 1: Show that $$M$$ is non-singular** A matrix is non-singular if its determinant is not zero. The determinant of $$M$$ is: $$\det(M) = (2)(15) - (5)(7) = 30 - 35 = -5$$ Since $$\det(M) = -5 \neq 0$$, matrix $$M$$ is non-singular. **Step 2: Find the inverse of $$M$$** The formula for the inverse of a 2x2 matrix $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ is: $$M^{-1} = \frac{1}{\det(M)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$ Applying this to $$M$$: $$M^{-1} = \frac{1}{-5} \begin{pmatrix} 15 & -5 \\ -7 & 2 \end{pmatrix} = \begin{pmatrix} \frac{15}{-5} & \frac{-5}{-5} \\ \frac{-7}{-5} & \frac{2}{-5} \end{pmatrix} = \begin{pmatrix} -3 & 1 \\ \frac{7}{5} & -\frac{2}{5} \end{pmatrix}$$ **Step 3: Write down the product $$M \times M^{-1}$$** By definition, $$M \times M^{-1} = I$$, the identity matrix: $$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ **Step 4: Pre-multiply both sides of $$M \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -3 \\ 17 \end{pmatrix}$$ by $$M^{-1}$$** $$M^{-1} M \begin{pmatrix} x \\ y \end{pmatrix} = M^{-1} \begin{pmatrix} -3 \\ 17 \end{pmatrix}$$ Since $$M^{-1} M = I$$: $$I \begin{pmatrix} x \\ y \end{pmatrix} = M^{-1} \begin{pmatrix} -3 \\ 17 \end{pmatrix}$$ So: $$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -3 & 1 \\ \frac{7}{5} & -\frac{2}{5} \end{pmatrix} \begin{pmatrix} -3 \\ 17 \end{pmatrix}$$ Calculate the product: $$x = (-3)(-3) + (1)(17) = 9 + 17 = 26$$ $$y = \frac{7}{5}(-3) + \left(-\frac{2}{5}\right)(17) = -\frac{21}{5} - \frac{34}{5} = -\frac{55}{5} = -11$$ **Final answers:** - $$M$$ is non-singular because $$\det(M) = -5 \neq 0$$. - $$M^{-1} = \begin{pmatrix} -3 & 1 \\ \frac{7}{5} & -\frac{2}{5} \end{pmatrix}$$. - $$M \times M^{-1} = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$. - Solution to the system is $$x = 26$$ and $$y = -11$$.