1. **QUESTION ONE** **a) Problem:** Determine values of $x$ such that matrix $$A = \begin{pmatrix}1 & 1 & x \\ 1 & x & x \\ x & x & x\end{pmatrix}$$ is invertible. - To be invertible, $\det(A) \neq 0$. - Calculate the determinant: $$\det(A) = 1 \cdot \det\begin{pmatrix} x & x \\ x & x \end{pmatrix} - 1 \cdot \det\begin{pmatrix} 1 & x \\ x & x \end{pmatrix} + x \cdot \det\begin{pmatrix} 1 & x \\ x & x \end{pmatrix}.$$ - Compute minors: $$\det\begin{pmatrix} x & x \\ x & x \end{pmatrix} = x \cdot x - x \cdot x = 0,$$ $$\det\begin{pmatrix} 1 & x \\ x & x \end{pmatrix} = 1 \cdot x - x \cdot x = x - x^{2} = x(1-x).$$ - Substitute back: $$\det(A) = 1 \cdot 0 - 1 \cdot x(1-x) + x \cdot x(1-x) = -x(1 - x) + x^{2}(1-x) = (1-x)(-x + x^{2}) = (1-x) x (x-1) = -x(1-x)^{2}.$$ - So, $$\det(A) = -x (1 - x)^2.$$ - The matrix is invertible when $\det(A) \neq 0$, so: $$-x (1 - x)^2 \neq 0 \implies x \neq 0 \text{ and } x \neq 1.$$ 2. **b) Problem:** Find $\det(A^{2} B^{-1} A^{-2} B^{2})$ given $$A = \begin{pmatrix} 1 & 0 & 3 \\ 4 & 5 & 6 \\ 7 & 0 & 9 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{pmatrix}.$$ - Use determinant properties: - $\det(A^{2} B^{-1} A^{-2} B^{2}) = \det(A^{2}) \det(B^{-1}) \det(A^{-2}) \det(B^{2})$ - Recall: $$\det(A^{k}) = [\det(A)]^{k}, \quad \det(B^{-1}) = \frac{1}{\det(B)}, \quad \det(A^{-2}) = \frac{1}{[\det(A)]^{2}}, \quad \det(B^{2}) = [\det(B)]^{2}.$$ - Substitute: $$\det(A^{2} B^{-1} A^{-2} B^{2}) = [\det(A)]^{2} \cdot \frac{1}{\det(B)} \cdot \frac{1}{[\det(A)]^{2}} \cdot [\det(B)]^{2} = \frac{[\det(B)]^{2}}{\det(B)} = \det(B).$$ - Calculate $\det(B)$: $$\det(B) = 2 \times 3 \times 4 = 24.$$ - So, $$\det(A^{2} B^{-1} A^{-2} B^{2}) = 24.$$ 3. **c) Problem:** Solve the system by finding inverse of matrix $A$ where $$A = \begin{pmatrix} -1 & -2 \\ 2 & 3 \end{pmatrix}$$ and system is $$-x - 2y = 1,$$ $$2x + 3y = -1.$$ - Write system in matrix form: $A \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix}1 \\ -1\end{pmatrix}$. - Find $\det(A)$: $$\det(A) = (-1)(3) - (-2)(2) = -3 + 4 = 1 \neq 0,$$ - Inverse of $A$: $$A^{-1} = \frac{1}{\det(A)} \begin{pmatrix}3 & 2 \\ -2 & -1 \end{pmatrix} = \begin{pmatrix}3 & 2 \\ -2 & -1 \end{pmatrix}.$$ - Find solution: $$\begin{pmatrix} x \\ y \end{pmatrix} = A^{-1} \begin{pmatrix}1 \\ -1 \end{pmatrix} = \begin{pmatrix}3 & 2 \\ -2 & -1 \end{pmatrix} \begin{pmatrix}1 \\ -1 \end{pmatrix} = \begin{pmatrix}3 \cdot 1 + 2 \cdot (-1) \\ -2 \cdot 1 + (-1) \cdot (-1) \end{pmatrix} = \begin{pmatrix}1 \\ -1 \end{pmatrix}.$$ - Thus solution: $$x = 1, \quad y = -1.$$ --- 4. **QUESTION TWO** **a) Given** $$v_1 = \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}, v_2 = \begin{pmatrix} 0 \\ 2 \\ -3 \end{pmatrix}, v_3 = \begin{pmatrix} 2 \\ 2 \\ 3 \end{pmatrix}.$$ Suppose $v_4$ is orthogonal to $v_1$ and $v_3$, and satisfies $v_2 \cdot v_4 = -3$. **i) Find** $v_1 \cdot v_2$ - Dot product: $$v_1 \cdot v_2 = (-1) \times 0 + 0 \times 2 + 2 \times (-3) = 0 + 0 - 6 = -6.$$ **ii) Find** $v_3 \cdot v_4$ - Since $v_4$ is orthogonal to $v_3$, $$v_3 \cdot v_4 = 0.$$ **iii) Find** $(2v_1 + 3v_2 - v_3) \cdot v_4$ - Use distributive property: $$(2v_1 + 3v_2 - v_3) \cdot v_4 = 2 (v_1 \cdot v_4) + 3 (v_2 \cdot v_4) - (v_3 \cdot v_4).$$ - We know: $v_1 \cdot v_4 = 0$ (since $v_4$ orthogonal to $v_1$), $v_2 \cdot v_4 = -3$, $v_3 \cdot v_4 = 0$. - Therefore: $2 \times 0 + 3 \times (-3) - 0 = -9.$ **iv) Find magnitudes** $\|v_1\|, \|v_2\|, \|v_3\|$ - $\|v_1\| = \sqrt{(-1)^2 + 0^2 + 2^2} = \sqrt{1 + 0 + 4} = \sqrt{5}.$ - $\|v_2\| = \sqrt{0^2 + 2^2 + (-3)^2} = \sqrt{0 + 4 + 9} = \sqrt{13}.$ - $\|v_3\| = \sqrt{2^2 + 2^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17}.$ **v) Distance between** $v_1$ and $v_2$ - Distance is norm of their difference: $$\|v_1 - v_2\| = \sqrt{(-1 - 0)^2 + (0 - 2)^2 + (2 - (-3))^2} = \sqrt{(-1)^2 + (-2)^2 + 5^2} = \sqrt{1 + 4 + 25} = \sqrt{30}.$$ 5. **b) Problem:** Find null space of $$A = \begin{pmatrix}1 & 2 & 2 \\ 2 & 3 & 2 \\ -1 & -3 & -4 \end{pmatrix}.$$ - We solve $A \mathbf{x} = \mathbf{0}$ for $\mathbf{x} = \begin{pmatrix}x \\ y \\ z \end{pmatrix}$. - Write system: $$1x + 2y + 2z = 0,$$ $$2x + 3y + 2z = 0,$$ $$-1x - 3y - 4z = 0.$$ - From first equation: $$x = -2y - 2z.$$ - Substitute into second: $$2(-2y - 2z) + 3y + 2z = -4y -4z + 3y + 2z = (-4y + 3y) + (-4z + 2z) = -y - 2z = 0 \implies y = -2z.$$ - Substitute $y = -2z$ back into equation for $x$: $$x = -2(-2z) - 2z = 4z - 2z = 2z.$$ - Check in third equation: $$-x - 3y - 4z = -2z - 3(-2z) - 4z = -2z + 6z - 4z = 0,$$ consistent. - Parameterize solution by $z = t$: $$\mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2t \\ -2t \\ t \end{pmatrix} = t \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}.$$ - Thus null space is all scalar multiples of $$\boxed{\begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}}.$$