Matrix Commute Gradient

1. **Problem:** Given matrices \(M = \begin{bmatrix}4 & 8 \\ 2 & x\end{bmatrix}\) and \(N = \begin{bmatrix}x^2 & 4 \\ 1 & 4x\end{bmatrix}\), find the positive value of \(x\) such that \(MN = NM\). 2. **Problem:** Find the gradient of the tangent to the curve \((x^2 - y^2 + 16)^2 - 2x = 46\) at the point \((1, -2)\). --- ### 1. Finding \(x\) for commutative matrices 1. The condition for commutativity is \(MN = NM\). 2. Compute \(MN\): $$MN = \begin{bmatrix}4 & 8 \\ 2 & x\end{bmatrix} \begin{bmatrix}x^2 & 4 \\ 1 & 4x\end{bmatrix} = \begin{bmatrix}4x^2 + 8 \cdot 1 & 4 \cdot 4 + 8 \cdot 4x \\ 2x^2 + x \cdot 1 & 2 \cdot 4 + x \cdot 4x\end{bmatrix} = \begin{bmatrix}4x^2 + 8 & 16 + 32x \\ 2x^2 + x & 8 + 4x^2\end{bmatrix}$$ 3. Compute \(NM\): $$NM = \begin{bmatrix}x^2 & 4 \\ 1 & 4x\end{bmatrix} \begin{bmatrix}4 & 8 \\ 2 & x\end{bmatrix} = \begin{bmatrix}x^2 \cdot 4 + 4 \cdot 2 & x^2 \cdot 8 + 4 \cdot x \\ 1 \cdot 4 + 4x \cdot 2 & 1 \cdot 8 + 4x \cdot x\end{bmatrix} = \begin{bmatrix}4x^2 + 8 & 8x^2 + 4x \\ 4 + 8x & 8 + 4x^2\end{bmatrix}$$ 4. Set \(MN = NM\) element-wise: - Top-left: \(4x^2 + 8 = 4x^2 + 8\) (always true) - Top-right: \(16 + 32x = 8x^2 + 4x\) - Bottom-left: \(2x^2 + x = 4 + 8x\) - Bottom-right: \(8 + 4x^2 = 8 + 4x^2\) (always true) 5. Simplify the two equations: - \(16 + 32x = 8x^2 + 4x \implies 8x^2 + 4x - 32x - 16 = 0 \implies 8x^2 - 28x - 16 = 0\) - \(2x^2 + x = 4 + 8x \implies 2x^2 + x - 8x - 4 = 0 \implies 2x^2 - 7x - 4 = 0\) 6. Solve the quadratic \(2x^2 - 7x - 4 = 0\) using the quadratic formula: $$x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2} = \frac{7 \pm \sqrt{49 + 32}}{4} = \frac{7 \pm \sqrt{81}}{4} = \frac{7 \pm 9}{4}$$ 7. Possible values: - \(x = \frac{7 + 9}{4} = \frac{16}{4} = 4\) - \(x = \frac{7 - 9}{4} = \frac{-2}{4} = -0.5\) 8. Check which satisfies the other quadratic \(8x^2 - 28x - 16 = 0\): - For \(x=4\): \(8(16) - 28(4) - 16 = 128 - 112 - 16 = 0\) ✓ - For \(x=-0.5\): \(8(0.25) - 28(-0.5) - 16 = 2 + 14 - 16 = 0\) ✓ 9. Both satisfy, but the problem asks for the positive value, so \(\boxed{4}\). --- ### 2. Gradient of tangent to the curve 1. Given curve: $$ (x^2 - y^2 + 16)^2 - 2x = 46 $$ 2. Differentiate implicitly with respect to \(x\): Let \(u = x^2 - y^2 + 16\), then the curve is \(u^2 - 2x = 46\). 3. Differentiate both sides: $$ 2u \cdot \frac{du}{dx} - 2 = 0 $$ 4. Find \(\frac{du}{dx}\): $$ \frac{du}{dx} = 2x - 2y \frac{dy}{dx} $$ 5. Substitute back: $$ 2u (2x - 2y \frac{dy}{dx}) - 2 = 0 $$ 6. Rearrange: $$ 4u x - 4u y \frac{dy}{dx} - 2 = 0 \implies -4u y \frac{dy}{dx} = -4u x + 2 $$ 7. Solve for \(\frac{dy}{dx}\): $$ \frac{dy}{dx} = \frac{4u x - 2}{4u y} $$ 8. Evaluate at \((x,y) = (1, -2)\): Calculate \(u\): $$ u = 1^2 - (-2)^2 + 16 = 1 - 4 + 16 = 13 $$ Calculate numerator: $$ 4 \cdot 13 \cdot 1 - 2 = 52 - 2 = 50 $$ Calculate denominator: $$ 4 \cdot 13 \cdot (-2) = -104 $$ 9. Therefore, $$ \frac{dy}{dx} = \frac{50}{-104} = -\frac{25}{52} $$ --- **Final answers:** 1. \(x = 4\) 2. Gradient of tangent at \((1, -2)\) is \(-\frac{25}{52}\).