1. **Problem statement:** Express the vectors as linear combinations of \(u = (2, 1, 4)\), \(v = (1, -1, 3)\), and \(w = (3, 2, 5)\). We want to find scalars \(a, b, c\) such that: $$a u + b v + c w = (x, y, z)$$ where \((x, y, z)\) is each given vector. 2. **Formula and approach:** We write the system: $$a(2,1,4) + b(1,-1,3) + c(3,2,5) = (x,y,z)$$ which gives the equations: $$2a + b + 3c = x$$ $$a - b + 2c = y$$ $$4a + 3b + 5c = z$$ We solve these linear systems for each vector. --- ### a. For \((-9, -7, -15)\): Equations: $$2a + b + 3c = -9$$ $$a - b + 2c = -7$$ $$4a + 3b + 5c = -15$$ From the second equation: $$a = -7 + b - 2c$$ Substitute into first: $$2(-7 + b - 2c) + b + 3c = -9$$ $$-14 + 2b -4c + b + 3c = -9$$ $$3b - c = 5$$ Substitute into third: $$4(-7 + b - 2c) + 3b + 5c = -15$$ $$-28 + 4b - 8c + 3b + 5c = -15$$ $$7b - 3c = 13$$ Now solve: $$3b - c = 5$$ $$7b - 3c = 13$$ Multiply first by 3: $$9b - 3c = 15$$ Subtract second: $$(9b - 3c) - (7b - 3c) = 15 - 13$$ $$2b = 2 \implies b = 1$$ From \(3b - c = 5\): $$3(1) - c = 5 \implies c = -2$$ From \(a = -7 + b - 2c\): $$a = -7 + 1 - 2(-2) = -7 + 1 + 4 = -2$$ **Answer:** \(a = -2, b = 1, c = -2\) --- ### b. For \((6, 11, 6)\): Equations: $$2a + b + 3c = 6$$ $$a - b + 2c = 11$$ $$4a + 3b + 5c = 6$$ From second: $$a = 11 + b - 2c$$ Substitute into first: $$2(11 + b - 2c) + b + 3c = 6$$ $$22 + 2b - 4c + b + 3c = 6$$ $$3b - c = -16$$ Substitute into third: $$4(11 + b - 2c) + 3b + 5c = 6$$ $$44 + 4b - 8c + 3b + 5c = 6$$ $$7b - 3c = -38$$ Solve system: $$3b - c = -16$$ $$7b - 3c = -38$$ Multiply first by 3: $$9b - 3c = -48$$ Subtract second: $$(9b - 3c) - (7b - 3c) = -48 - (-38)$$ $$2b = -10 \implies b = -5$$ From \(3b - c = -16\): $$3(-5) - c = -16 \implies -15 - c = -16 \implies c = 1$$ From \(a = 11 + b - 2c\): $$a = 11 - 5 - 2(1) = 11 - 5 - 2 = 4$$ **Answer:** \(a = 4, b = -5, c = 1\) --- ### c. For \((0, 0, 0)\): Equations: $$2a + b + 3c = 0$$ $$a - b + 2c = 0$$ $$4a + 3b + 5c = 0$$ From second: $$a = b - 2c$$ Substitute into first: $$2(b - 2c) + b + 3c = 0$$ $$2b - 4c + b + 3c = 0$$ $$3b - c = 0$$ Substitute into third: $$4(b - 2c) + 3b + 5c = 0$$ $$4b - 8c + 3b + 5c = 0$$ $$7b - 3c = 0$$ From \(3b - c = 0\), we get \(c = 3b\). Substitute into \(7b - 3c = 0\): $$7b - 3(3b) = 7b - 9b = -2b = 0 \implies b = 0$$ Then \(c = 3(0) = 0\), and \(a = b - 2c = 0 - 0 = 0\). **Answer:** \(a = 0, b = 0, c = 0\) --- **Summary:** - a) \((-9, -7, -15) = -2u + 1v - 2w\) - b) \((6, 11, 6) = 4u - 5v + 1w\) - c) \((0, 0, 0) = 0u + 0v + 0w\)