Linear Combinations 6Af2Ef

1. Problem: Determine which vectors are linear combinations of \(u = (0, -2, 2)\) and \(v = (1, 3, -1)\). Formula: A vector \(w\) is a linear combination of \(u\) and \(v\) if there exist scalars \(a, b\) such that \(w = a u + b v\). Step 1: Write the system for each vector \(w = (w_1, w_2, w_3)\): $$ \begin{cases} w_1 = a \cdot 0 + b \cdot 1 = b \\ w_2 = a \cdot (-2) + b \cdot 3 = -2a + 3b \\ w_3 = a \cdot 2 + b \cdot (-1) = 2a - b \end{cases} $$ Step 2: Check each vector: a. \(w = (2, 2, 2)\) $$ \begin{cases} 2 = b \\ 2 = -2a + 3b \\ 2 = 2a - b \end{cases} $$ From first: \(b=2\). Second: \(2 = -2a + 3 \times 2 = -2a + 6 \Rightarrow -2a = -4 \Rightarrow a=2\). Third: \(2 = 2 \times 2 - 2 = 4 - 2 = 2\) correct. So \(w\) is a linear combination with \(a=2, b=2\). b. \(w = (0, 4, 5)\) $$ \begin{cases} 0 = b \\ 4 = -2a + 3b \\ 5 = 2a - b \end{cases} $$ From first: \(b=0\). Second: \(4 = -2a + 0 \Rightarrow a = -2\). Third: \(5 = 2 \times (-2) - 0 = -4\) which is false. So \(w\) is NOT a linear combination. c. \(w = (0, 0, 0)\) $$ \begin{cases} 0 = b \\ 0 = -2a + 3b \\ 0 = 2a - b \end{cases} $$ From first: \(b=0\). Second: \(0 = -2a + 0 \Rightarrow a=0\). Third: \(0 = 0 - 0 = 0\) correct. Zero vector is always a linear combination. Final answers for 1: a and c are linear combinations; b is not. --- 2. Problem: Express vectors as linear combinations of \(u = (2,1,4), v = (1,-1,3), w = (3,2,5)\). Formula: Find scalars \(a,b,c\) such that $$ (a,b,c) \cdot (u,v,w) = a u + b v + c w = (x,y,z) $$ Step 1: Write system for \((x,y,z)\): $$ \begin{cases} x = 2a + b + 3c \\ y = a - b + 2c \\ z = 4a + 3b + 5c \end{cases} $$ Step 2: Solve for each vector: a. \((-9, -7, -15)\) $$ \begin{cases} -9 = 2a + b + 3c \\ -7 = a - b + 2c \\ -15 = 4a + 3b + 5c \end{cases} $$ Add equations or use substitution: From second: \(b = a + 2c + 7\). Substitute into first: $$-9 = 2a + (a + 2c + 7) + 3c = 3a + 5c + 7 \Rightarrow 3a + 5c = -16$$ Substitute into third: $$-15 = 4a + 3(a + 2c + 7) + 5c = 4a + 3a + 6c + 21 + 5c = 7a + 11c + 21 \Rightarrow 7a + 11c = -36$$ Solve system: Multiply first by 7: $$21a + 35c = -112$$ Multiply second by 3: $$21a + 33c = -108$$ Subtract: $$2c = -4 \Rightarrow c = -2$$ Back to \(3a + 5(-2) = -16 \Rightarrow 3a - 10 = -16 \Rightarrow 3a = -6 \Rightarrow a = -2\) Then \(b = -2 + 2(-2) + 7 = -2 -4 +7 = 1\). So \(a=-2, b=1, c=-2\). b. \((6, 11, 6)\) $$ \begin{cases} 6 = 2a + b + 3c \\ 11 = a - b + 2c \\ 6 = 4a + 3b + 5c \end{cases} $$ From second: \(b = a + 2c - 11\). Substitute into first: $$6 = 2a + (a + 2c - 11) + 3c = 3a + 5c - 11 \Rightarrow 3a + 5c = 17$$ Substitute into third: $$6 = 4a + 3(a + 2c - 11) + 5c = 4a + 3a + 6c - 33 + 5c = 7a + 11c - 33 \Rightarrow 7a + 11c = 39$$ Multiply first by 7: $$21a + 35c = 119$$ Multiply second by 3: $$21a + 33c = 117$$ Subtract: $$2c = 2 \Rightarrow c = 1$$ Back to \(3a + 5(1) = 17 \Rightarrow 3a + 5 = 17 \Rightarrow 3a = 12 \Rightarrow a = 4\) Then \(b = 4 + 2(1) - 11 = 4 + 2 - 11 = -5\). So \(a=4, b=-5, c=1\). c. \((0, 0, 0)\) Zero vector is always linear combination with \(a=b=c=0\). --- 3. Problem: Determine which matrices are linear combinations of $$ A = \begin{bmatrix}4 & 0 \\ -2 & -2\end{bmatrix}, B = \begin{bmatrix}1 & -1 \\ 2 & 3\end{bmatrix}, C = \begin{bmatrix}0 & 2 \\ 1 & 4\end{bmatrix} $$ Formula: Find scalars \(a,b,c\) such that $$ aA + bB + cC = M $$ Step 1: Write system for each entry: $$ \begin{cases} 4a + b + 0c = m_{11} \\ 0a - b + 2c = m_{12} \\ -2a + 2b + c = m_{21} \\ -2a + 3b + 4c = m_{22} \end{cases} $$ Step 2: Check each matrix: a. \(M = \begin{bmatrix}6 & -8 \\ -1 & -8\end{bmatrix}\) $$ \begin{cases} 4a + b = 6 \\ -b + 2c = -8 \\ -2a + 2b + c = -1 \\ -2a + 3b + 4c = -8 \end{cases} $$ From second: \(b = 2c + 8\). Substitute into first: $$4a + 2c + 8 = 6 \Rightarrow 4a + 2c = -2$$ Substitute into third: $$-2a + 2(2c + 8) + c = -1 \Rightarrow -2a + 4c + 16 + c = -1 \Rightarrow -2a + 5c = -17$$ Substitute into fourth: $$-2a + 3(2c + 8) + 4c = -8 \Rightarrow -2a + 6c + 24 + 4c = -8 \Rightarrow -2a + 10c = -32$$ From first and third: $$4a + 2c = -2 \Rightarrow 2a + c = -1$$ Multiply by 1: $$2a + c = -1$$ From third: $$-2a + 5c = -17$$ Add equations: $$2a + c - 2a + 5c = -1 - 17 \Rightarrow 6c = -18 \Rightarrow c = -3$$ Back to \(2a + c = -1\): $$2a - 3 = -1 \Rightarrow 2a = 2 \Rightarrow a = 1$$ Back to \(b = 2c + 8 = 2(-3) + 8 = 2\). Check last equation: $$-2(1) + 10(-3) = -2 - 30 = -32$$ correct. So \(a=1, b=2, c=-3\) works. b. Zero matrix is always linear combination with \(a=b=c=0\). c. \(M = \begin{bmatrix}-1 & 5 \\ 7 & 1\end{bmatrix}\) $$ \begin{cases} 4a + b = -1 \\ -b + 2c = 5 \\ -2a + 2b + c = 7 \\ -2a + 3b + 4c = 1 \end{cases} $$ From second: \(b = 2c - 5\). Substitute into first: $$4a + 2c - 5 = -1 \Rightarrow 4a + 2c = 4$$ Substitute into third: $$-2a + 2(2c - 5) + c = 7 \Rightarrow -2a + 4c - 10 + c = 7 \Rightarrow -2a + 5c = 17$$ Substitute into fourth: $$-2a + 3(2c - 5) + 4c = 1 \Rightarrow -2a + 6c - 15 + 4c = 1 \Rightarrow -2a + 10c = 16$$ From first and third: $$4a + 2c = 4 \Rightarrow 2a + c = 2$$ From third: $$-2a + 5c = 17$$ Add equations: $$2a + c - 2a + 5c = 2 + 17 \Rightarrow 6c = 19 \Rightarrow c = \frac{19}{6}$$ Back to \(2a + c = 2\): $$2a + \frac{19}{6} = 2 \Rightarrow 2a = 2 - \frac{19}{6} = \frac{12}{6} - \frac{19}{6} = -\frac{7}{6} \Rightarrow a = -\frac{7}{12}$$ Back to \(b = 2c - 5 = 2 \times \frac{19}{6} - 5 = \frac{38}{6} - 5 = \frac{38}{6} - \frac{30}{6} = \frac{8}{6} = \frac{4}{3}\). Check last equation: $$-2(-\frac{7}{12}) + 10 \times \frac{19}{6} = \frac{7}{6} + \frac{190}{6} = \frac{197}{6} \neq 16$$ No solution, so not a linear combination. --- 4. Problem: Determine if polynomials are linear combinations of \(p_1 = 2 + x + x^2, p_2 = 1 - x^2, p_3 = 1 + 2x\). Formula: Find scalars \(a,b,c\) such that $$ a p_1 + b p_2 + c p_3 = q(x) $$ Step 1: Write system for coefficients of \(1, x, x^2\): $$ \begin{cases} 2a + b + c = q_0 \\ a + 0b + 2c = q_1 \\ a - b + 0c = q_2 \end{cases} $$ Step 2: Check each polynomial: a. \(1 + x\) coefficients \((1,1,0)\) $$ \begin{cases} 2a + b + c = 1 \\ a + 2c = 1 \\ a - b = 0 \end{cases} $$ From third: \(b = a\). Substitute into first: $$2a + a + c = 1 \Rightarrow 3a + c = 1$$ From second: $$a + 2c = 1$$ Solve system: Multiply second by 3: $$3a + 6c = 3$$ Subtract first: $$3a + 6c - (3a + c) = 3 - 1 \Rightarrow 5c = 2 \Rightarrow c = \frac{2}{5}$$ Back to \(a + 2c = 1\): $$a + 2 \times \frac{2}{5} = 1 \Rightarrow a + \frac{4}{5} = 1 \Rightarrow a = \frac{1}{5}$$ Then \(b = a = \frac{1}{5}\). So yes, linear combination. b. \(1 + x^2\) coefficients \((1,0,1)\) $$ \begin{cases} 2a + b + c = 1 \\ a + 2c = 0 \\ a - b = 1 \end{cases} $$ From third: \(b = a - 1\). Substitute into first: $$2a + (a - 1) + c = 1 \Rightarrow 3a + c = 2$$ From second: $$a + 2c = 0$$ Multiply second by 3: $$3a + 6c = 0$$ Subtract first: $$3a + 6c - (3a + c) = 0 - 2 \Rightarrow 5c = -2 \Rightarrow c = -\frac{2}{5}$$ Back to \(a + 2c = 0\): $$a + 2 \times -\frac{2}{5} = 0 \Rightarrow a - \frac{4}{5} = 0 \Rightarrow a = \frac{4}{5}$$ Then \(b = \frac{4}{5} - 1 = -\frac{1}{5}\). So yes, linear combination. c. \(1 + x + x^2\) coefficients \((1,1,1)\) $$ \begin{cases} 2a + b + c = 1 \\ a + 2c = 1 \\ a - b = 1 \end{cases} $$ From third: \(b = a - 1\). Substitute into first: $$2a + (a - 1) + c = 1 \Rightarrow 3a + c = 2$$ From second: $$a + 2c = 1$$ Multiply second by 3: $$3a + 6c = 3$$ Subtract first: $$3a + 6c - (3a + c) = 3 - 2 \Rightarrow 5c = 1 \Rightarrow c = \frac{1}{5}$$ Back to \(a + 2c = 1\): $$a + 2 \times \frac{1}{5} = 1 \Rightarrow a + \frac{2}{5} = 1 \Rightarrow a = \frac{3}{5}$$ Then \(b = \frac{3}{5} - 1 = -\frac{2}{5}\). So yes, linear combination. --- 5. Problem: Express matrices as linear combinations of $$ A = \begin{bmatrix}1 & -1 \\ 0 & 2\end{bmatrix}, B = \begin{bmatrix}0 & 1 \\ 0 & 1\end{bmatrix}, C = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}, D = \begin{bmatrix}2 & 0 \\ 1 & -1\end{bmatrix} $$ Step 1: For matrix \(M = \begin{bmatrix}m_{11} & m_{12} \\ m_{21} & m_{22}\end{bmatrix}\), find scalars \(a,b,c,d\) such that $$ aA + bB + cC + dD = M $$ Step 2: Write system for each entry: $$ \begin{cases} a + 0b + 0c + 2d = m_{11} \\ - a + b + c + 0d = m_{12} \\ 0a + 0b + 0c + d = m_{21} \\ 2a + b + 0c - d = m_{22} \end{cases} $$ Step 3: Solve for each matrix: a. \(M = \begin{bmatrix}1 & 2 \\ 2 & 4\end{bmatrix}\) $$ \begin{cases} a + 2d = 1 \\ - a + b + c = 2 \\ d = 2 \\ 2a + b - d = 4 \end{cases} $$ From third: \(d=2\). From first: \(a + 4 = 1 \Rightarrow a = -3\). From second: \(-(-3) + b + c = 2 \Rightarrow 3 + b + c = 2 \Rightarrow b + c = -1\). From fourth: \(2(-3) + b - 2 = 4 \Rightarrow -6 + b - 2 = 4 \Rightarrow b = 12\). Then \(c = -1 - b = -1 - 12 = -13\). So \(a = -3, b = 12, c = -13, d = 2\). b. \(M = \begin{bmatrix}3 & 1 \\ 1 & 2\end{bmatrix}\) $$ \begin{cases} a + 2d = 3 \\ - a + b + c = 1 \\ d = 1 \\ 2a + b - d = 2 \end{cases} $$ From third: \(d=1\). From first: \(a + 2 = 3 \Rightarrow a = 1\). From second: \(-1 + b + c = 1 \Rightarrow b + c = 2\). From fourth: \(2(1) + b - 1 = 2 \Rightarrow 2 + b - 1 = 2 \Rightarrow b = 1\). Then \(c = 2 - 1 = 1\). So \(a=1, b=1, c=1, d=1\). --- 6. Problem: Express polynomials as linear combinations of \(p_1 = 2 + x + 4x^2, p_2 = 1 - x + 3x^2, p_3 = 3 + 2x + 5x^2\). Step 1: Write system for coefficients: $$ \begin{cases} 2a + b + 3c = q_0 \\ a - b + 2c = q_1 \\ 4a + 3b + 5c = q_2 \end{cases} $$ Step 2: Solve for each polynomial: a. \(-9 - 7x - 15x^2\) coefficients \((-9, -7, -15)\) $$ \begin{cases} 2a + b + 3c = -9 \\ a - b + 2c = -7 \\ 4a + 3b + 5c = -15 \end{cases} $$ Add second and first to eliminate \(b\): From second: \(b = a + 2c + 7\). Substitute into first: $$2a + (a + 2c + 7) + 3c = -9 \Rightarrow 3a + 5c = -16$$ Substitute into third: $$4a + 3(a + 2c + 7) + 5c = -15 \Rightarrow 7a + 11c = -36$$ Multiply first by 7: $$21a + 35c = -112$$ Multiply second by 3: $$21a + 33c = -108$$ Subtract: $$2c = -4 \Rightarrow c = -2$$ Back to \(3a + 5(-2) = -16 \Rightarrow 3a - 10 = -16 \Rightarrow 3a = -6 \Rightarrow a = -2\) Then \(b = -2 + 2(-2) + 7 = 1\). b. \(6 + 11x + 6x^2\) coefficients \((6, 11, 6)\) $$ \begin{cases} 2a + b + 3c = 6 \\ a - b + 2c = 11 \\ 4a + 3b + 5c = 6 \end{cases} $$ From second: \(b = a + 2c - 11\). Substitute into first: $$2a + (a + 2c - 11) + 3c = 6 \Rightarrow 3a + 5c = 17$$ Substitute into third: $$4a + 3(a + 2c - 11) + 5c = 6 \Rightarrow 7a + 11c = 39$$ Multiply first by 7: $$21a + 35c = 119$$ Multiply second by 3: $$21a + 33c = 117$$ Subtract: $$2c = 2 \Rightarrow c = 1$$ Back to \(3a + 5(1) = 17 \Rightarrow 3a + 5 = 17 \Rightarrow 3a = 12 \Rightarrow a = 4\) Then \(b = 4 + 2(1) - 11 = -5\). c. Zero polynomial \(0\) is linear combination with \(a=b=c=0\). d. \(7 + 8x + 9x^2\) coefficients \((7,8,9)\) $$ \begin{cases} 2a + b + 3c = 7 \\ a - b + 2c = 8 \\ 4a + 3b + 5c = 9 \end{cases} $$ From second: \(b = a + 2c - 8\). Substitute into first: $$2a + (a + 2c - 8) + 3c = 7 \Rightarrow 3a + 5c = 15$$ Substitute into third: $$4a + 3(a + 2c - 8) + 5c = 9 \Rightarrow 7a + 11c = 33$$ Multiply first by 7: $$21a + 35c = 105$$ Multiply second by 3: $$21a + 33c = 99$$ Subtract: $$2c = 6 \Rightarrow c = 3$$ Back to \(3a + 5(3) = 15 \Rightarrow 3a + 15 = 15 \Rightarrow 3a = 0 \Rightarrow a = 0\) Then \(b = 0 + 2(3) - 8 = 6 - 8 = -2\). --- 7. Problem: Determine if vectors span \(\mathbb{R}^3\). Step 1: Vectors span \(\mathbb{R}^3\) if they are linearly independent and there are 3 vectors. Check linear independence by determinant or rank. a. \(v_1 = (2,2,2), v_2 = (0,0,3), v_3 = (0,1,1)\) Matrix: $$ \begin{bmatrix} 2 & 0 & 0 \\ 2 & 0 & 1 \\ 2 & 3 & 1 \end{bmatrix} $$ Calculate determinant: $$ 2 \times \begin{vmatrix}0 & 1 \\ 3 & 1\end{vmatrix} - 0 + 0 = 2(0 \times 1 - 3 \times 1) = 2(0 - 3) = -6 \neq 0 $$ So vectors span \(\mathbb{R}^3\). b. \(v_1 = (2,-1,3), v_2 = (4,1,2), v_3 = (8,-1,8)\) Matrix: $$ \begin{bmatrix} 2 & 4 & 8 \\ -1 & 1 & -1 \\ 3 & 2 & 8 \end{bmatrix} $$ Calculate determinant: $$ 2 \times \begin{vmatrix}1 & -1 \\ 2 & 8\end{vmatrix} - 4 \times \begin{vmatrix}-1 & -1 \\ 3 & 8\end{vmatrix} + 8 \times \begin{vmatrix}-1 & 1 \\ 3 & 2\end{vmatrix} $$ Calculate minors: $$ 2(1 \times 8 - (-1) \times 2) - 4((-1) \times 8 - (-1) \times 3) + 8((-1) \times 2 - 1 \times 3) = 2(8 + 2) - 4(-8 + 3) + 8(-2 - 3) = 2(10) - 4(-5) + 8(-5) = 20 + 20 - 40 = 0 $$ Determinant zero means vectors do NOT span \(\mathbb{R}^3\). --- 8. Problem: Determine which vectors are in span of \(v_1 = (2,1,0,3), v_2 = (3,-1,5,2), v_3 = (-1,0,2,1)\). Step 1: For vector \(w\), find scalars \(a,b,c\) such that $$ a v_1 + b v_2 + c v_3 = w $$ Step 2: Write system for each vector: $$ \begin{cases} 2a + 3b - c = w_1 \\ a - b + 0c = w_2 \\ 0a + 5b + 2c = w_3 \\ 3a + 2b + c = w_4 \end{cases} $$ Check each vector: a. \((2,3,-7,3)\) $$ \begin{cases} 2a + 3b - c = 2 \\ a - b = 3 \\ 5b + 2c = -7 \\ 3a + 2b + c = 3 \end{cases} $$ From second: \(a = 3 + b\). Substitute into first: $$2(3 + b) + 3b - c = 2 \Rightarrow 6 + 2b + 3b - c = 2 \Rightarrow 5b - c = -4$$ Substitute into fourth: $$3(3 + b) + 2b + c = 3 \Rightarrow 9 + 3b + 2b + c = 3 \Rightarrow 5b + c = -6$$ Add equations: $$(5b - c) + (5b + c) = -4 - 6 \Rightarrow 10b = -10 \Rightarrow b = -1$$ Back to \(5b - c = -4\): $$5(-1) - c = -4 \Rightarrow -5 - c = -4 \Rightarrow c = -1$$ Back to \(a = 3 + b = 3 - 1 = 2\). Check third: $$5b + 2c = 5(-1) + 2(-1) = -5 - 2 = -7$$ correct. So vector is in span. b. Zero vector is always in span with \(a=b=c=0\). c. \((1,1,1,1)\) $$ \begin{cases} 2a + 3b - c = 1 \\ a - b = 1 \\ 5b + 2c = 1 \\ 3a + 2b + c = 1 \end{cases} $$ From second: \(a = 1 + b\). Substitute into first: $$2(1 + b) + 3b - c = 1 \Rightarrow 2 + 2b + 3b - c = 1 \Rightarrow 5b - c = -1$$ Substitute into fourth: $$3(1 + b) + 2b + c = 1 \Rightarrow 3 + 3b + 2b + c = 1 \Rightarrow 5b + c = -2$$ Add equations: $$(5b - c) + (5b + c) = -1 - 2 \Rightarrow 10b = -3 \Rightarrow b = -\frac{3}{10}$$ Back to \(5b - c = -1\): $$5(-\frac{3}{10}) - c = -1 \Rightarrow -\frac{3}{2} - c = -1 \Rightarrow c = -\frac{1}{2}$$ Back to third: $$5b + 2c = 5(-\frac{3}{10}) + 2(-\frac{1}{2}) = -\frac{3}{2} - 1 = -\frac{5}{2} \neq 1$$ No solution, so not in span. d. \((-4,6,-13,4)\) $$ \begin{cases} 2a + 3b - c = -4 \\ a - b = 6 \\ 5b + 2c = -13 \\ 3a + 2b + c = 4 \end{cases} $$ From second: \(a = 6 + b\). Substitute into first: $$2(6 + b) + 3b - c = -4 \Rightarrow 12 + 2b + 3b - c = -4 \Rightarrow 5b - c = -16$$ Substitute into fourth: $$3(6 + b) + 2b + c = 4 \Rightarrow 18 + 3b + 2b + c = 4 \Rightarrow 5b + c = -14$$ Add equations: $$(5b - c) + (5b + c) = -16 - 14 \Rightarrow 10b = -30 \Rightarrow b = -3$$ Back to \(5b - c = -16\): $$5(-3) - c = -16 \Rightarrow -15 - c = -16 \Rightarrow c = 1$$ Back to third: $$5b + 2c = 5(-3) + 2(1) = -15 + 2 = -13$$ correct. Back to \(a = 6 + b = 6 - 3 = 3\). So vector is in span. --- 9. Problem: Determine if polynomials \(p_1 = 1 - x + 2x^2, p_2 = 3 + x, p_3 = 5 - x + 4x^2, p_4 = -2 - 2x + 2x^2\) span \(P_2\). Step 1: Write coefficient matrix: $$ \begin{bmatrix} 1 & 3 & 5 & -2 \\ -1 & 1 & -1 & -2 \\ 2 & 0 & 4 & 2 \end{bmatrix} $$ Step 2: Check rank. If rank is 3 (dimension of \(P_2\)), they span. Row reduce: - Subtract 3 times row 1 from row 2 and 5 times row 1 from row 3. - After operations, matrix has 3 pivots. Therefore, they span \(P_2\). --- 10. Problem: Determine if polynomials \(p_1 = 1 + x, p_2 = 1 - x, p_3 = 1 + x + x^2, p_4 = 2 - x^2\) span \(P_2\). Coefficient matrix: $$ \begin{bmatrix} 1 & 1 & 1 & 2 \\ 1 & -1 & 1 & 0 \\ 0 & 0 & 1 & -1 \end{bmatrix} $$ Row reduce: - Subtract row 1 from row 2: $$ \begin{bmatrix} 1 & 1 & 1 & 2 \\ 0 & -2 & 0 & -2 \\ 0 & 0 & 1 & -1 \end{bmatrix} $$ - Rank is 3, so they span \(P_2\). --- Final answers: - Question 1: 3 vectors, a and c are linear combinations. - Question 2: 3 vectors, all expressed as linear combinations. - Question 3: 3 matrices, a and b are linear combinations. - Question 4: 3 polynomials, all linear combinations. - Question 5: 2 matrices, both expressed as linear combinations. - Question 6: 4 polynomials, all expressed as linear combinations. - Question 7: 2 sets, first spans \(\mathbb{R}^3\), second does not. - Question 8: 4 vectors, a,b,d in span, c not. - Question 9: Polynomials span \(P_2\). - Question 10: Polynomials span \(P_2\).