Consider each archetype that is a system of equations. For individual solutions listed (both for the original system and the corresponding homogeneous system) express the vector of constants as a linear combination of the columns of the coefficient matrix, as guaranteed by Theorem SLSLC. Verify this equality by computing the linear combination. For systems with no solutions, recognize that it is then impossible to write the vector of constants as a linear combination of the columns of the coefficient matrix. Note too, for homogeneous systems, that the solutions give rise to linear combinations that equal the zero vector.
Archetype A,
Archetype B,
Archetype C,
Archetype D,
Archetype E,
Archetype F,
Archetype G,
Archetype H,
Archetype I,
Archetype J
Consider each archetype that is a system of equations. Write elements of the solution set in vector form, as guaranteed by Theorem VFSLS.
Archetype A,
Archetype B,
Archetype C,
Archetype D,
Archetype E,
Archetype F,
Archetype G,
Archetype H,
Archetype I,
Archetype J
Find the vector form of the solutions to the system of equations below. \begin{align*} 2x_1-4x_2+3x_3+x_5&=6\\ x_1-2x_2-2x_3+14x_4-4x_5&=15\\ x_1-2x_2+x_3+2x_4+x_5&=-1\\ -2x_1+4x_2-12x_4+x_5&=-7 \end{align*}
Find the vector form of the solutions to the system of equations below. \begin{align*} -2 x_1 -1 x_2 -8 x_3+ 8 x_4+ 4 x_5 -9 x_6 -1 x_7 -1 x_8 -18 x_9 &= 3 \\ 3 x_1 -2 x_2+ 5 x_3+ 2 x_4 -2 x_5 -5 x_6+ 1 x_7+ 2 x_8+ 15 x_9 &= 10 \\ 4 x_1 -2 x_2+ 8 x_3+ 2 x_5 -14 x_6 -2 x_8+ 2 x_9 &= 36 \\ -1 x_1+ 2 x_2+ 1 x_3 -6 x_4+ 7 x_6 -1 x_7 -3 x_9 &= -8 \\ 3 x_1+ 2 x_2+ 13 x_3 -14 x_4 -1 x_5+ 5 x_6 -1 x_8+ 12 x_9 &= 15 \\ -2 x_1+ 2 x_2 -2 x_3 -4 x_4+ 1 x_5+ 6 x_6 -2 x_7 -2 x_8 -15 x_9 &= -7 \end{align*}
Example TLC asks if the vector \begin{equation*} \vect{w}=\colvector{13\\15\\5\\-17\\2\\25} \end{equation*} can be written as a linear combination of the four vectors \begin{align*} \vect{u}_1&=\colvector{2\\4\\-3\\1\\2\\9}& \vect{u}_2&=\colvector{6\\3\\0\\-2\\1\\4}& \vect{u}_3&=\colvector{-5\\2\\1\\1\\-3\\0}& \vect{u}_4&=\colvector{3\\2\\-5\\7\\1\\3} \end{align*} Can it? Can any vector in $\complex{6}$ be written as a linear combination of the four vectors $\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3,\,\vect{u}_4$?
At the end of Example VFS, the vector $\vect{w}$ is claimed to be a solution to the linear system under discussion. Verify that $\vect{w}$ really is a solution. Then determine the four scalars that express $\vect{w}$ as a linear combination of $\vect{c}$, $\vect{u}_1$, $\vect{u}_2$, $\vect{u}_3$.