Question

Express v=(2,-5,3) in R³ as a linear combination of the vectors U₁ =(1,-3,2) , U₂=(2,-4,-1), U₃ =(1,-5,7)

Answer 1

Rewriting the equation in the form,

$\begin{bmatrix} 2 \\ -5\\ 3 \end{bmatrix}= x\begin{bmatrix} 1 \\ -3\\ 2 \end{bmatrix}+y\begin{bmatrix} 2 \\ -4\\ -1 \end{bmatrix}+z\begin{bmatrix} 1 \\ -5\\ 7 \end{bmatrix}\\ \begin{bmatrix} x + 2y + z \\ -3x -4y -5z \\ 2x -y +7z \end{bmatrix}=\begin{bmatrix} 2 \\ -5\\ 3 \end{bmatrix}\\ \text{reducing the matrix from the back yields}\\. \begin{bmatrix} x + 2y + z=2 \\ -3x -4y -5z=-5 \\ 2x -y +7z=3 \end{bmatrix}\\ -2R_1+R_3 \implies R_3\\ 3R_1+R_2 \implies R_2\\ \begin{bmatrix} x + 2y + z=2 \\ 2y -2z=1 \\ -5y +5z=-1 \end{bmatrix}\\ \frac{5}{2}R_2+R_3 \implies R_3\\ \\ \begin{bmatrix} x + 2y + z=2 \\ 2y -2z=1 \\ 0=\frac{3}{2} \end{bmatrix}\\ \text{The system has no solution.}\\ \text{Thus it is impossible to write the vector v as a linear combination of}\\u_1, u_2, u_3$