Sum of two solutions
Task number: 4567
Find at least one solution to the system \(\boldsymbol{Ax}=\boldsymbol b\) and at least one non-trivial solution \(\boldsymbol y\) to the system \(\boldsymbol{Ax}=\mathbf 0\) and verify that \(\boldsymbol x+\boldsymbol y\) is a solution to the system \(\boldsymbol{Ax}=\boldsymbol b\). Consider the following matrix and vector:
\(\boldsymbol A= \begin{pmatrix} 2 & 3 & -1 & -4 \\ 1 & -1 & 7 & 0 \\ 2 & -4 & 20 & 2 \\ 1 & 0 & 4 & -1\end{pmatrix} \), \(\boldsymbol b=(-9, 5, 18, 1)^{\mathrm T}\).Solution
We transform the augmented matrix of the system to a reduced echelon form using Gaussian–Jordan elimination: \((\boldsymbol A|\boldsymbol b)= \left( \begin{array}{cccc|c} 2 & 3 & -1 & -4 & -9 \\ 1 & -1 & 7 & 0 & 5 \\ 2 & -4 & 20 & 2 & 18 \\ 1 & 0 & 4 & -1 & 1 \end{array}\right) \sim\sim \left( \begin{array}{cccc|c} 1 & 0 & 4 & 0 & 2 \\ 0 & 1 & -3 & 0 & -3 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right) \)
The solution \(\boldsymbol x=(2, -3, 0, 1)^{\mathrm T}\) is in the last column and the sulution \(\boldsymbol y=(4, -3, -1, 0)^{\mathrm T}\) corresponds to the third column by subtitution \(y_3=-1\).
The verification of the solution \(\boldsymbol x+\boldsymbol y=(-2, 0, 1, 1)^{\mathrm T}\) can be performed by calculating the matrix product \(\boldsymbol A(\boldsymbol x+\boldsymbol y)\).
