Question-82
Consider the following system:
\[ \begin{equation*} \begin{aligned} x+2y-2z & =b_{1}\\ 2x+5y-4z & =b_{2}\\ 4x+9y-8z & =b_{3} \end{aligned} \end{equation*} \]
What conditions should \(\displaystyle b_{1} ,b_{2} ,b_{3}\) satisfy so that the system has at least one solution?
Row reduce the augmented matrix and enforce consistency
The augmented matrix:
\[ \begin{equation*} \begin{bmatrix} 1 & 2 & -2 & & b_{1}\\ 2 & 5 & -4 & & b_{2}\\ 4 & 9 & -8 & & b_{3} \end{bmatrix} \end{equation*} \]
We first compute the row echelon form of this matrix:
\[ \begin{equation*} \begin{bmatrix} 1 & 2 & -2 & & b_{1}\\ 2 & 5 & -4 & & b_{2}\\ 4 & 9 & -8 & & b_{3} \end{bmatrix}\rightarrow \begin{bmatrix} 1 & 2 & -2 & & b_{1}\\ 0 & 1 & 0 & & b_{2} -2b_{1}\\ 4 & 9 & -8 & & b_{3} \end{bmatrix} \end{equation*} \]
\[ \begin{equation*} \begin{bmatrix} 1 & 2 & -2 & & b_{1}\\ 0 & 1 & 0 & & b_{2} -2b_{1}\\ 4 & 9 & -8 & & b_{3} \end{bmatrix}\rightarrow \begin{bmatrix} 1 & 2 & -2 & & b_{1}\\ 0 & 1 & 0 & & b_{2} -2b_{1}\\ 0 & 1 & 0 & & b_{3} -4b_{1} \end{bmatrix} \end{equation*} \]
\[ \begin{equation*} \begin{bmatrix} 1 & 2 & -2 & & b_{1}\\ 0 & 1 & 0 & & b_{2} -2b_{1}\\ 0 & 1 & 0 & & b_{3} -4b_{1} \end{bmatrix}\rightarrow \begin{bmatrix} 1 & 2 & -2 & & b_{1}\\ 0 & 1 & 0 & & b_{2} -2b_{1}\\ 0 & 0 & 0 & & b_{3} -b_{2} -2b_{1} \end{bmatrix} \end{equation*} \]
For this system to have at least one solution, the last row should be a zero row, from which we get:
\[ \begin{equation*} 2b_{1} +b_{2} -b_{3} =0 \end{equation*} \]