# Chapter 4 - Vector Spaces - 4.6 Rank of a Matrix and Systems of Linear Equations - 4.6 Exercises - Page 200: 39

(a) A basis for the solution space is \left[\begin{aligned}-\frac{1}{2}\\-\frac{3}{2} \\1 \end{aligned}\right]. (b) The dimension of the solution space is $1$.

#### Work Step by Step

The coefficient matrix is given by $$\left[ \begin {array}{ccc} -1&1&1\\ 3&-1&0 \\ 2&-4&-5\end {array} \right] .$$ The reduced row echelon form is $$\left[ \begin {array}{ccc} 1&0&\frac{1}{2}\\ 0&1&\frac{3}{2} \\ 0&0&0\end {array} \right] .$$ The corresponding system is \begin{aligned} x+\frac{1}{2}z &=0\\ y+\frac{3}{2}z &=0\\ \end{aligned}. The solution of the above system is $x=-\frac{1}{2}t$,$y=-\frac{3}{2}t$, $z=t$. This means that the solution space of $Ax = 0$ consists of the vectors on the following form x= \left[\begin{aligned} x\\ y\\z \end{aligned}\right]= \left[\begin{aligned}-\frac{1}{2}t\\-\frac{3}{2}t \\t \end{aligned}\right]=t \left[\begin{aligned}-\frac{1}{2}\\-\frac{3}{2} \\1 \end{aligned}\right] . (a) A basis for the solution space is \left[\begin{aligned}-\frac{1}{2}\\-\frac{3}{2} \\1 \end{aligned}\right]. (b) The dimension of the solution space is $1$.

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