## Linear Algebra and Its Applications (5th Edition)

(a) $h=2$ and $k=0$ (other values for $k$ will work) (b) $h=0$ and $k=0$ (other values for $h$ and $k$ will work) (c) $h=2$ and $k=8$
We know we can find how many solutions a system has by looking at the echelon form of the corresponding augmented matrix. The augmented matrix of our system is $$\left[ \begin{array}{ccc} 1 & h & 2\\ 4 & 8 & k \end{array} \right].$$ To convert this to echelon form, we replace row 2 with _4*(row 1)+(row 2), which gives $$\left[ \begin{array}{ccc} 1 & h & 2\\ 0 & -4h+8 & -8+k \end{array} \right].$$ Now we can answer (a), (b), and (c). (a) For the system to have no solutions (i.e., to be inconsistent), the last column of the augmented matrix must be a pivot column. Therefore, we must have $$-4h+8=0$$ and $$-8+k\neq0.$$ Hence, we must have $$h=2$$ and $$k\neq 8$$ Since any value for $k$ other than $8$ will work, we can choose $k=0.$ Then for $h=2$ and $k=0$, the second row of our matrix is [0 0 -8], which corresponds to the equation $0=-8,$ which means the system is inconsistent. (b) For the system to have a unique solution, it must be consistent, which means the last column of the corresponding augmented matrix cannot be a pivot column, and have no free variables. Now, looking at the echelon form of our augmented matrix, we can see the system will not have any free variables if $-4h+8 \neq0$. Note that if $-4h+8\neq 0$ or equivalently $h\neq2$, then the last column is not a pivot column for any value of $k.$ Since we can choose any value other than 2 for $h$ and any value for $k$, we choose $h=0$ and $k=0.$ Then our matrix becomes $$\left[ \begin{array}{ccc} 1 & 0 & 2\\ 0 & 8 & -8 \end{array} \right],$$ which corresponds to the system $$x_1=2\\ 8x_2=-8.$$ Solving for $x_2$, we get $x_2=-1$. Hence our unique solution is $(x_1, x_2)=(2,-1)$. (c) For the system to have many solutions, it must be consistent and have a free variable. Looking again at the echelon form of our matrix, we see the system will have a free variable if $-4h+8=0$, or equivalently $h=2$. Now if $h=2$, then in order for the system to be consistent (i.e., in order for the last column of our matrix not to be a pivot column), we must have $-8+k=0$, or equivalently $k=8$. So for $h=2$ and $k=8$ our matrix becomes $$\left[ \begin{array}{ccc} 1 & 2 & 2\\ 0 & 0 & 0 \end{array} \right],$$ Which means our system is consistent and has infinitely many solutions.