Answer
$a. x=\frac{8}{3}-\frac{5}{6}t,\:y=-\frac{8}{3}+\frac{5}{6}t,\:z=t\:$
$b.x=\frac{18}{7}-\frac{11}{14}t,\:y=-\frac{20}{7}+\frac{13}{14}t,\:z=t$
$c.x=3-t,\:y=-3+t,\:z=t$
$d.Each\:system\:has\:an\:infinite\:number\:of\:solutions.$
Work Step by Step
$a.\:The\:augmented\:matrix\:corresponding\:to\:this\:system\:of\:equations\:is\\\begin{bmatrix}4&-2&5&16\\ 1&1&0&0\:\:\:\end{bmatrix}\\Putting\:into\:row-echelon\:form\\\begin{bmatrix}4&-2&5&16\\1&1&0&0\end{bmatrix}\:-4R_2+R_1\rightarrow R_1\\\begin{bmatrix}0&-6&5&16\\ \:\:1&1&0&0\:\:\:\end{bmatrix}\:R_1\leftrightarrow R_2\\\begin{bmatrix}1&1&0&0\\ 0&-6&5&16\end{bmatrix}-\frac{1}{6}R_2\rightarrow R_2\\\begin{bmatrix}1&1&0&0\\ \:0&1&-\frac{5}{6}&-\frac{8}{3}\end{bmatrix}\\We\:can\:see\:that\:this\:in\:row-echelon\:form.\:The\:corresponding\:system\:of\:equations\:is\:x+y=0\\y-\frac{5}{6}z=-\frac{8}{3}\\We\:can\:write\:the\:solution\:to\:this\:system\:parametrically\:by\:letting\:z\:as\:a\:free\:variable\:t\:from\:the\:second\:equation\:we\:can\:solve\:for\:y\\y-\frac{5}{6}t=-\frac{8}{3}\\\Rightarrow \:y=-\frac{8}{3}+\frac{5}{6}t\\Substituting\:this\:into\:the\:first\:equation\\x-\frac{8}{3}+\frac{5}{6}t=0\\\Rightarrow \:\:x=\frac{8}{3}-\frac{5}{6}t\\Therefore,\:the\:solution\:is\:x=\frac{8}{3}-\frac{5}{6}t,\:y=-\frac{8}{3}+\frac{5}{6}t,\:z=t\:\\$
$\\b.\:The\:augmented\:matrix\:corresponding\:to\:this\:system\:of\:equations\:is\:\\\begin{bmatrix}4&-2&5&16\\-1&-3&2&6\:\:\:\end{bmatrix}\\Putting\:into\:row-echelon\:form\\\begin{bmatrix}4&-2&5&16\\ -1&-3&2&6\:\:\:\end{bmatrix}\:4R_2+R_1\rightarrow R_1\\\begin{bmatrix}0&-14&13&40\\ \:\:-1&-3&2&6\:\:\:\end{bmatrix}\:R_1\leftrightarrow R_2\\\begin{bmatrix}-1&-3&2&6\\ 0&-14&13&40\end{bmatrix}-R_1\rightarrow R_1\\\begin{bmatrix}1&3&-2&-6\\ \:0&-14&13&40\end{bmatrix}\:-\frac{1}{14}R_2\rightarrow R_2\\\begin{bmatrix}1&3&-2&-6\\ \:\:0&1&-\frac{13}{14}&-\frac{20}{7}\end{bmatrix}\\We\:can\:see\:that\:this\:in\:row-echelon\:form.\:The\:corresponding\:system\:of\:equations\:is\:x+3y-2z=-6\\y-\frac{13}{14}z=-\frac{20}{7}\\We\:can\:write\:the\:solution\:to\:this\:system\:parametrically\:by\:letting\:z\:as\:a\:free\:variable\:t\:from\:the\:second\:equation\:we\:can\:solve\:for\:y\\y-\frac{13}{14}t=-\frac{20}{7}\\\Rightarrow \:y=-\frac{20}{7}+\frac{13}{14}t\\Substituting\:this\:into\:the\:first\:equation\\x-3\left(-\frac{20}{7}+\frac{13}{14}\right)-2t=-6\\\Rightarrow \:\:x-\frac{60}{7}+\frac{39}{14}t-2t=-6\\\Rightarrow \:\:\:x-\frac{60}{7}+\frac{11}{14}t=-6\\\Rightarrow \:\:\:x=\frac{18}{7}+\frac{11}{14}t\\Therefore,\:the\:solution\:is\:x=\frac{18}{7}-\frac{11}{14}t,\:y=-\frac{20}{7}+\frac{13}{14}t,\:z=t\\$
$\\c.\:The\:augmented\:matrix\:corresponding\:to\:this\:system\:of\:equations\:is\:\:\\\begin{bmatrix}1&1&0&0\\ -1&-3&2&6\:\:\:\end{bmatrix}\\Putting\:into\:row-echelon\:form\\\begin{bmatrix}1&1&0&0\\ -1&-3&2&6\:\:\:\end{bmatrix}\:R_1+R_2\rightarrow R_2\\\begin{bmatrix}1&1&0&0\\ 0&-2&2&6\end{bmatrix}\:-\frac{1}{2}R_2\leftrightarrow R_2\\\begin{bmatrix}1&1&0&0\\ 0&1&-1&-3\end{bmatrix}\\We\:can\:see\:that\:this\:in\:row-echelon\:form.\:The\:system\:of\:equations\:described\:by\:this\:matrix\:is\\x+y=0\\y-z=-3\\We\:can\:write\:the\:solution\:to\:this\:system\:parametrically\:by\:letting\:z\:as\:a\:free\:variable\:t\:from\:the\:second\:equation\:we\:can\:solve\:for\:y\|y-t=-3\\\Rightarrow \:y=-3+t\\Substituting\:this\:into\:the\:first\:equation\\x-3+t=0\\x=3-t\\Therefore,\:the\:solution\:is\:x=3-t,\:y=-3+t,\:z=t\\$
$\\d.\:Each\:system\:has\:an\:infinite\:number\:of\:solutions.$
