[amp_mcq option1=”-4″ option2=”-3″ option3=”3″ option4=”4″ correct=”option3″]<\/p>\n
The set of equations $x + y + z = 1$, $ax – ay + 3z = 5$, and $5x – 3y + az = 6$ has infinite solution if $a = 3$.<\/p>\n
To see this, we can first express the equations in terms of $x$, $y$, and $z$:<\/p>\n
\\begin{align} Subtracting the second equation from the third equation, we get:<\/p>\n \\begin{align} Substituting this into the first equation, we get:<\/p>\n \\begin{align} Substituting this into the second equation, we get:<\/p>\n \\begin{align} Therefore, the solution to the system of equations is:<\/p>\n \\begin{align} Note that this solution is valid for any value of $a$, including $a = 3$. Therefore, the set of equations has infinite solution if $a = 3$.<\/p>\n For the other options, $a = -4$, $a = -3$, and $a = 4$, the system of equations has no solution. This can be seen by substituting these values of $a$ into the equations and trying to solve for $x$, $y$, and $z$. In each case, we will get an equation that is inconsistent with the other two equations.<\/p>\n","protected":false},"excerpt":{"rendered":" [amp_mcq option1=”-4″ option2=”-3″ option3=”3″ option4=”4″ correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[489],"tags":[],"class_list":["post-20001","post","type-post","status-publish","format-standard","hentry","category-linear-algebra","no-featured-image-padding"],"yoast_head":"\n
\nx &= 1 – y – z \\
\nax – ay + 3z &= 5 \\
\n5x – 3y + az &= 6
\n\\end{align<\/em>}<\/p>\n
\n4x – 2y &= 1 \\
\n\\Rightarrow x – \\frac{y}{2} &= \\frac{1}{4}
\n\\end{align<\/em>}<\/p>\n
\n1 – y – z &= 1 – \\frac{y}{2} – z \\
\n\\Rightarrow z &= \\frac{1}{2}
\n\\end{align<\/em>}<\/p>\n
\n3 \\left( \\frac{1}{2} \\right) – ay + 3z &= 5 \\
\n\\Rightarrow -ay + 3 &= 5 \\
\n\\Rightarrow -ay &= 2 \\
\n\\Rightarrow y &= -\\frac{2}{a}
\n\\end{align<\/em>}<\/p>\n
\nx &= \\frac{1}{4} \\
\ny &= -\\frac{2}{a} \\
\nz &= \\frac{1}{2}
\n\\end{align<\/em>}<\/p>\n