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For a certain value of , the system

 

\begin{align*}
x + y + 3z &= 10, \\
-4x + 2y + 5z &= 7, \\
kx + z &= 3
\end{align*}

 

has no solutions. What is this value of ?

 Sep 19, 2019
 #1
avatar+6179 
+2

\(\text{putting this into augmented matrix form}\\ \begin{pmatrix} 1&1&3&|&10\\ -4&2&5&|&7\\ k&0&1&|&3 \end{pmatrix}\)

 

\(\text{Now we row reduce this a bit}\\ \begin{pmatrix} 1&1&3&|&10\\ 0&6&17&|&47\\ 0&-k&1-3k&|&3-10k \end{pmatrix}\\ \begin{pmatrix} 1&1&3&|&10\\ 0&1&\frac{17}{6}&|&\frac{47}{6}\\ 0&0&1-\frac{k}{6}&|&3-\frac{10k}{6} \end{pmatrix}\)

 

\(\text{Now we see from the bottom row that if we let $k=6$ we have $0x+0y+0z = -7$}\\ \text{This is clearly impossible and thus $k=6$ is the number we are after}\)

.
 Sep 20, 2019
 #2
avatar+24378 
+2

For a certain value of \(k\), the system
\(\begin{align*} x + y + 3z &= 10, \\ -4x + 2y + 5z &= 7, \\ kx + z &= 3 \end{align*}\)
has no solutions. What is this value of  \(k\)?

 

answer see: https://web2.0calc.com/questions/help_12179

 

laugh

 Sep 20, 2019

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