Question 1209084
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Answer: <font color=red>k = -16/9</font>


Explanation


The given system is
{{{system(x + y + 3z = 10,4x + 5y + 6z = 7,kx - 3y + 2z = 3)}}}


It converts to this matrix
<table border = "1" cellpadding = "5"><tr><td>1</td><td>1</td><td>3</td><td>10</td></tr><tr><td>4</td><td>5</td><td>6</td><td>7</td></tr><tr><td>k</td><td>-3</td><td>2</td><td>3</td></tr></table>
I placed the matrix in a table or grid format to separate out each entry. 


From here we can apply row operations to get the matrix into Row Echelon Form (REF).
This is where we turn each pivot into a 1, and zero out each item below the pivots. 
The pivots are along the northwest diagonal.


Here's what the steps look like.
<table border = "1" cellpadding = "5"><tr><td>1</td><td>1</td><td>3</td><td>10</td></tr><tr><td>4</td><td>5</td><td>6</td><td>7</td></tr><tr><td>k</td><td>-3</td><td>2</td><td>3</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>1</td><td>3</td><td>10</td><td></td></tr><tr><td>0</td><td>1</td><td>-6</td><td>-33</td><td>R2 - 4*R1 --> R2</td></tr><tr><td>k</td><td>-3</td><td>2</td><td>3</td><td></td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>1</td><td>3</td><td>10</td><td></td></tr><tr><td>0</td><td>1</td><td>-6</td><td>-33</td><td></td></tr><tr><td>0</td><td>-3-k</td><td>2-3k</td><td>3-10k</td><td>R3 - k*R1 --> R3</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>1</td><td>3</td><td>10</td><td></td></tr><tr><td>0</td><td>1</td><td>-6</td><td>-33</td><td></td></tr><tr><td>0</td><td>0</td><td>-16-9k</td><td>-96-43k</td><td>R3 - (-3-k)*R2 --> R3</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>1</td><td>3</td><td>10</td><td></td></tr><tr><td>0</td><td>1</td><td>-6</td><td>-33</td><td></td></tr><tr><td>0</td><td>0</td><td>1</td><td>(-96-43k)/(-16-9k)</td><td>(1/(-16-9k))*R3 --> R3</td></tr></table>
At this point we can stop since the pivots are 1 and each item below the pivot is 0.
The third row leads to the equation {{{0x+0y+1z = (-96-43k)/(-16-9k)}}} or {{{z = (-96-43k)/(-16-9k)}}}


Set the denominator equal to zero and isolate k.
-16-9k = 0
-9k = 16
<font color=red>k = -16/9</font>


If <font color=red>k = -16/9</font>, then the denominator of {{{z = (-96-43k)/(-16-9k)}}} is zero, and it would mean there are no solutions to the original system.
For any other value of k, that fraction is some real number; and we can use that value of z to find y, and then find x using back substitution.


Verification using <a href="https://www.wolframalpha.com/input?i=x+%2B+y+%2B+3z+%3D+10%2C4x+%2B+5y+%2B+6z+%3D+7%2C%28-16%2F9%29x+-+3y+%2B+2z+%3D+3">WolframAlpha</a>


Somewhat similar questions are found <a href="https://www.algebra.com/algebra/homework/Systems-of-equations/Systems-of-equations.faq.question.1206698.html">here</a> and <a href="https://www.algebra.com/algebra/college/linear/Linear_Algebra.faq.question.1204100.html">here</a>
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