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Linear Algebra
- Thread starter Gc1991
- Start date
D
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Gc1991 said:How to I solve the following for K?
x+y+4z+-2
x+2y-5z+1
3x+7y+kz+7
when I use a matrix I tend to get down to
1 0 13...-5
0 1 -9...3
0 0 24+k...1
Then Im lost
Please check your post for accuracy.
You did not post any equation to solve - your expressions do not have any "equal to" (=) sign.
According to your question, only 3rd equation is required; assuming you mean =7:Gc1991 said:How to I solve the following for K?
x+y+4z+-2
x+2y-5z+1
3x+7y+kz+7
3x + 7y + kz = 7
kz = 7 - 3x - 7y
k = (7 - 3x - 7y) / z
Hello, Gc1991!
I get a strange result . . . Did I make some blunders?
\(\displaystyle \text{Keep going . . .}\)
\(\displaystyle \text{We have: }\;\left|\begin{array}{ccc|c}1 & 0 & 13 & \text{-}5 \\ 0 & 1 & \text{-}9 & 3 \\ 0 & 1 & k+24 & 1 \end{array}\right|\)
. . \(\displaystyle \begin{array}{c}\\ \\ \frac{1}{k+24}R_3 \end{array} \left| \begin{array}{ccc|c}1 & 0 & 13 & \text{-}5 \\ 0 & 1 & \text{-}9 & 3 \\ 0 & 0 & 1 & \frac{1}{k+24} \end{array}\right|\)
\(\displaystyle \begin{array}{c}R_1-13R_3 \\ \\[-3mm] R_2 + 9R_3 \\ \\\end{array} \left|\begin{array}{ccc|c}1 & 0 & 0 & \frac{-5k-133}{k+24} \\ \\[-3mm] 0 & 1 & 0 & \frac{3k+81}{k+24} \\ \\[-3mm] 0 & 0 & 1 & \frac{1}{k+24} \end{array}\right|\)
\(\displaystyle \text{Therefore: }\;\begin{Bmatrix}x &=& \dfrac{-5k-133}{k+24} \\ \\[-3mm] y &=& \dfrac{3k+81}{k+24} \\ \\[-3mm] z &=& \dfrac{1}{k+24} \end{Bmatrix}\)
\(\displaystyle \text{We have an infinite number of possible values for }k.\)
. . . . \(\displaystyle (k \in \mathbb{R})\,\wedge\,(k \ne \text{-}24)\)
I get a strange result . . . Did I make some blunders?
How to I solve the following for \(\displaystyle k\)?
. . \(\displaystyle \begin{array}{ccc}x+y+4z &=&-2 \\ x+2y-5z &=& 1 \\ 3x+7y+kz &=& 7 \end{array}\)
When I use a matrix, I tend to get down to:
. . \(\displaystyle \left| \begin{array}{ccc|c} 1& 0 &13 & \text{-}5 \\ 0 & 1 & \text{-}9 & 3 \\ 0 & 0 & 24+k & 1 \end{array}\right|\) . RIGHT!
Then I'm lost.
\(\displaystyle \text{Keep going . . .}\)
\(\displaystyle \text{We have: }\;\left|\begin{array}{ccc|c}1 & 0 & 13 & \text{-}5 \\ 0 & 1 & \text{-}9 & 3 \\ 0 & 1 & k+24 & 1 \end{array}\right|\)
. . \(\displaystyle \begin{array}{c}\\ \\ \frac{1}{k+24}R_3 \end{array} \left| \begin{array}{ccc|c}1 & 0 & 13 & \text{-}5 \\ 0 & 1 & \text{-}9 & 3 \\ 0 & 0 & 1 & \frac{1}{k+24} \end{array}\right|\)
\(\displaystyle \begin{array}{c}R_1-13R_3 \\ \\[-3mm] R_2 + 9R_3 \\ \\\end{array} \left|\begin{array}{ccc|c}1 & 0 & 0 & \frac{-5k-133}{k+24} \\ \\[-3mm] 0 & 1 & 0 & \frac{3k+81}{k+24} \\ \\[-3mm] 0 & 0 & 1 & \frac{1}{k+24} \end{array}\right|\)
\(\displaystyle \text{Therefore: }\;\begin{Bmatrix}x &=& \dfrac{-5k-133}{k+24} \\ \\[-3mm] y &=& \dfrac{3k+81}{k+24} \\ \\[-3mm] z &=& \dfrac{1}{k+24} \end{Bmatrix}\)
\(\displaystyle \text{We have an infinite number of possible values for }k.\)
. . . . \(\displaystyle (k \in \mathbb{R})\,\wedge\,(k \ne \text{-}24)\)