Linear Algebra Examples

Determine the Value of k for Which the System Has No Solutions
x+y+4z=2 , x+2y-4z=1 , 3x+8y+kz=2
Step 1
Write the system of equations in matrix form.
[114212-4138k2]
Step 2
Find the reduced row echelon form.
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Step 2.1
Perform the row operation R2=R2-R1 to make the entry at 2,1 a 0.
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Step 2.1.1
Perform the row operation R2=R2-R1 to make the entry at 2,1 a 0.
[11421-12-1-4-41-238k2]
Step 2.1.2
Simplify R2.
[114201-8-138k2]
[114201-8-138k2]
Step 2.2
Perform the row operation R3=R3-3R1 to make the entry at 3,1 a 0.
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Step 2.2.1
Perform the row operation R3=R3-3R1 to make the entry at 3,1 a 0.
[114201-8-13-318-31k-342-32]
Step 2.2.2
Simplify R3.
[114201-8-105k-12-4]
[114201-8-105k-12-4]
Step 2.3
Perform the row operation R3=R3-5R2 to make the entry at 3,2 a 0.
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Step 2.3.1
Perform the row operation R3=R3-5R2 to make the entry at 3,2 a 0.
[114201-8-10-505-51k-12-5-8-4-5-1]
Step 2.3.2
Simplify R3.
[114201-8-100k+281]
[114201-8-100k+281]
Step 2.4
Multiply each element of R3 by 1k+28 to make the entry at 3,3 a 1.
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Step 2.4.1
Multiply each element of R3 by 1k+28 to make the entry at 3,3 a 1.
[114201-8-10k+280k+28k+28k+281k+28]
Step 2.4.2
Simplify R3.
[114201-8-10011k+28]
[114201-8-10011k+28]
Step 2.5
Perform the row operation R2=R2+8R3 to make the entry at 2,3 a 0.
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Step 2.5.1
Perform the row operation R2=R2+8R3 to make the entry at 2,3 a 0.
[11420+801+80-8+81-1+81k+280011k+28]
Step 2.5.2
Simplify R2.
[1142010-k+20k+280011k+28]
[1142010-k+20k+280011k+28]
Step 2.6
Perform the row operation R1=R1-4R3 to make the entry at 1,3 a 0.
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Step 2.6.1
Perform the row operation R1=R1-4R3 to make the entry at 1,3 a 0.
[1-401-404-412-41k+28010-k+20k+280011k+28]
Step 2.6.2
Simplify R1.
[1102(k+26)k+28010-k+20k+280011k+28]
[1102(k+26)k+28010-k+20k+280011k+28]
Step 2.7
Perform the row operation R1=R1-R2 to make the entry at 1,2 a 0.
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Step 2.7.1
Perform the row operation R1=R1-R2 to make the entry at 1,2 a 0.
[1-01-10-02(k+26)k+28+k+20k+28010-k+20k+280011k+28]
Step 2.7.2
Simplify R1.
[1003(k+24)k+28010-k+20k+280011k+28]
[1003(k+24)k+28010-k+20k+280011k+28]
[1003(k+24)k+28010-k+20k+280011k+28]
Step 3
Since 1k+28 is undefined when k=-28, then k=-28 makes the system have no solution.
k=-28
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 [x2  12  π  xdx ] 
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