Singular 3x3 Matrix: Solving & Understanding

[Seneka]

Homework Statement

The problem and the statement are attached. I found the solution as shown in the attached files but I don't understand why in the solutions they added R1 and R3 to get a row equivalent to R1. Problem is Q4.

Relevant Equations

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[Mark44]

In the solution, they are taking the determinant to see if the matrix is singular. One of the three row operations you can do on determinants is replacing a row by the sum of itself and another row. The other two row operations are switching two rows, and replacing a row by a multiple of itself.
The reason they replaced R1 by R1 + R3 was to get a zero entry in the first row, which makes finding the determinant easier.

 

[Seneka]

In the solution, they are taking the determinant to see if the matrix is singular. One of the three row operations you can do on determinants is replacing a row by the sum of itself and another row. The other two row operations are switching two rows, and replacing a row by a multiple of itself.
The reason they replaced R1 by R1 + R3 was to get a zero entry in the first row, which makes finding the determinant easier.

But won't that change the value of R1 which will be equal to 3 instead of 2.

 

[RPinPA]

But won't that change the value of R1 which will be equal to 3 instead of 2.

Yes. The two equations combined to give an equation which now says 0 = 3, with a right hand side of 3.

The right-hand side doesn't matter in deciding if the determinant is 0, but it does matter if determining whether the system has zero or infinitely many solutions. 0 = 3 is a contradiction, so this system is inconsistent, no solutions. If you got 0 = 0 would be true for all x, y, z.

 

[Seneka]

Thanks @RPinPA @Mark44

I was more confused as to why row switching worked. A nice explanation I found was by looking at what the determinant does. By looking at a simple two by two the determinant will give you an area of a parallelogram. When you add a row to another one that creates a parallelogram with the same area and therefore doesn't change the value of the determinant.