# What is m if det A = 0? A is matrix of coefficients of variables from equations: x+y+z=2 2x-y-2z=-2 x+4y+mz=8

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First we will write the coofficient matrix:

det (A)= 1 1 1

2 -1 -2

1 4 m

Now we will us ethe formula to calculate the determinant:

==> det(A) = 1*-1*m + 1*-2*1 + 1*2*4 - 1*-1*1 - 2*1*m - 1*-2*4

= -m -2 + 8 +1 - 2m + 8

Noe combine like terms:

= -3m -15

But given that det(A) = 0

==> -3m -15 = 0

Add 15 to both sides:

==> -3m = 15

Now divide by -3:

**==> m = -5**

From the given details,

A = [ (1 1 1), (2 - 2 -2), (1 4 , m)]

Therefore det A = R1 +(1/2)R2 in row1 |[(2 0 0 ), ( 2 -2 -2) ( 1 4 m)]|

Therefore det A is now expanded from the fist row:

2 (-2*m +4) +0*(..) + 0*(..) should be zero.

- 2m+4 = 0

Therefore -2m = -4.

m = -4/-2 = 2.

m = 2.

Therefore if m = 0, detA is zero.

det A = 2

First, we'll create the determinant:

|1 1 1|

det A =|2 -1 -2|

|1 4 m|

We'll calculate the determinant:

detA = 1*(-1)*m + 2*4*1 + 1*1*(-2) - 1*1*(-1) - 1*4*(-2) - 2*1*m

We'll remove the brackets:

det A = -m + 8 - 2 + 1 + 2 - 2m

We'll combine and eliminate like terms:

det A = -3m + 9

But det A = 0 => -3m + 9 = 0

We'll subtract 9 both sides:

-3m = -9

**m = 3**

**For det A = 0, the value of m has to be: m = 3**