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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hala718 | High School Teacher | (Level 1) Educator Emeritus

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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

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neela | High School Teacher | (Level 3) Valedictorian

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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

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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

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