SOLUTION: Find the value(s)of b for which Z = 6 −b −4 b 0 1 1 −2 1 is singular i worked it but not sure of the answer, please show me the st

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Question 1055331: Find the value(s)of b for which Z =
6 −b −4
b 0 1
1 −2 1
is singular
i worked it but not sure of the answer,
please show me the steps to get the answer.
thank you

Found 3 solutions by stanbon, Boreal, ikleyn:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find the value(s)of b for which
Z =
6 −b −4
b 0 1
1 −2 1
is singular.
--------------------
Determine the determinant of the matrix.
(6*0*1 + b*-2*-4 + 1*1*b) - (-4*0*1 + 1*-2*6 + 1*b*-b)
---------
(0 + 8b + b) - (0 - 12 - b^2)
---------
Determinant = 9b + 12 + b^2
----
Matrix is singular iff determinant = 0
Solve::
b^2 + 9b + 12 = 0
b = [-9 +- sqrt(81-4*12)]/2
b = [-9 +- sqrt(33)]/2
------------
Cheers,
Stan H.


Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
6===-b==-4== 6==-b
b====0==1===b===0
1===-2===1====1===-2
Need the determinant to be 0. I added the first and second columns to the right side, since sometimes that makes it easier to find the determinant.
8b-b-(-12-b^2)=0. Watch the signs here. We go left to right with 6*0*1=0 + b*-2*-4=8b+1*-b*1=-b. That is 7b. Then we subtract the quantity 1*0*4+-2*1*6+-1*b*-b=-12-b^2. That is adding 12+b^2. Then we have the quadratic equation below
7b+12+b^2=0
b^2+7b+12=0
(b+4)(b+3)=0
b=-4,-3
The matrix is
6===4===-4
-4==0====1
1===-2===1, and that determinant is 28-28
and
6===3===-4
-3===0===1
1===-2===1, and that determinant is -21-(-21)



Answer by ikleyn(53763) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find the value(s)of b for which Z =
6 −b −4
b 0 1
1 −2 1
is singular
i worked it but not sure of the answer,
please show me the steps to get the answer.
thank you
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Z = %28matrix%283%2C3%2C+6%2C-b%2C-4%2C++b%2C0%2C1%2C++1%2C-2%2C1%29%29.


det(Z) = [6*0*1 + (-b)*1*1 + b*(-2)*(-4)] - [1*0*(-4) + b*(-b)*1 + 6*(-2)*1] = 

          0 - b + 8b + 0 + b^2 + 12 = b^2 + 7b + 12.

The matrix is singular if and only if det(Z) = 0, i.e.

b^2 + 7b + 12 = 0.

Factor the polynomial in the left side:

(b-3)*(b-4) = 0.

The roots are b = 3  and  b = 4.

The matrix Z is singular if and only if b = 3  or  b = 4.

On calculating determinants of 3x3 matrices see the lesson
    - Determinant of a 3x3 matrix
in this site.


Also, you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic
     "3x3-Matrices, determinants, Cramer's rule for systems in three unknowns"