Please Help with Matrices - I am so confused.?

Question

x – y + 2z = 13
2x + xy – z = -6
-x + 3y + z = -7

need coefficent matrix
inverse
transpose
determinant

need explanation so I can do it myself

the 2nd equation should be

2x + 2y-z +6

Answer 1

Hi,

The coefficient matrix for your equations is this 3 x 3 matrix:

[1..-1...2]
[2...1..-1] I assumed the x wasn't supposed to be in 2nd term.
[-1..3...1]

The transpose of a matrix is when the rows and columns are reversed. So a 2 x 4 matrix would have a 4 x 2 transpose matrix. The transpose of the matrix above is:

[1...2..-1]
[-1..1...3]
[2..-1...1]

A determinant is the numerical value that can be found for any square matrix. Do these on your graphing calculator. Enter your matrix above as a 3 x 3 matrix as [A]. Then enter det[A] to get your answer of 19.

To do this by hand, take your matrix
[1..-1...2]
[2...1..-1]
[-1..3...1]

Recopy first 2 columns to the right.
[1..-1...2]..1..-1
[2...1..-1]..2...1
[-1..3...1].-1..3
Now start in the upper left corner and multiply each diagonal together. Add those results.
1(1)(1) + (-1)(-1)(-1) + 2(2)(3) = 1 + -1 + 12 = 12
Now start in the BOTTOM left corner and multiply each diagonal together. Add those results.
(-1)(1)(2) + 3(-1)(1) + 1(2)(-1) = -2 + -3 + -2 = -7

Now take the first total MINUS the second total. This gives you the value of the determinant. 12 - (-7) = 19

I hope that helps you!! :-)

Answer 2

The coefficient matrix is made of the coefficients of x, y, z:

[ 1 -1 2 ]
[ 2 ? -1 ]
[ -1 3 1 ]

The ? is a possible typo: did you mean something other than xy in the second equation?

The transpose of a matrix is obtained swapping rows by columns. In this case,

[ 1 2 -1 ]
[ -1 ? 3 ]
[ 2 -1 1 ]

For matrix inversion, determinants of matrices, and general information abut matrices, please see the sources.

Answer 3

Ok go

-------Coefficent-------------

..x .– y + 2z = 13
2x + 2y – z = 6
-x + 3y + z = -7
------------------------


│.1.....-1.....2....│.x = ¦...13 ¦
│.2......2....-1....│.y = ¦.....6 ¦
│-1......3.....1....│.z = ¦....-7 ¦


Now

│.1.....-1.....2....│.
│.2......2....-1....│=D
│-1......3.....1....│

D= 22
--------------¦


For "X"

│13....-1.....2....│.
│.6......2....-1....│
│-7......3.....1....│
-----------------------...= D(X)
..........D

X= 128 / 22.......> X= 5,82
--------------------------------------...


For "Y"

│.1.....13.....2....│.
│.2.......6....-1....│
│-1......-7.....1....│
------------------------...= D(Y)
..........D

Y= -30 / 22 =......> Y= -1,36
--------------------------------------...


For "Z"

│.1.....-1...13....│.
│.2......2.....6....│
│-1......3....-7....│
-----------------------...= D(Z)
..........D

Z= 64 / 22......> Z= 2,91
------------------------------------¦


Los coefficent are: X = 5,82; Y = -1,36; Z= 2,91
______________________________________...

---------Inverse--------

¦ .1...-1....2 ¦
¦ .2....2...-1 ¦ = A
¦ -1....3....1 ¦


¦ .1...-1....2 ¦
¦ .2....2...-1 ¦ => D(A) = 22
¦ -1....3....1 ¦


¦ .2...-1 ¦
¦ .3....1 ¦....¦ a11 ¦= 5

¦ .2...-1 ¦
¦ -1....1 ¦....-¦ a12 ¦ = -1

¦ .2....2 ¦
¦ -1....3 ¦....¦ a13 ¦= 8

¦ -1....2 ¦
¦ .3....1 ¦....-¦ a21¦ = 7

¦ .1....2 ¦
¦ -1....1 ¦....¦ a22 ¦= 3

¦ .1...-1 ¦
¦ -1....3 ¦....-¦ a23 ¦ = -2

¦ -1.....2 ¦
¦ .2....-1 ¦....¦ a31 ¦= -3

¦ .1.....2 ¦
¦ .2....-1 ¦....-¦ a32 ¦ = 5

¦ 1....-1 ¦
¦ 2.....2 ¦....¦ a33 ¦= 4


Now A^-1


.........¦..5.....7....-3 ¦
.1/22..¦.-1.....3.....5 ¦
.........¦..8....-2.....4 ¦

______________________________________...

----transpose----

¦ .1...-1....2 ¦
¦ .2....2...-1 ¦ = A
¦ -1....3....1 ¦


Now A^T

¦ .1....2...-1 ¦
¦ -1....2....3 ¦ = T
¦ .2...-1....1 ¦

______________________________________...

-----Determinat------

│.1.....-1.....2....│= (1x2x1)+(2x3x2)+(-1x-1-1)=2+12-1= 13
│.2......2....-1....│= -(2x2x-1)-(-1x3x1)-(1x-1x2)= 4+3+2=9
│-1......3.....1....│= 13 +9 =22

D = 22
-----------¦


Bye.Bye

Answer 4

The xy term on the middle equation does not fit with using a matrix.

Answer 5

equation (2) is a nonlinear eqn., so
you cannot use matrices , instead ,

eliminate x from (1) & (3) , eliminate z
from (1) & (3) , then get ,

y = (6-3z)/4
y = (2x-27)/4
use the above forms in eqn. (2)