Solve for X. [[-4,-2],[2,0]] dot product X + [[1,1],[0,-4]] = X dot product [[0,1],[1,2]]   X = ??

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rakesh05's profile pic

rakesh05 | High School Teacher | (Level 1) Assistant Educator

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Given system is

           `[[-4,-2],[2,0]].X+[[1,1],[0,-4]]=X.[[0,1],[1,2]]`   ...... (1)

Because matrices of same type can be added or subtracted. So the product matrix `[[-4,-2],[2,0]].X`  must be a  `2xx2` matrix. Also matrices of same type can be compared. So the product matrix `X.[[0,1],[1,2]]`  must be a   `2xx2`  matrix. Which is possible only when `X`  itself is a `2xx2` matrix.

Let  `X=[[x_1,x_2],[x_3,x_4]]` .

Now the system (1) can be written as

        `[[-4,-2],[2,0]].[[x_1,x_2],[x_3,x_4]]+[[1,1],[0,-4]]=[[x_1,x_2],[x_3,x_4]].[[0,1],[1,2]]`

or,

       `[[-4x_1-2x_3, -4x_2-2x_4],[2x_1+0x_3,2x_2+0x_4]]+[[1,1],[0,-4]]=[[0x_1+x_2,x_1+2x_2],[0x_3+x_4,x_3+2x_4]]`

or,

`[[-4x_1-2x_3,-4x_2-2x_4],[2x_1,2x_2]]-[[x_2,x_1+2x_2],[x_4,x_3+2x_4]]=-[[1,1],[0,-4]]`

or,

`[[-4x_1-2x_3-x_2, -4x_2-2x_4-x_1-2x_2],[2x_1-x_4, 2x_2-x_3-2x_4]]=[[-1,-1],[0,4]]`

Comparing the corresponding elements of the matrices we get the following system of linear equations

`-4x_1-2x_3-x_2=-1`

or,     `4x_1+x_2+2x_3=1`     ........     (2)

`-6x_2-2x_4-x_1=-1`

or,     `x_1+6x_2+2x_4=1`     ............  (3)

                     `2x_1-x_4=0 `    ...........    (4)

` 2x_2-x_3-2x_4=4`              ...............(5)

Solving equations (2),(3),(4) and (5) we get

`x_1=-1` ,   `x_2=1`  ,   `x_3=2`     and `x_4=-2` .

So,      `X=[[-1,1],[2,-2]]` .     Answer.

pramodpandey's profile pic

pramodpandey | College Teacher | (Level 3) Valedictorian

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[[-4,-2],[2,0]] dot product X + [[1,1],[0,-4]] = X dot product [[0,1],[1,2]]

Let `X=[[a,b],[c,d]]`

`[[-4,-2],[2,0]].[[a,b],[c,d]]+[[1,1],[0,-4]]=[[a,b],[c,d]].[[0,1],[1,2]]`

`[[-4a-2c,-4b-2d],[2a,2b]]+[[1,1],[0,-4]]=[[b,a+2b],[d,c+2d]]`

`[[-4a-2c+1,-4b-2d+1],[2a,2b-4]]=[[b,a+2b],[d,c+2d]]`

comparing elements in matrices.

-4a-2c+1=b         (i)

-4b-2d+1=a+2b    (ii)

2a=d                   (iii)

2b-4=c+2d             (iv)

from (ii) and (i) ,(iii),(iv) we have 

6b+5a=1

b+4a+2c=1

2b-4a-c=4

Solving system of equation ,we have

b=1,a=-1,d=-2 and c=2

Thus `X=[[-1,1],[2,-2]]`

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