Matrix transformation f:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3} \(\begin{pmatrix}4 & 1 & 3\\ 2 &-1 &

Emmy Combs

Emmy Combs

Answered question

2022-01-29

Matrix transformation
f:R3R3
(413213220)
Establish x,y and z such that,
f(xyz)
Do I just need to multiply the values of f for the 3×3 matrix? What does this mean overall?

Answer & Explanation

dodato0n

dodato0n

Beginner2022-01-30Added 9 answers

Step 1 You need to solve the non-homogeneous system (413213220)(xyz)=(451) Form then the augmented coefficients matrix and reduce it by rows: (413:4213:5220:1)R1R3(12)(110:1/2213:5413:4)R22R1R34R1 (110:1/2033:6033:6)R3R2(110:1/2033:6000:0) Well, there are infinite solutions to the above system: begin with row 3 , and remember the columns represent x,y,z from left to right, so: R2:3y+3z=6y=z2R1:x+y=12x=12y=12z+2=32z Now just choose a nice z , say z=0y=2,x=32, and one solution to your problem is (xyz)=(3/220) as you can easily check.
kumewekwah0

kumewekwah0

Beginner2022-01-31Added 14 answers

Step 1
You are given that
(xyz)=(413213220)(xyz)
Of course,
(413213220)(xyz)=(4x+y+3z2xy+3z2x+2y)
So we have
(xyz)=(4x+y+3z2xy+3z2x+2y)
That gives us three equations:
x=4x+y+3zwhich reduces to3x+y+3z=0.
y=2xy+3zwhich reduces to2x2y+3z=0.
z=2x+2ywhich reduces to2x+2yz=0.
An obvious solution to that is x=y=z=0. Notice that the determinant is |313223221|=3|2321||2321|+3|2222|=3(26)(26)+3(4+4)=12+8+24=20
Since that is not 0,x=y=z=0 is the only solution.

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