Matrix transformation f:R3→R3 (4132−13220) 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?

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Matrix transformation  f:R3→R3  (4132−13220)  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?

dodato0n · Accepted Answer

Step 1 You need to solve the non-homogeneous system (4132−13220)(xyz)=(45−1) Form then the augmented coefficients matrix and reduce it by rows: (413:42−13:5220:−1)⟶R1↔R3⋅(12)(110:−1/22−13:5413:4)⟶R2−2R1R3−4R1 (110:−1/20−33:60−33:6)⟶R3−R2(110:−1/20−33:6000:0) 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=6⟹y=z−2R1:x+y=−12⟹x=−12−y=−12−z+2=32−z Now just choose a nice z , say z=0⟹y=−2,x=32, and one solution to your problem is (xyz)=(3/2−20) as you can easily check.

kumewekwah0 · Accepted Answer

Step 1  You are given that  (xyz)=(4132−13220)(xyz)  Of course,  (4132−13220)(xyz)=(4x+y+3z2x−y+3z2x+2y)  So we have  (xyz)=(4x+y+3z2x−y+3z2x+2y)  That gives us three equations:  x=4x+y+3zwhich reduces to3x+y+3z=0.  y=2x−y+3zwhich reduces to2x−2y+3z=0.  z=2x+2ywhich reduces to2x+2y−z=0.  An obvious solution to that is x=y=z=0. Notice that the determinant is  |3132−2322−1|=3|−232−1|−|232−1|+3|2−222|=3(2−6)−(−2−6)+3(4+4)=−12+8+24=20  Since that is not 0,x=y=z=0 is the only solution.

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

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