suppose L:R3to R3 is a linear operator and L[1,0,0]=[]-2,1,0 L=[0,1,0]=[-3,-2,1] L[0,0,1]=[0,-1,3] find L[-3,2,4]. give a formula for L[x,y,z] for all [x,y,z] belongs to R3 find basis

Q) Suppose L:R3to R3 is a linear operator and L[1,0,0]=[-2,1,0], L=[0,1,0]=[-3,-2,1], L[0,0,1]=[0,-1,3]  find basis

Find L[-3,2,4]. give a formula for L[x,y,z] for all [x,y,z] belongs to R3

Given that $L$L is a linear operator and $L[1,0,0]=[-2,1,0]$L[1,0,0]=[−2,1,0], $L[0,1,0]=[-3,-2,1]$L[0,1,0]=[−3,−2,1], and $L[0,0,1]=[0,-1,3]$L[0,0,1]=[0,−1,3], we can write:

[−3,2,4]​=−3[1,0,0]+2[0,1,0]+4[0,0,1]=−3L[1,0,0]+2L[0,1,0]+4L[0,0,1]​

Using the given values for $L[1,0,0]$L[1,0,0], $L[0,1,0]$L[0,1,0], and $L[0,0,1]$L[0,0,1], we can compute:

−3L[1,0,0]2L[0,1,0]4L[0,0,1]​=−3[−2,1,0]=[6,−3,0]=2[−3,−2,1]=[−6,−4,2]=4[0,−1,3]=[0,−4,12]​

Now, add these vectors:

[−3,2,4]​=[6,−3,0]+[−6,−4,2]+[0,−4,12]=[6−6+0,−3−4−4,0+2+12]=[0,−11,14]​

So, $L[-3,2,4]=[0,-11,14]$L[−3,2,4]=[0,−11,14].

General Formula for $L[x,y,z]$L[x,y,z]

To find a general formula for $L[x,y,z]$L[x,y,z] for all $[x,y,z]\in R^3$[x,y,z]∈R3, notice that $L$L is a linear transformation, and it's uniquely defined by its action on the standard basis vectors:

L[x,y,z]​=xL[1,0,0]+yL[0,1,0]+zL[0,0,1]=x[−2,1,0]+y[−3,−2,1]+z[0,−1,3]=[−2x−3y,x−2y−z,y+3z]​

So, the general formula for $L[x,y,z]$L[x,y,z] is:

$L[x,y,z]=[-2x-3y,x-2y-z,y+3z]$L[x,y,z]=[−2x−3y,x−2y−z,y+3z]