suppose L:R3-> R3 is a linear operator and L([1,0,0])=[-3,2,4], l=([0,1,0]) = [5,-1,3] and L([0,0,1])=[-4,0,-2]find L([6,2,-7]) and find L([x,y,z]) for any [x,y,z] belongs to R3

Q) Suppose L:R3-> R3 is a linear operator  and L([1,0,0])=[-3,2,4], l=([0,1,0]) = [5,-1,3] and L([0,0,1])=[-4,0,-2]find L([6,2,-7]) and find L([x,y,z]) for any [x,y,z] belongs to R3.

To find $L\left(\right[6,2,-7\left]\right)$L([6,2,−7]), we'll use linearity:

Given:

• $L\left(\right[1,0,0\left]\right)=[-3,2,4]$L([1,0,0])=[−3,2,4]
• $L\left(\right[0,1,0\left]\right)=[5,-1,3]$L([0,1,0])=[5,−1,3]
• $L\left(\right[0,0,1\left]\right)=[-4,0,-2]$L([0,0,1])=[−4,0,−2]

We have $[6,2,-7]=6[1,0,0]+2[0,1,0]-7[0,0,1]$[6,2,−7]=6[1,0,0]+2[0,1,0]−7[0,0,1].

Since $L$L is linear, we can apply it to each term separately and then add the results:

$L\left(\right[6,2,-7\left]\right)=6L\left(\right[1,0,0\left]\right)+2L\left(\right[0,1,0\left]\right)-7L\left(\right[0,0,1\left]\right)$L([6,2,−7])=6L([1,0,0])+2L([0,1,0])−7L([0,0,1])

$=6[-3,2,4]+2[5,-1,3]-7[-4,0,-2]$=6[−3,2,4]+2[5,−1,3]−7[−4,0,−2]

$=[-18,12,24]+[10,-2,6]-[-28,0,14]$=[−18,12,24]+[10,−2,6]−[−28,0,14]

$=[-18+10+28,12-2,24+6-14]$=[−18+10+28,12−2,24+6−14]

$=[20,10,16]$=[20,10,16]

So, $L\left(\right[6,2,-7\left]\right)=[20,10,16]$L([6,2,−7])=[20,10,16].

Now, to find $L\left(\right[x,y,z\left]\right)$L([x,y,z]) for any $[x,y,z]$[x,y,z] belonging to $R^3$R3, we can use the linearity property of $L$L:

$L\left(\right[x,y,z\left]\right)=xL\left(\right[1,0,0\left]\right)+yL\left(\right[0,1,0\left]\right)+zL\left(\right[0,0,1\left]\right)$L([x,y,z])=xL([1,0,0])+yL([0,1,0])+zL([0,0,1])

$=x[-3,2,4]+y[5,-1,3]+z[-4,0,-2]$=x[−3,2,4]+y[5,−1,3]+z[−4,0,−2]

$=[-3x+5y-4z,2x-y,4x+3y-2z]$=[−3x+5y−4z,2x−y,4x+3y−2z]

So, $L\left(\right[x,y,z\left]\right)=[-3x+5y-4z,2x-y,4x+3y-2z]$L([x,y,z])=[−3x+5y−4z,2x−y,4x+3y−2z] for any $[x,y,z]$[x,y,z] belonging to $R^3$R3.3.5