Question

If F(x,y,z)=(2x + y, x+z, y-z), G (x,y,z)=(x-z, y-z, x+2y) and the matrix of F o G in the canonical basis is *IMAGE*, then a+b+c+d+e is:

Answer 1

Solution

We are given

$\begin{gathered} F(x,y,z)=(2x+y,x+z,y-z) \\ Writing\text{ out the associated matrix with respect to the canonical basis.} \\ Let\text{ the matrix be A} \\ \\ \\ A=\begin{bmatrix}{2} & {1} & {0} \\ {1} & {0} & {1} \\ {0} & {1} & {-1}\end{bmatrix} \end{gathered}$

The function

$\begin{gathered} G(x,y,z)=(x-z,y-z,x+2y) \\ Writing\text{ the associated matrix for G} \\ Let\text{ the matrix be B} \\ \\ B=\begin{bmatrix}{1} & {0} & {-1} \\ {0} & {1} & {-1} \\ {1} & {2} & {0}\end{bmatrix} \end{gathered}$

Thus, to find F o G

$\begin{gathered} FoG=AB \\ \\ FoG=\begin{bmatrix}{2} & {1} & {0} \\ {1} & {0} & {1} \\ {0} & {1} & {-1}\end{bmatrix}\begin{bmatrix}{1} & {0} & {-1} \\ {0} & {1} & {-1} \\ {1} & {2} & {0}\end{bmatrix} \\ \\ FoG=\begin{bmatrix}{2} & {1} & {-3} \\ {2} & {2} & {-1} \\ {-1} & {-1} & {-1}\end{bmatrix} \end{gathered}$

Comparing with M

It follows that

$\begin{gathered} a=2 \\ b=2 \\ c=-1 \\ d=-1 \\ e=-1 \end{gathered}$

Therefore, the answer is

$\begin{gathered} a+b+c+d+e=2+2-1-1-1 \\ \\ a+b+c+d+e=1 \end{gathered}$