If matrix $A$ has orthogonal columns, then what is solution of $Ax=b$ Let $A\in M^4(\mathbb R)$ is matrix which column is $a_1,a_2,a_3,a_4$ they are orthogonal and length is $2,1,3,2$.
a) Write an explicit form for the solution of $Ax=b$
b) Write matrix as sum of four matrix which rank is $1$.
c) Let $B=AS$ where $S=\operatorname{diag}(3,6,2,3)$. Find $B^TB$ and $\lVert Bx\rVert_2/\lVert x\rVert_2$ if $x\ne 0$
Since $\operatorname{rank}A=4$ then $Ax=b$ has a unique solution, then the solution can be expressed using base $x=\frac{(a_1,x)a_1}{2}+\frac{(a_2,x)a_2}{1}+\frac{(a_3,x)a_3}{3}+\frac{(a_4,x)a_4}{2}$, for b) I only think that you can write as four matrices that first column is $a_1$, and other zero and the same for second, third and fourth matrix, I do not see that I can use something specific, for c) $B^TB=(AS)^T(AS)=S^TA^TAS=S^TS=\operatorname{diag}(9,36,4,9)$ and $\lVert Bx\rVert_2/\lVert x\rVert_2=\lVert Sx\rVert_2/\lVert x\rVert$
 A: $(a)$ The columns of $A$ are orthogonal so in particular they are linearly independent. Hence $A$ is invertible so the unique solution is given by $x = A^{-1}b$.
Your $(b)$ is correct.
$(c)$ Notice that 
$$(A^TA)_{ij} = \sum_{k=1}^n (A^T)_{ik}A_{kj} = \sum_{k=1}^n A_{ki}A_{kj} = \langle a_i, a_j\rangle = \|a_i\|^2\delta_{ij}$$ so $A^TA = \operatorname{diag}(\|a_1\|^2, \|a_2\|^2, \|a_3\|^2, \|a_4\|^2) = \operatorname{diag}(4, 1, 9, 4)$. Therefore
$$B^TB = S^TA^TAS = \operatorname{diag}(3, 6, 2, 3) \operatorname{diag}(4, 1, 9, 4)\operatorname{diag}(3, 6, 2, 3) = \operatorname{diag}(36, 36, 36, 36) = 36I$$
So
$$\frac{\|Bx\|^2}{\|x\|^2} = \frac{\langle Bx, Bx\rangle}{\|x\|^2} = \frac{\langle B^TBx, x\rangle}{\|x\|^2} = 36$$

Alternative solution for $(a)$:
If $Ax = b$ then also $$\begin{bmatrix} 4x_1 \\ x_2 \\ 9x_3\\ 4x_4\end{bmatrix} = \operatorname{diag}(4, 1, 9, 4) x = A^TAx = A^Tb = \begin{bmatrix} a_1 \\ a_2 \\ a_3\\ a_4\end{bmatrix}b = \begin{bmatrix} \langle a_1, b\rangle \\ \langle a_2, b\rangle \\ \langle a_3, b\rangle \\ \langle a_4, b\rangle\end{bmatrix}$$
so $$x = \begin{bmatrix} \frac{1}4\langle a_1, b\rangle \\ \langle a_2, b\rangle \\ \frac{1}9\langle a_3, b\rangle \\ \frac{1}4\langle a_4, b\rangle\end{bmatrix}$$
A: Recall that if the columns of $A$ are $b_1,\dot,b_4$ then $$Ax=x_1b_1+\dots+x_4b_4.$$So  you need to solve $$x_1b_1+\dots+x_4b_4=b,$$which is easy: Take the inner product of both sides with $b_k$.
