Suppose I am given a matrix $A_{1×3}$ and another matrix $C$. I have to find the best possible solution of the form $(A + B)\cdot X = C$. But the solution will be considered best only if
- $X_{3×3}$ is a diagonal matrix, more better if it is an identity matrix of the form constant$\cdot I_{3×3}$
- And matrix $B_{1×3}$ is such that $B = [b_1 \space \space b_2 \space \space b_3]$ where either $b_1=b_2$ or $b_2=b_3$ or $b_1=b_3$ or $b_1=b_2=b_3$ (best).
Here is an example: $A = [2 \space \space 8 \space \space 45]$, $C = [9 \space \space 27 \space \space 46]$
So in this case the best solution will be $B = [1 \space \space 1 \space \space 1]$ and
$$X=\begin{bmatrix}
3&0&0\\
0&3&0\\
0&0&1
\end{bmatrix}$$
In matrix $X$ also we should try to make $a_{1,1}=a_{2,2}=a_{3,3}$. The attempt should be made to find values such that $b = [0 \space \space 0 \space \space 0]$ and $X = k \cdot I_{3×3}$
or $B = [n \space \space n \space \space n]$ and $X = I_{3×3}$.
Let $x_1, x_2, x_3$ be the diagonal elements of $X$. Assuming that $b_1 = b_2 = b_3 =b$, this reduces to the system of equations $$(a_1 + b)x_1 = c_1\\(a_2 + b)x_2 = c_2\\(a_3 + b)x_3 = c_3$$
This is three equations in four unknowns. If you pick any $b$ other than $b = -a_1, b= -a_2, b= -a_3$, you will have a unique solution with $$x_i = \frac {c_i}{a_i + b}. \quad i = 1,2,3$$
Now if you also want $x_1 = x_2 = x_3 = x$, then instead we get three equations in only two unknowns, which can only have solutions when the equations are not independent. In particular when these ratios are equal: $$\frac {c_1 - c_2}{a_1 - a_2} = \frac {c_2 - c_3}{a_2 - a_3} = \frac {c_3 - c_1}{a_3 - a_1}$$ $x$ is then the common value, and $b = \frac {c_i}x - a_i$ for any $i$.