# Hint for a problem on condition number

I would like to know if the second part of this question is asking something different.

Problem: Consider the linear system $19x_1+20x_2=b_1, 20x_1+21x_2=b_2$. Compute the condition number of the coefficient matrix. Is the system well-conditioned with respect to perturbations of the right-handside constants ${b_1,b_2}$?

Do I need to introduce a $\delta$ into the right-handside, or is computing the coefficient number enough to conjecture about the condition of the right-handside constants?

Thanks.

This is partial answer,
As Sasha emphasized the condition number of a matrix is dependent on matrix norm. If we denote p-norm of $A$ by $\left \| A \right \| _p$ for $p=1$ and $p=\infty$ the norms can be computed as:

$\left \| A \right \| _1 = \max \limits _{1 \leq j \leq n} \sum _{i=1} ^m | a_{ij} |$, the maximum absolute column sum of the matrix (wikipedia)

$\left \| A \right \| _\infty = \max \limits _{1 \leq i \leq m} \sum _{j=1} ^n | a_{ij} |,$ the maximum absolute row sum of the matrix (wikipedia)

$$A=\left(\begin{array}{cc} 19 & 20\\ 20 & 21 \end{array}\right) \Rightarrow A^{-1}=\left(\begin{array}{cc} -21 & 20\\ 20 & -19 \end{array}\right)$$

$$A=\left(\begin{array}{cc} 19 & 20\\ 20 & 21 \end{array}\right) \Rightarrow \left \| A \right \| _1 = \max(|19|+|20|,|20|+|21|)=41$$

$$A^{-1}=\left(\begin{array}{cc} -21 & 20\\ 20 & -19 \end{array}\right)\Rightarrow \left \| A^{-1} \right \| _1 = \max(|-21|+|20|,|20|+|-19|)=41$$ $$\kappa(A) = \| A^{-1} \|_1 \cdot \| A \|_1=41^2$$

$$A=\left(\begin{array}{cc} 19 & 20\\ 20 & 21 \end{array}\right) \Rightarrow \left \| A \right \| _\infty = \max(|19|+|20|,|20|+|21|)=41$$

$$A^{-1}=\left(\begin{array}{cc} -21 & 20\\ 20 & -19 \end{array}\right)\Rightarrow \left \| A^{-1} \right \| _\infty = \max(|-21|+|20|,|20|+|-19|)=41$$ $$\kappa(A) = \| A^{-1} \|_\infty \cdot \| A \|_\infty=41^2$$

The condition number of matrix $A$ is dependent on the choice of the matrix norm $A \mapsto \|A\|$ and is defined as $$\kappa(A) = \| A^{-1} \| \cdot \| A \|$$ In the case when the norm is $\| \cdot \|_2$ the condition number equals the ratio of the largest and the smallest in absolute values eigenvalues. Since $A$ is two by two matrix, it only has two eigenvalues.

Here is the solution using Mathematica:

In[72]:= Abs[Last[#]/First[#]] &[
SortBy[Eigenvalues[{{19, 20}, {20, 21}}], Abs]] // Simplify

Out[72]= (20 + Sqrt[401])^2