Consider the following $3$x$3$ matrix which contains matrix elements:
$$A = \begin{bmatrix} M & N & 0 \\ N & M & N \\ 0 & N & M \end{bmatrix} \tag 1$$
where:
$$M = \begin{bmatrix} 1-2a-2b & a & 0 \\ a & 1-2a-2b & a \\ 0 & a & 1-2a-2b \end{bmatrix} \tag 2$$
and:
$$N = \begin{bmatrix} b & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & b \end{bmatrix} \tag 3$$
where $a$ and $b$ are known non-zero positive constants. The infinity norm of matrix $A$ is the maximum sum of one of its rows. The way I understand this, the infinity norm of matrix $A$ is:
$$ ||A||_{\infty}=|N|+|M|+|N| = 2|N| + |M| \tag 4 $$
On this link, I found out that the absolute value of a matrix is actually its determinant.
According to the Sarrus rule, the determinant of matrix $M$ is:
$$|M| = \det(M) = -4a^3 + (10-20b)a^2 - (1-2b)6a - (2b - 1)^3 \tag 5$$
and the determinant of matrix $N$ is:
$$|N| = \det(N) = b^3 \tag 6$$
By inserting expressions $(5)$ and $(6)$ into equation $(4)$, I have obtained:
$$||A||_{\infty}= -4a^3 + 10a^2-6a -6b^3+12b^2-6b - 20a^2b - 24ab^2 +24ab + 1 \tag7$$
I found out on page $[6]$ of this paper, that if:
$$ 1 - 2a - 2b \geq 0 \tag8$$
then the infinity norm of matrix $A$ is:
$$||A||_{\infty} = 1 \tag9$$
However, when I insert $b=0.5-a$ into equation $(7)$, I do not get that the infinity norm of matrix $A$ is one.
My question is, what am I doing wrong here? How can I prove that if $ (1 - 2a - 2b) \geq 0 $, then the infinity norm of matrix $A$ is one?
inf-norm of a matrix is maximum of row sums, i.e.: $$\| A\|_\infty = \max_{1\leq i \leq n} \sum_{j=1}^n |A_{i,j}| $$ and there's no relation like, $\|A\|_\infty = |N|+|M|+|N|$. and showing determinant of matrix ($det(A)$) as absolute value ($|A|$) is just a symbol, and it's not even common.
look for row of $A$. because $|a|$ and $|b|$ are positive having more of these elements increase row sum. element $|1-2a-2b|$ appears on all rows exactly once. so the maximum value happens in middle row of $A$ ,middle row of $M$ (because both $a$ and $b$ appears twice): $$ \| A \|_\infty = |1-2a-2b|+2|a|+2|b| $$ $a$ and $b$ are positive and if $1-2a-2b \geq 0$ : $$ \| A \|_\infty = 1-2a-2b+2a+2b = 1. $$