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Suppose $$A = \begin{pmatrix} 1+a_{1}+a_{1}b_{1}+b_{2} & 1+a_{1} & 1 & 0\\ a_{2}+a_{2}b_{1}+b_{3} & 1+a_{2} & 1 & 1\\ a_{3}+a_{3}b_{1} + b_{4} & a_{3} & 1 & 1\\ a_{4} + a_{4}b_{1} & a_{4} & 0 & 1\end{pmatrix}$$ Show that $$\det(A) = \det\begin{pmatrix} b_{1}-b_{2}+b_{4} & 1+a_{1}-a_{3}+a_{4}\\ -1-b_{2}+b_{3} & a_{1}-a_{2}+a_{4}\end{pmatrix}$$

I am not sure how to show this. I tried to perform some row and column operations but could simplify matrix $A$.

Note that for a general matrix of size $(3k+1)\times (3k+1)$, the determinant of $A$ can be written in a similar form. Say for a $7 \times 7$ matrix,

$$A_{7 \times 7} = \begin{pmatrix} 1+a_1+a_1b_1+b_2 & 1+a_1 & 1& 0 &0 &0 &0\\ a_2 + a_2b_1 + b_3 & 1+a_2 & 1 & 1&0&0&0\\ a_3+a_3b_1+b_4 & a_3 & 1 & 1 & 1& 0 &0\\ a_4+a_4b_1+b_5 & a_4 & 0 & 1 & 1&1 &0\\ a_5+a_5b_1+b_6 & a_5 & 0 & 0 & 1&1 &1\\ a_6+a_6b_1+b_7 & a_6 & 0 & 0 & 0&1 &1\\ a_7 + a_7b_1 & a_7 & 0 & 0 & 0&0 &1 \end{pmatrix}$$

$$\det(A) = \det\begin{pmatrix} b_{1}-b_{2}+b_{4}-b_{5}+b_{7} & 1+a_{1}-a_{3}+a_{4}-a_{6}+a_{7}\\ -1-b_{2}+b_{3}-b_{5}+b_{6} & a_{1}-a_{2}+a_{4}-a_{5}+a_{7}\end{pmatrix}$$

Any thoughts on how to show this in general?

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    $\begingroup$ Oh ok..I have written down the $7 \times 7$ matrix $A$, i see there is a pattern to it $\endgroup$ Commented Sep 10, 2022 at 8:56
  • $\begingroup$ Thanks for your quick reaction. It's clear now. $\endgroup$ Commented Sep 10, 2022 at 9:03
  • $\begingroup$ Remark: Subtracting column 2 to column 1 doesn't change the determinant and simplifies the matrix. $\endgroup$ Commented Sep 10, 2022 at 9:12
  • $\begingroup$ I wonder what argument can we use for a general matrix of size $3k+1 \times 3k+1$. I was thinking to use induction but which row and column to take in a general matrix? $\endgroup$ Commented Sep 10, 2022 at 9:33
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    $\begingroup$ Actually it is analysing the determinants of tridiagonal matrix, when columns are added later to it..it was an observation i saw.. see here: math.stackexchange.com/questions/4522806/… also observe tha tthe later right part of matrix is tridiagonal $\endgroup$ Commented Sep 10, 2022 at 10:23

3 Answers 3

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First, for any integer $n\geq 1$ and any numbers $A_1,...,A_n,B_1,...,B_n$ we have $$ \det\left(I_n+\begin{pmatrix}A_1 & B_1 & 0 & \cdots &0 \\ A_2 & B_2 & 0 & \cdots &0 \\ \vdots & \vdots & \vdots & \ddots &\vdots \\ A_n & B_n & 0 & \cdots &0 \\\end{pmatrix}\right)=\det\begin{pmatrix}1+A_1 & B_1 \\ A_2 & 1+B_2\end{pmatrix}. $$

The determinant under consideration (for all $n$) is $$ \det(M+N)=\det M\cdot\det(I_n+M^{-1}N) $$ where $$ M=\begin{pmatrix} 1 & 1 & 1 & 0 & \cdots &\cdots\\ 0 & 1 & 1 & 1 & \cdots&\cdots \\ 0 & 0 & 1 & 1 & \cdots&\cdots \\ 0 & 0 & 0 & 1 & \cdots&\cdots \\ \vdots & \vdots& \vdots& \vdots& \ddots&\vdots\\ 0&0&0&0&\cdots&1\end{pmatrix},\quad N= \begin{pmatrix} a_1(1+b_1)+b_2 & a_1 & 0 & \cdots &0 \\ a_2(1+b_1)+b_3 & a_2 & 0 & \cdots &0\\ a_3(1+b_1)+b_4 & a_3 & 0 & \cdots &0\\ \vdots & \vdots & \vdots & \ddots &\vdots\\ a_n(1+b_1)+b_{n+1} & a_n & 0 & \cdots &0 \end{pmatrix} $$ We have $\det M=1$ and $M^{-1}$ is the matrix whose first row is $(1,-1,0,1,-1,0,1,-1,0,...)$ and the $(k+1)$th row is obtained from the first by adding $k$ zeros to the left and truncating it to the right: $$ M^{-1}=\begin{pmatrix} 1&-1&0&1&-1&0&\cdots\\ 0&1&-1&0&1&-1&\cdots\\ 0&0&1&-1&0&1&\cdots\\ 0&0&0&1&-1&0&\cdots\\ 0&0&0&0&1&-1&\cdots\\ 0&0&0&0&0&1&\cdots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\\ 0&0&0&0&0&0&\cdots&1 \end{pmatrix} $$ By inspection, $M^{-1}N$ has the 3rd, 4th,... columns equal to the zero vector (as $N$ does so) and therefore we can apply the formula stated in the beginning to obtain $$ \det(M+N)=\det(I_n+M^{-1}N)=\det\begin{pmatrix}1+A_1 & B_1 \\ A_2 & 1+B_2\end{pmatrix} $$ with \begin{align} A_1&=\sum_{j=1}^n i_j(a_j(1+b_1)+b_{j+1}),& B_1&=\sum_{j=1}^n i_ja_j,\\ A_2&=\sum_{j=1}^n i_{j-1}(a_j(1+b_1)+b_{j+1}),& B_2&=\sum_{j=1}^n i_{j-1}a_j, \end{align} where we set $i_0:=0,i_1:=1,i_2:=-1$ and $i_{j+3k}:=i_j$ for all integers $k$. (To match the matrix in the original question, $b_{n+1}=0$, but the formula is valid for any value of $b_{n+1}$.)

Finally, the constraint on the remainder mod $3$ of the size is not necessary, the formula holds for all $n$.

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  • $\begingroup$ Yes but the inverse of tridiagonal matrix doesnot exist if $n$ is of the form $3k-1$, see here: math.stackexchange.com/questions/4522806/… $\endgroup$ Commented Sep 10, 2022 at 10:58
  • $\begingroup$ there is no tridiagonal matrix involved here as far as I can tell $\endgroup$ Commented Sep 10, 2022 at 10:59
  • $\begingroup$ I can't see the relation with the link, the matrix involved in my answer is identity + strictly upper diagonal (and so, invertible, with explicit inverse) and it is not tridiagonal $\endgroup$ Commented Sep 10, 2022 at 16:06
  • $\begingroup$ Is the first equation of the answer obvious or is an important consequence of a theorem? $\endgroup$ Commented Sep 11, 2022 at 5:16
  • $\begingroup$ I think you can still simplify the 2 by 2 matrix at the last line..like $C_1 \rightarrow C_1 - (1+b_1) C_2$ $\endgroup$ Commented Sep 11, 2022 at 6:24
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Remember that $\det$ is a multi linear functional and that if the columns of a matrix $A$ are linear dependent, then $\det A=0$. Hence

$$\det \begin{pmatrix} 1+a_{1}+a_{1}b_{1}+b_{2} & 1+a_{1} & 1 & 0\\ a_{2}+a_{2}b_{1}+b_{3} & 1+a_{2} & 1 & 1\\ a_{3}+a_{3}b_{1} + b_{4} & a_{3} & 1 & 1\\ a_{4} + a_{4}b_{1} & a_{4} & 0 & 1\end{pmatrix}=$$

$$=\det \begin{pmatrix} 1+a_{1}& 1+a_{1} & 1 & 0\\ 1+a_{2} & 1+a_{2} & 1 & 1\\ a_{3}& a_{3} & 1 & 1\\ a_{4} & a_{4} & 0 & 1\end{pmatrix}+\det \begin{pmatrix} a_{1}b_{1}+b_{2} & 1+a_{1} & 1 & 0\\ -1+a_2b_1+b_3 & 1+a_{2} & 1 & 1\\ a_{3}b_{1} + b_{4} & a_{3} & 1 & 1\\ a_{4}b_{1} & a_{4} & 0 & 1\end{pmatrix}=$$

$$=0+\det\begin{pmatrix} a_{1}b_{1}+b_{2} & 1+a_{1} & 1 & 0\\ -1+a_{2}b_{1}+b_{3} & 1+a_{2} & 1 & 1\\ a_{3}b_{1} + b_{4} & a_{3} & 1 & 1\\ a_{4}b_{1} & a_{4} & 0 & 1\end{pmatrix}=$$

$$=\det \begin{pmatrix} (1+a_{1})b_{1} & 1+a_{1} & 1 & 0\\ (1+a_{2})b_{1} & 1+a_{2} & 1 & 1\\ a_{3}b_{1} & a_{3} & 1 & 1\\ a_{4}b_{1} & a_{4} & 0 & 1\end{pmatrix}+\det\begin{pmatrix} -b_{1}+b_{2} & 1+a_{1} & 1 & 0\\ -1-b_{1}+b_{3} & 1+a_{2} & 1 & 1\\ b_{4} & a_{3} & 1 & 1\\ 0 & a_{4} & 0 & 1\end{pmatrix}$$

$$=0+\det\begin{pmatrix} b_{4}& 1+a_{1} & 1 & 0\\ b_4& 1+a_{2} & 1 & 1\\ b_{4} & a_{3} & 1 & 1\\ 0 & a_{4} & 0 & 1\end{pmatrix}+\det\begin{pmatrix} -b_{1}+b_{2} & -a_4& 1 & 0\\ -1-b_{1}+b_{3} & 0 & 1 & 1\\ b_{4} & 0 & 1 & 1\\ 0 & a_{4} & 0 & 1\end{pmatrix}+\det\begin{pmatrix} -b_{1}+b_{2}-b_4 & 1+a_{1}+a_4 & 1 & 0\\ -1-b_{1}+b_{3}-b_4 & 1+a_{2} & 1 & 1\\ 0 & a_{3}& 1 & 1\\ 0 & 0 & 0 & 1\end{pmatrix}=$$

$$=0+0+\det \begin{pmatrix} -b_{1}+b_{2}-b_4 & a_3 & 1 & 0\\ -1-b_{1}+b_{3}-b_4 & a_3 & 1 & 1\\ 0 & a_{3}& 1 & 1\\ 0 & 0& 0 & 1\end{pmatrix}+\det \begin{pmatrix} -b_{1}+b_{2}-b_4 & 1+a_{1}+a_4-a_3 & 1 & 0\\ -1-b_{1}+b_{3}-b_4 & 1+a_{2}-a_3 & 1 & 1\\ 0 & 0& 1 & 1\\ 0 & 0& 0 & 1\end{pmatrix}=$$

$$=\det \begin{pmatrix} -b_{1}+b_{2}-b_4 & 1+a_{1}+a_4-a_3 \\ -1-b_{1}+b_{3}-b_4 & 1+a_{2}-a_3 \\ \end{pmatrix}=$$

$$=0+\det \begin{pmatrix} -b_{1}+b_{2}-b_4 & 1+a_{1}+a_4-a_3 \\ -1-b_{1}+b_{3}-b_4-(-b_1+b_2-b_4) & 1+a_{2}-a_3-(1+a_1+a_4-a_3) \\ \end{pmatrix}=$$

$$= \det \begin{pmatrix} -b_{1}+b_{2}-b_4 & 1+a_{1}+a_4-a_3 \\ -1+b_{3}-b_2 & a_{2}-a_1-a_4 \\ \end{pmatrix}=$$

$$= \det \begin{pmatrix} b_{1}-b_{2}+b_4 & 1+a_{1}-a_3+a_4 \\ -1-b_2+b_3 & a_1-a_2+a_4 \\ \end{pmatrix}$$

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When looking for transformations to come from \begin{align*} \det\begin{pmatrix} 1+a_{1}+a_{1}b_{1}+b_{2} & 1+a_{1} & 1 & 0\\ a_{2}+a_{2}b_{1}+b_{3} & 1+a_{2} & 1 & 1\\ a_{3}+a_{3}b_{1} + b_{4} & a_{3} & 1 & 1\\ a_{4} + a_{4}b_{1} & a_{4} & 0 & 1\end{pmatrix}\tag{1} \end{align*} to \begin{align*} \det\begin{pmatrix} b_{1}-b_{2}+b_{4} & 1+a_{1}-a_{3}+a_{4}\\ -1-b_{2}+b_{3} & a_{1}-a_{2}+a_{4}\end{pmatrix}\tag{2} \end{align*} we observe that we don't have any product $a_kb_1$ in (2).

  • So, it's plausible to first try to remove the products by subtracting from column $1$, $b_1$ times column $2$.

  • We can simplify column $1$ even more by additionally subtracting column $2$ from it.

  • And since it's often convenient to increase zero entries we also subtract column $4$ from column $3$.

We obtain from (1) \begin{align*} \color{blue}{\det}&\color{blue}{\begin{pmatrix} 1+a_{1}+a_{1}b_{1}+b_{2} & 1+a_{1} & 1 & 0\\ a_{2}+a_{2}b_{1}+b_{3} & 1+a_{2} & 1 & 1\\ a_{3}+a_{3}b_{1} + b_{4} & a_{3} & 1 & 1\\ a_{4} + a_{4}b_{1} & a_{4} & 0 & 1\end{pmatrix}}\\\\ &\qquad\qquad\qquad C_1\leftarrow C_1-C_2-b_1C_2\\ &\qquad\qquad\qquad C_3\leftarrow C_3-C_4\\\\ &=\det\begin{pmatrix} -b_1+b_2 & 1+a_{1} & 1 & 0\\ -1-b_1+b_3 & 1+a_{2} & 0 & 1\\ b_{4} & a_{3} & 0 & 1\\ 0 & a_{4} & -1 & 1\end{pmatrix}\\\\ &\qquad\qquad\qquad C_2\leftarrow C_2+a_4C_3\\ &\qquad\qquad\qquad C_4\leftarrow C_4+C_3\\\\ &=\det\begin{pmatrix} -b_1+b_2 & 1+a_{1}+a_4 & 1 & 1\\ -1-b_1+b_3 & 1+a_{2} & 0 & 1\\ b_{4} & a_{3} & 0 & 1\\ \color{blue}{0} & \color{blue}{0} & \color{blue}{-1} & \color{blue}{0}\end{pmatrix}\\ &=\det\begin{pmatrix} -b_1+b_2 & 1+a_{1}+a_4 & 1\\ -1-b_1+b_3 & 1+a_{2} & 1\\ b_{4} & a_{3} & 1\end{pmatrix}\\\\ &\qquad\qquad\qquad R_1\leftarrow R_1-b_4R_3\\ &\qquad\qquad\qquad R_2\leftarrow R_2-a_3R_3\\\\ &=\det\begin{pmatrix} -b_1+b_2 -b_4& 1+a_{1}+a_4-a_3&1\\ -1-b_1+b_3-b_4 & 1+a_{2}-a_3&1\\\color{blue}{0}&\color{blue}{0}&\color{blue}{1}\end{pmatrix}\\ &=\det\begin{pmatrix} -b_1+b_2 -b_4& 1+a_{1}+a_4-a_3\\ -1-b_1+b_3-b_4 & 1+a_{2}-a_3\end{pmatrix}\\\\ &\qquad\qquad\qquad R_2\leftarrow R_2-R_1\\\\ &=\det\begin{pmatrix} -b_1+b_2 -b_4& 1+a_{1}+a_4-a_3\\ -1-b_2+b_3& -a_1+a_{2}-a_4\end{pmatrix}\\ &=\color{blue}{\det\begin{pmatrix} b_1-b_2 +b_4& 1+a_{1}-a_3+a_4\\ -1-b_2+b_3& a_1-a_{2}+a_4\end{pmatrix}}\\ \end{align*} and the claim (2) follows.

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