Answer to Question #243467 in Linear Algebra for Buzz

Question

Select 3 different digits from these numbers (987621). Use only these numbers as coefficient and create a matrix of 5x5 (with all coefficient non zero) that has rank 3. Also, explain why the rank is equal to 3.

Answer 1

Let the $5\times 5$ matrix made from the digits $1,2,6.$

Let $\mathbf{A}=\left(\begin{array}{lllll} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 & 2 \\ 2 & 2 & 2 & 2 & 2 \\ 1 & 1 & 6 & 6 & 6 \\ 6 & 6 & 6 & 6 & 6 \end{array}\right)$

To calculate matrix rank transform matrix to ụpper triangular form, using elementary row operations.

$\mathrm{R}_{2}-1 \mathrm{R}_{1} \rightarrow \mathrm{R}_{2} (multiply \ 1 \ row \ by \ 1 \ and \ subtract \ it \ from \ 2 \ row) ;\\ \mathrm{R}_{3}-2 \mathrm{R}_{1} \rightarrow \mathrm{R}_{3} (multiply \ 1 \ row \ by \ 2 \ and \ subtract \ it \ from \ 3 \ row);\\ \mathrm{R}_{4}-1 \mathrm{R}_{1} \rightarrow \mathrm{R}_{4} (multiply \ 1 \ row \ by \ 1 \ and \ subtract \ it \ from \ 4 \ row ) ;\\ \mathrm{R}_{5}-6 \mathrm{R}_{1} \rightarrow \mathrm{R}_{5} (multiply \ 1 \ row \ by \ 6 \ and \ subtract \ it \ from \ 5 \ row )$

$\left(\begin{array}{lllll} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 5 & 5 & 5 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right)\\ \mathrm{R}_{3} \leftrightarrow \ \mathrm{R}_{4} (interchange \ the \ 3 \ and \ 4 \ rows)\\ \left(\begin{array}{lllll} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 5 & 5 & 5 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right)\\$

$\begin{aligned} &\mathrm{R}_{3} / 5 \rightarrow \mathrm{R}_{3} \text { (divide the } 3 \text { row by } 5 \text { ) } \\ &\left(\begin{array}{lllll} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right) \end{aligned}$

Answer. Since there is 3 non-zero rows, then $\operatorname{Rank}(\mathbf{A})=3 .$