Question #243467

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.




1
Expert's answer
2021-09-29T06:31:49-0400

Let the 5×55\times 5 matrix made from the digits 1,2,6.1,2,6.

Let A=(1111112222222221166666666)\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.

R21R1R2(multiply 1 row by 1 and subtract it from 2 row);R32R1R3(multiply 1 row by 2 and subtract it from 3 row);R41R1R4(multiply 1 row by 1 and subtract it from 4 row);R56R1R5(multiply 1 row by 6 and subtract it from 5 row)\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 )

(1111101111000000055500000)R3 R4(interchange the 3 and 4 rows)(1111101111005550000000000)\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)\\

R3/5R3 (divide the 3 row by 5 ) (1111101111001110000000000)\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 Rank(A)=3.\operatorname{Rank}(\mathbf{A})=3 .  


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