Let the 5 × 5 5\times 5 5 × 5 matrix made from the digits 1 , 2 , 6. 1,2,6. 1 , 2 , 6.
Let A = ( 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 ) \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) A = ⎝ ⎛ 1 1 2 1 6 1 2 2 1 6 1 2 2 6 6 1 2 2 6 6 1 2 2 6 6 ⎠ ⎞
To calculate matrix rank transform matrix to ụpper triangular form, using elementary row operations.
R 2 − 1 R 1 → R 2 ( m u l t i p l y 1 r o w b y 1 a n d s u b t r a c t i t f r o m 2 r o w ) ; R 3 − 2 R 1 → R 3 ( m u l t i p l y 1 r o w b y 2 a n d s u b t r a c t i t f r o m 3 r o w ) ; R 4 − 1 R 1 → R 4 ( m u l t i p l y 1 r o w b y 1 a n d s u b t r a c t i t f r o m 4 r o w ) ; R 5 − 6 R 1 → R 5 ( m u l t i p l y 1 r o w b y 6 a n d s u b t r a c t i t f r o m 5 r o w ) \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 ) R 2 − 1 R 1 → R 2 ( m u lt i pl y 1 ro w b y 1 an d s u b t r a c t i t f ro m 2 ro w ) ; R 3 − 2 R 1 → R 3 ( m u lt i pl y 1 ro w b y 2 an d s u b t r a c t i t f ro m 3 ro w ) ; R 4 − 1 R 1 → R 4 ( m u lt i pl y 1 ro w b y 1 an d s u b t r a c t i t f ro m 4 ro w ) ; R 5 − 6 R 1 → R 5 ( m u lt i pl y 1 ro w b y 6 an d s u b t r a c t i t f ro m 5 ro w )
( 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 ) R 3 ↔ R 4 ( i n t e r c h a n g e t h e 3 a n d 4 r o w s ) ( 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 ) \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)\\ ⎝ ⎛ 1 0 0 0 0 1 1 0 0 0 1 1 0 5 0 1 1 0 5 0 1 1 0 5 0 ⎠ ⎞ R 3 ↔ R 4 ( in t erc han g e t h e 3 an d 4 ro w s ) ⎝ ⎛ 1 0 0 0 0 1 1 0 0 0 1 1 5 0 0 1 1 5 0 0 1 1 5 0 0 ⎠ ⎞
R 3 / 5 → R 3 (divide the 3 row by 5 ) ( 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 ) \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} R 3 /5 → R 3 (divide the 3 row by 5 ) ⎝ ⎛ 1 0 0 0 0 1 1 0 0 0 1 1 1 0 0 1 1 1 0 0 1 1 1 0 0 ⎠ ⎞
Answer. Since there is 3 non-zero rows, then Rank ( A ) = 3. \operatorname{Rank}(\mathbf{A})=3 . Rank ( A ) = 3.
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