Problems from Rank of a matrix: Gaussian method

Create a $4 \times 4$ matrix, with the peculiarity that one of its rows is a linear combination of the other $3$ . Calculate the rank of the matrix.

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Development:

We create the $4 \times 4$ matrix: $A = (\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 2 & - 1 & 2 \\ - 1 & - 1 & 1 & 1 \\ 0 & 1 & 1 & 3 \end{array})$ As we can see $r 4 = r 1 + r 2 + r 3$ .

Let's calculate the rank.

The first thing is to remove the linear row combination of others, that is, remove row $4$ .

-Obviously there is nonempty submatrix $1 \times 1$ (all nonzero elements are).

-Are there any nonempty $2 \times 2$ submatrix? Yes $| \begin{array}{cc} 1 & 0 \\ 0 & 2 \end{array} | = 2 \neq 0$

-Are there any nonempty $3 \times 3$ submatrixes? Yes $| \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 2 & - 1 \\ - 1 & - 1 & 1 \end{array} | = - 1 \neq 0$

$| \begin{array}{ccc} 0 & 1 & 0 \\ 2 & - 1 & 2 \\ - 1 & 1 & 1 \end{array} | = - 2 - 2 = - 4 \neq 0$

As the $4 \times 4$ matrix is empty (the row $4$ is a linear combination of the other $3$ rows) the order of the biggest nonempty square submatrix is $3$ .

Therefore, rank $(A) = 3$ .

Solution:

$(\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 2 & - 1 & 2 \\ - 1 & - 1 & 1 & 1 \\ 0 & 1 & 1 & 3 \end{array})$

rank $(A) = 3$

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