Problems from Rank of a matrix: Gaussian method

Create a $$4\times4$$ 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 $$4\times4$$ matrix: $$$A=\left( \begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 2 & -1 & 2 \\ -1 & -1 & 1 & 1 \\ 0 & 1 & 1 & 3 \end{array} \right)$$$ As we can see $$r4=r1+r2+r3$$.

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 $$1\times1$$ (all nonzero elements are).

-Are there any nonempty $$2\times2$$ submatrix? Yes $$$\left| \begin{array}{cc} 1 & 0 \\ 0 & 2 \end{array} \right| = 2 \neq 0 $$$

-Are there any nonempty $$3\times3$$ submatrixes? Yes $$$\left| \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 2 & -1 \\ -1 & -1 & 1 \end{array} \right| = -1 \neq 0 $$$

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

As the $$4\times4$$ 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:

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

rank$$(A)=3$$

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