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We will therefore look for a square sub-matrix of order 3, contained in B, whose determinant is other than zero. Only if we find one, the range of B, will already be equal to 3.
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As you can see, the procedure for calculating the range of an array is always the same. The following exercise is a little different, which will allow you to understand this concept a little more.
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However, if the rest of the elements of the third row were all zeros, the value of “a” would be decisive. Before we tell you why, let’s make that element zero.
In theory, the matrix is already triangulated. I say in theory, because we do not have a zero in the third element of the first column, but we have parameter “a”.
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We are left with a row full of zeros. Therefore, this matrix has two linearly independent vectors and as a consequence, the other vector is dependent on the other two.
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In the cases we have solved here, when triangulating the matrix we have 2 zeros in the first column and 1 zero in the second column, since the matrices had 3 rows. Remember that to triangulate the matrix is to make zeros below the main diagonal and that in each case it will be different, depending on the number of rows in the matrix.