### Video Transcript

Find the rank of the following
matrix.

Recall that the rank of a matrix π΄
is the number of rows or columns in the largest square submatrix of π΄ with a
nonzero determinant. Recall also that the rank of π΄ is
greater than or equal to zero and less than or equal to the minimum of π and π,
where π is the number of rows in π΄ and π is the number of columns in π΄. Since π΄ in this case is a
three-by-three matrix, the rank of π΄ is between zero and three. Recall also that the rank of π΄ is
equal to zero if and only if π΄ is the zero matrix. This matrix clearly isnβt the zero
matrix. Therefore, its rank cannot be
zero.

The largest possible square
submatrix of this matrix is just itself, a three-by-three matrix. Taking the determinant of the
matrix by expanding along the top row, we get a result of zero. This is the only possible
three-by-three submatrix of π΄, and it has a determinant of zero. Therefore, the rank of π΄ cannot be
three. We now seek a two-by-two submatrix
of π΄ with a nonzero determinant. This is a problem because there are
nine possible two-by-two submatrices of π΄, and we may need to check every single
one of them.

Consider, for instance, if the
original matrix had looked like this. The only two-by-two submatrix we
can select that might have a nonzero determinant is this one. In this hypothetical example, the
choice is clear, but it might not be for our question. If we look at the original matrix,
we can see that the bottom row is an exact scalar multiple of the top row. Any two-by-two submatrix selected
from these two rows will have determinant of zero. And we might suppose from this that
this means that there are no two-by-two submatrices with a nonzero determinant.

We might suppose from this that
there are no two-by-two submatrices of π΄ with a nonzero determinant. However, if we select a two-by-two
submatrix that doesnβt come from just these two rows that are scalar multiples of
each other, for example, by removing the bottom row and the right-most column, we
get a nonzero determinant. We have therefore found a
two-by-two submatrix of π΄ with a nonzero determinant. Therefore, the rank of π΄ is
two.