### Video Transcript

Find the number of solutions for
the following system of linear equations: 20, negative 19, negative 17, 17, four,
negative 19, negative 16, nine, 15 multiplied by π₯, π¦, π§ is equal to negative 13,
negative 20, negative seven.

We begin by recalling that the
RouchΓ©βCapelli theorem states that a system of linear equations has solutions if and
only if the rank of its coefficient matrix is equal to the rank of its augmented
matrix. In our question, the coefficient
matrix, which we will call π΄, is the three-by-three matrix on the left-hand side of
our equation. The augmented matrix, written π΄
stroke π, is formed by adding the solution matrix negative 13, negative 20,
negative seven to the coefficient matrix. This gives us a three-by-four
matrix as shown.

Next, we recall that the rank of a
matrix π΄, written π
π of π΄, is the number of rows or columns, π, of the largest
π-by-π square submatrix of π΄ for which the determinant is nonzero. In this question, the square
submatrices weβll begin considering are three by three. At this point, it is worth
recalling a simple process we can follow when trying to identify the rank of a
three-by-three matrix.

The flowchart begins by considering
whether the matrix is the zero matrix. This is not true of matrix π΄ or
any three-by-three submatrix of π΄ stroke π. Likewise, neither of our matrices
have rows or columns that are nonzero scalar multiples. Our sole focus will therefore be
whether the determinant of the three-by-three matrix is equal to zero, as this will
tell us whether the rank of that matrix is equal to two or three.

Beginning with matrix π΄, the only
three-by-three submatrix of this is the coefficient matrix itself. We will therefore begin by finding
the determinant of this matrix. Expanding along the top row, the
determinant of matrix π΄ is equal to 20 multiplied by the determinant of the
two-by-two matrix four, negative 19, nine, 15 minus negative 19 multiplied by the
determinant of 17, negative 19, negative 16, 15 plus negative 17 multiplied by the
determinant of 17, four, negative 16, nine.

Recalling that to find the
determinant of a two-by-two matrix, we find the product of the elements in the top
left and bottom right and subtract the product of the elements in the top right and
bottom left, the determinant of π΄ is equal to 20 multiplied by 231 plus 19
multiplied by negative 49 minus 17 multiplied by 217. This is equal to zero. The rank of matrix π΄ is therefore
equal to two. And we note that two is the number
of rows and columns of the largest submatrix of π΄ with a nonzero determinant.

Our next step is to find the rank
of the augmented matrix. And if this rank is equal to the
rank of the coefficient matrix, by the RouchΓ©βCapelli theorem, the system of
equations will have solutions. Recalling that the augmented matrix
is a three-by-four matrix, we begin by identifying a three-by-three submatrix. One such submatrix can be found by
eliminating the fourth column and giving us the three-by-three coefficient
matrix. However, we have already shown that
this has a determinant equal to zero. We therefore need to identify a
different submatrix by eliminating either the first, second, or third column.

Eliminating the first column, we
have the three-by-three submatrix negative 19, negative 17, negative 13, four,
negative 19, negative 20, nine, 15, negative seven. We will call this matrix capital
π΅. Calculating the determinant by
expanding over the first row once again, we have negative 19 multiplied by 433 minus
negative 17 multiplied by 152 plus negative 13 multiplied by 231. This is equal to negative 8646. The determinant of the submatrix π΅
is therefore not equal to zero. And from our definition of the rank
of a matrix, the rank of π΄ stroke π is equal to three.

We can therefore conclude that the
rank of matrix π΄ is not equal to the rank of matrix π΄ stroke π. That is, the rank of the
coefficient matrix is not equal to the rank of the augmented matrix. And the system of linear equations
therefore has no solutions.