# Question Video: Finding The Number of Solutions to a System of Linear Equations Mathematics

Find the number of solutions for the following system of linear equations: [20, β19, β17 and 17, 4, β19 and β16, 9, 15][π₯ and π¦ and π§] = [β13 and β20 and β7].

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### 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©βCapelli 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 weβll 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©βCapelli 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.