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

Find the number of solutions of the following system of linear equations: [12, 6, 20 and 1, 20, 8 and βˆ’11, 14, βˆ’12][π‘₯ and 𝑦 and 𝑧] = [5 and 16 and 11].

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Video Transcript

Find the number of solutions of the following system of linear equations: 12, six, 20, one, 20, eight, negative 11, 14, negative 12 multiplied by π‘₯, 𝑦, 𝑧 is equal to 5, 16, 11.

In this question, we’re given a system of three linear equations in three unknowns written in matrix form. We have a three-by-three coefficient matrix, which we will call 𝐴, multiplied by a variables matrix that is equal to a matrix of constants, which we will call 𝑏. In order to find the number of solutions, we begin by recalling the Rouché–Capelli theorem. This 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.

More specifically, we have three possible scenarios. Firstly, if the rank of the coefficient matrix 𝐴 is not equal to the rank of the augmented matrix 𝐴 stroke 𝑏, then there are no solutions. Secondly, if the rank of the coefficient matrix is equal to the rank of the augmented matrix, which is equal to 𝑛, we have one unique solution, where 𝑛 is the number of variables in the system of linear equations. Finally, if the rank of the coefficient matrix is equal to the rank of the augmented matrix, but this is not equal to 𝑛, we have infinitely many solutions.

As already mentioned, the coefficient matrix 𝐴 is equal to 12, six, 20, one, 20, eight, negative 11, 14, negative 12. And the augmented matrix 𝐴 stroke 𝑏 is the three-by-four matrix shown. We add the constants matrix to the right-hand side of the coefficient matrix. Next, we recall that the rank of a matrix is the number of rows or columns π‘˜ of the largest π‘˜-by-π‘˜ square submatrix for which the determinant is nonzero. We notice that our matrix 𝐴 has one row that may be formed from a linear combination of the other two. Specifically, row two is equal to the sum of row one and row three. Since one row is a linear combination of the other two, the rank of 𝐴 must be less than three. And we can verify this directly using determinants.

The only three-by-three submatrix of the coefficient matrix is matrix 𝐴 itself. Taking the determinant of 𝐴 by expanding along the top row, we have 12 multiplied by the determinant of the two-by-two matrix 20, eight, 14, negative 12 minus six multiplied by determinant of one, eight, negative 11, negative 12 plus 20 multiplied by the determinant of one, 20, negative 11, 14. The determinant of the two-by-two matrix π‘Ž, 𝑏, 𝑐, 𝑑 is π‘Žπ‘‘ minus 𝑏𝑐. We find the product of the elements in the top left and bottom right and then subtract the product of the elements in the top right and bottom left.

The determinant of 𝐴 is therefore equal to 12 multiplied by negative 352 minus six multiplied by 76 plus 20 multiplied by 234. This is equal to zero, so the rank of matrix 𝐴 is not equal to three. When calculating the determinant of 𝐴, we have seen that the largest submatrix of 𝐴 with nonzero determinant is a two-by-two matrix. Therefore, the rank of the coefficient matrix 𝐴 is two.

Our next step is to find the rank of the augmented matrix 𝐴 stroke 𝑏. As already mentioned, this is a three-by-four matrix. So in order to find a three-by-three square submatrix, we need to delete one of the columns. Deleting the fourth column would leave us with the coefficient matrix. And we have already found that the determinant of this is equal to zero. We therefore need to consider the other three-by-three submatrices that are formed by deleting columns one, two, and three, respectively. Deleting the first column, we have a three-by-three submatrix six, 20, five, 20, eight, 16, 14, negative 12, 11.

By expanding over any row or column, we find that the determinant of this matrix is also equal to zero. The same is true when we eliminate the second column of the augmented matrix. The determinant of 12, 20, five, one, eight, 16, negative 11, negative 12, 11 is zero. Finally, when we delete the third column of the augmented matrix, the determinant of the remaining three-by-three square submatrix is also equal to zero. We have therefore proved that the determinant of any three-by-three submatrix of 𝐴 stroke 𝑏 is equal to zero. And as a result, the rank of 𝐴 stroke 𝑏 is equal to two. This is because the largest submatrix of the augmented matrix 𝐴 stroke 𝑏 with a nonzero determinant is a two-by-two matrix.

Using the Rouché–Capelli theorem, we see that the rank of the coefficient matrix is equal to the rank of the augmented matrix. However, these are not equal to 𝑛 as the coefficient matrix was a three-by-three matrix. We can therefore conclude that the system of linear equations has an infinite number of solutions.

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