Prompt Title: Rank and determinant of matrices.

When dealing with linear equation matrices, the rank of a matrix is the number of linearly independent rows or columns, and the determinant is a scalar value that can be computed from the elements of a square matrix, which provides information about the matrix's invertibility.

Let's consider a system of linear equations in the form Ax = b, where A is the coefficient matrix, x is the vector of variables, and b is the constant vector.

1. No solution (inconsistent system): In this case, the system is inconsistent, meaning that there is no combination of the variables that will satisfy all the equations simultaneously. The rank of the coefficient matrix (A) is less than the rank of the augmented matrix (A|b) i.e., rank(A) < rank(A|b).

Determinants are only computed for square matrices, so it's not directly related to the existence of solutions in the system. However, if the coefficient matrix (A) is a square matrix and its determinant is non-zero, the system is guaranteed to have a unique solution. Otherwise, the system might have no solutions or infinite solutions.

2. One solution (consistent and unique solution): In this case, the system has a unique solution, meaning there is only one combination of the variables that will satisfy all the equations. The rank of the coefficient matrix (A) is equal to the rank of the augmented matrix (A|b) and is also equal to the number of variables, i.e., rank(A) = rank(A|b) = n (where n is the number of variables). If A is a square matrix, its determinant will be non-zero (det(A) ≠ 0).

3. Infinite solutions (consistent and dependent system): In this case, the system has infinitely many solutions, meaning there are infinitely many combinations of the variables that will satisfy all the equations. The rank of the coefficient matrix (A) is equal to the rank of the augmented matrix (A|b) but less than the number of variables, i.e., rank(A) = rank(A|b) < n (where n is the number of variables). If A is a square matrix, its determinant will be zero (det(A) = 0).

To summarize:

• No solution: rank(A) < rank(A|b)
• One solution: rank(A) = rank(A|b) = n, and if A is a square matrix, det(A) ≠ 0
• Infinite solutions: rank(A) = rank(A|b) < n, and if A is a square matrix, det(A) = 0

To find the rank of a matrix, you can use the Gaussian elimination (also known as row reduction) method to convert the matrix to its row echelon form or its reduced row echelon form. The rank of the matrix is the number of non-zero rows in its row echelon form or the number of leading ones (pivots) in its reduced row echelon form. Here are the steps to find the rank of a matrix using Gaussian elimination:

1. Write down the given matrix.
2. Perform row operations to transform the matrix into row echelon form or reduced row echelon form: a. Interchange rows if necessary to have a non-zero element in the top-left corner of the matrix (the first pivot). b. Replace a row with the sum of itself and a multiple of another row such that the element below the pivot becomes zero. c. Repeat steps a and b for the remaining rows until you reach the bottom row of the matrix. d. If you want to obtain the reduced row echelon form, divide each row by its pivot and use row operations to make all elements above the pivots zero as well.
3. Count the number of non-zero rows in the row echelon form (or the number of leading ones in the reduced row echelon form). This count is the rank of the matrix.

Keep in mind that the rank of a matrix is the number of linearly independent rows or columns. So, after you obtain the row echelon form or reduced row echelon form, you can determine the linearly independent rows or columns more easily.

In row-reduced echelon form (also known as reduced row echelon form, or RREF), a matrix has the following properties:

1. All rows with all zero elements are at the bottom of the matrix.
2. The first non-zero element in each non-zero row (called the leading entry or pivot) is 1.
3. Each pivot is to the right of the pivot in the row above it.
4. All elements above and below a pivot are 0.

Here are some examples of matrices in row-reduced echelon form:

Example 1:

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[1  0  0  4]
[0  1  0  2]
[0  0  1 -3]
[0  0  0  0]

Example 2:

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[1  0 -2]
[0  1  3]
[0  0  0]

Example 3:

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[1  2  0  0]
[0  0  1 -1]
[0  0  0  0]
[0  0  0  0]

Example 4:

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[1  0  0]
[0  1  0]
[0  0  1]

In these examples, the matrices are in reduced row echelon form. You can observe that each row with all zero elements is at the bottom, each pivot is 1, and there are no non-zero elements above or below the pivots.

Apologies for any confusion. My previous examples were not perfectly clear for demonstrating infinitely many solutions. In order to determine whether a matrix has infinitely many solutions, we need to consider an augmented matrix formed by appending the constants vector b to the coefficient matrix A. In the examples I provided earlier, I did not include the constants vector b.

Here are corrected examples of augmented matrices in reduced row echelon form (RREF) for systems of linear equations:

Example 1 (unique solution):

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[1  0  0 | 4]
[0  1  0 | 2]
[0  0  1 |-3]

Example 2 (infinitely many solutions):

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[1  0 -2 | 1]
[0  1  3 | 4]
[0  0  0 | 0]

In Example 1, the matrix represents a consistent system with a unique solution, as the rank of the coefficient matrix (A) is equal to the rank of the augmented matrix (A|b) and is also equal to the number of variables