91Ó°ÊÓ

An augmented matrix is a compact representation of a system of linear equations, where the matrix includes both the coefficients of the variables and the constant terms from the equations. For a system of equations like the one in our exercise, where we have variables x, y, and z, the augmented matrix would basically align the coefficients into columns according to which variable they multiply and the constants into an additional column. This effective organization allows for the application of various matrix operations to solve the system efficiently.

• First Column: Coefficients of x
• Second Column: Coefficients of y
• Third Column: Coefficients of z
• Last Column: Constants (Right-hand side of equations)

In this case, we have the augmented matrix \[A|B\] where \(A = \left[\begin{array}{ccc}2 & -1 & -1\ 4 & -1 & 1\ 1 & -3 & -4\end{array}\right]\) representing the coefficients and \(B = \left[\begin{array}{c}1\ -5\ 2\end{array}\right]\) representing the constants.

The determinant of a matrix is a special number that provides valuable information about the matrix. Specifically for a square matrix, the determinant helps to determine if the matrix is invertible, which is a prerequisite for solving a system of equations using matrix inversion. The determinant is calculated using a specific formula depending on the size of the matrix.

For a 3x3 matrix like matrix \(A\) in our exercise, the determinant is found using a cross-multiplication method that takes into account all the elements of the matrix. If the determinant is zero, the matrix does not have an inverse, and the system of equations may be dependent or inconsistent. However, in our exercise, the determinant of \(A\) is calculated to be -13, which is non-zero, indicating that \(A\) is invertible and the system can be solved by finding the inverse of \(A\).

The inverse of a matrix \(A\), denoted as \(A^{-1}\), is a matrix that, when multiplied with the original \(A\), yields the identity matrix. Finding the inverse of a matrix is essential when we want to solve a system of linear equations using matrix methods. The process to find the inverse includes several steps:

• Finding the matrix of minors.
• Applying a checkerboard pattern of signs to create the cofactor matrix.
• Transposing the cofactor matrix to get the adjugate matrix.
• Finally, dividing the adjugate matrix by the determinant of \(A\) to get \(A^{-1}\).

In our given exercise, after performing these steps, the inverse matrix \(A^{-1}\) is \(\left[\begin{array}{ccc}-4/13 & -3/13 & 1/13\ 5/13 & 0 & -2/13\ -6/13 & 1/13 & 1/13\end{array}\right]\). This inverse is used to solve the system by multiplying it with the constants matrix \(B\).

Matrix multiplication is a way of combining two matrices to produce a third matrix. Unlike element-wise multiplication, matrix multiplication involves taking the rows of the first matrix and the columns of the second matrix to calculate the elements of the resulting matrix. The key requirement is that the number of columns in the first matrix must match the number of rows in the second matrix.

In the context of solving a system of equations, after computing the inverse matrix \(A^{-1}\), we multiply it by matrix \(B\), which contains the constant terms. This operation will yield the solution matrix \(X\), which contains the values for our variables x, y, and z. The calculated solution matrix represents each variable's value in the original system of equations. In the current exercise, this multiplication results in the final values for the variables, effectively solving the system of linear equations presented at the beginning.