Augmented matrix

Let A be the square 2×2 matrix A = \begin{bmatrix} 1 & 3 \\ -5 & 0 \end{bmatrix}. To find the inverse of A we form the augmented matrix (A \vert \mathbf{I}_2) where \mathbf{I}_2 is the 2\times 2 identity matrix. We then reduce the part of (A \vert \mathbf{I}_2) corresponding to A to the identity matrix using elementary row operations on (A \vert \mathbf{I}_2) . (A \vert \mathbf{I}_2) = \left[\begin{array}{cc|cc} 1 & 3 & 1 & 0\\ -5 & 0 & 0 & 1 \end{array}\right] (I|A^{-1}) = \left[\begin{array}{cc|cc} 1 & 0 & 0 & -\frac{1}{5} \\ 0 & 1 & \frac{1}{3} & \frac{1}{15} \end{array}\right], the right part of which is the inverse A^{-1}. ==Existence and number of solutions==

Consider the system of equations \begin{align} x + y + 2z &= 2 \\ x + y + z &= 3 \\ 2x + 2y + 2z &= 6. \end{align} The coefficient matrix is A = \begin{bmatrix} 1 & 1 & 2 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \\ \end{bmatrix}, and the augmented matrix is (A|B) = \left[\begin{array}{ccc|c} 1 & 1 & 2 & 2\\ 1 & 1 & 1 & 3 \\ 2 & 2 & 2 & 6 \end{array}\right]. Since both of these have the same rank, namely 2, there exists at least one solution; and since their rank is less than the number of unknowns, the latter being 3, there are an infinite number of solutions. In contrast, consider the system \begin{align} x + y + 2z &= 3 \\ x + y + z &= 1 \\ 2x + 2y + 2z &= 5. \end{align} The coefficient matrix is A = \begin{bmatrix} 1 & 1 & 2 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \\ \end{bmatrix}, and the augmented matrix is (A|B) = \left[\begin{array}{ccc|c} 1 & 1 & 2 & 3 \\ 1 & 1 & 1 & 1 \\ 2 & 2 & 2 & 5 \end{array}\right]. In this example the coefficient matrix has rank 2 while the augmented matrix has rank 3; so this system of equations has no solution. Indeed, an increase in the number of linearly independent rows has made the system of equations inconsistent. ==Solution of a linear system==

As used in linear algebra, an augmented matrix is used to represent the coefficients and the solution vector of each equation set. For the set of equations \begin{align} x + 2y + 3z &= 0 \\ 3x + 4y + 7z &= 2 \\ 6x + 5y + 9z &= 11 \end{align} the coefficients and constant terms give the matrices A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 4 & 7 \\ 6 & 5 & 9 \end{bmatrix} , \quad B = \begin{bmatrix} 0 \\ 2 \\ 11 \end{bmatrix}, and hence give the augmented matrix (A|B) = \left[\begin{array}{ccc|c} 1 & 2 & 3 & 0 \\ 3 & 4 & 7 & 2 \\ 6 & 5 & 9 & 11 \end{array}\right]. Note that the rank of the coefficient matrix, which is 3, equals the rank of the augmented matrix, so at least one solution exists; and since this rank equals the number of unknowns, there is exactly one solution. To obtain the solution, row operations can be performed on the augmented matrix to obtain the identity matrix on the left side, yielding \left[\begin{array}{ccc|r} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -2 \\ \end{array}\right], so the solution of the system is . ==References==