For each of the two matrices \[ A = \begin{bmatrix} 2 & 5 & -1 \\ 4 & -1 & 2 \\ 6 & 4 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 2 & 4 \\ 0 & 1 & -2 \end{bmatrix} \] use elementary row operations to discover whether it is invertible, and to find the inverse in case it is.
Solution
Goal: Apply Gauss-Jordan elimination to augmented matrices \([M \mid I_3]\) to determine invertibility and compute matrix inverses.
1. Analysis of Matrix \(A = \begin{bmatrix} 2 & 5 & -1 \\ 4 & -1 & 2 \\ 6 & 4 & 1 \end{bmatrix}\):Proof:
1.1. Form the augmented matrix \([A \mid I_3]\):\[\left[\begin{array}{ccc|ccc} 2 & 5 & -1 & 1 & 0 & 0 \\ 4 & -1 & 2 & 0 & 1 & 0 \\ 6 & 4 & 1 & 0 & 0 & 1 \end{array}\right].\]
1.2. Perform row operations to eliminate the first column:- \(R_2 \gets R_2 - 2R_1\):\[\left[\begin{array}{ccc|ccc} 2 & 5 & -1 & 1 & 0 & 0 \\ 0 & -11 & 4 & -2 & 1 & 0 \\ 6 & 4 & 1 & 0 & 0 & 1 \end{array}\right].\]- \(R_3 \gets R_3 - 3R_1\):\[\left[\begin{array}{ccc|ccc} 2 & 5 & -1 & 1 & 0 & 0 \\ 0 & -11 & 4 & -2 & 1 & 0 \\ 0 & -11 & 4 & -3 & 0 & 1 \end{array}\right].\]
1.3. Perform row operation on row 3:- \(R_3 \gets R_3 - R_2\):\[\left[\begin{array}{ccc|ccc} 2 & 5 & -1 & 1 & 0 & 0 \\ 0 & -11 & 4 & -2 & 1 & 0 \\ 0 & 0 & 0 & -1 & -1 & 1 \end{array}\right].\]
1.4. The left \(3 \times 3\) block has a row of zeros (row rank 2), so \(A\) cannot be reduced to the identity matrix \(I_3\).
1.5. Therefore, the matrix \(A\) is not invertible (singular).
2. Analysis of Matrix \(B = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 2 & 4 \\ 0 & 1 & -2 \end{bmatrix}\):Proof:
2.1. Form the augmented matrix \([B \mid I_3]\):\[\left[\begin{array}{ccc|ccc} 1 & -1 & 2 & 1 & 0 & 0 \\ 3 & 2 & 4 & 0 & 1 & 0 \\ 0 & 1 & -2 & 0 & 0 & 1 \end{array}\right].\]
2.2. Eliminate the first column below row 1:- \(R_2 \gets R_2 - 3R_1\):\[\left[\begin{array}{ccc|ccc} 1 & -1 & 2 & 1 & 0 & 0 \\ 0 & 5 & -2 & -3 & 1 & 0 \\ 0 & 1 & -2 & 0 & 0 & 1 \end{array}\right].\]
2.3. Swap \(R_2 \leftrightarrow R_3\):\[\left[\begin{array}{ccc|ccc} 1 & -1 & 2 & 1 & 0 & 0 \\ 0 & 1 & -2 & 0 & 0 & 1 \\ 0 & 5 & -2 & -3 & 1 & 0 \end{array}\right].\]
2.4. Eliminate the second column:- \(R_1 \gets R_1 + R_2\):\[\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & -2 & 0 & 0 & 1 \\ 0 & 5 & -2 & -3 & 1 & 0 \end{array}\right].\]- \(R_3 \gets R_3 - 5R_2\):\[\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & -2 & 0 & 0 & 1 \\ 0 & 0 & 8 & -3 & 1 & -5 \end{array}\right].\]
2.5. Scale \(R_3 \gets \frac{1}{8} R_3\):\[\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & -2 & 0 & 0 & 1 \\ 0 & 0 & 1 & -\frac{3}{8} & \frac{1}{8} & -\frac{5}{8} \end{array}\right].\]
2.6. Eliminate the third column in \(R_2\):- \(R_2 \gets R_2 + 2R_3\):- Entry \((2, 1)\): \(0 + 2(-3/8) = -3/4 = -6/8\).- Entry \((2, 2)\): \(0 + 2(1/8) = 1/4 = 2/8\).- Entry \((2, 3)\): \(1 + 2(-5/8) = 1 - 5/4 = -1/4 = -2/8\).\[\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & -\frac{3}{4} & \frac{1}{4} & -\frac{1}{4} \\ 0 & 0 & 1 & -\frac{3}{8} & \frac{1}{8} & -\frac{5}{8} \end{array}\right].\]
2.7. Writing in common denominator \(\frac{1}{8}\):\[B^{-1} = \frac{1}{8} \begin{bmatrix} 8 & 0 & 8 \\ -6 & 2 & -2 \\ -3 & 1 & -5 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1 \\ -\frac{3}{4} & \frac{1}{4} & -\frac{1}{4} \\ -\frac{3}{8} & \frac{1}{8} & -\frac{5}{8} \end{bmatrix}.\]
2.8. Verification:\[\begin{bmatrix} 1 & -1 & 2 \\ 3 & 2 & 4 \\ 0 & 1 & -2 \end{bmatrix} \cdot \frac{1}{8} \begin{bmatrix} 8 & 0 & 8 \\ -6 & 2 & -2 \\ -3 & 1 & -5 \end{bmatrix} = \frac{1}{8} \begin{bmatrix} 8+6-6 & 0-2+2 & 8+2-10 \\ 24-12-12 & 0+4+4 & 24-4-20 \\ 0-6+6 & 0+2-2 & 0-2+10 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.\]
3. Conclusion:\(A\) is not invertible; \(B\) is invertible with inverse \(B^{-1} = \frac{1}{8}\begin{bmatrix} 8 & 0 & 8 \\ -6 & 2 & -2 \\ -3 & 1 & -5 \end{bmatrix}\). Q.E.D.