Question

Find the inverse of the matrix A =[1 - 1 1, 1 - 2 4, 1 2 2] by gauss Jordan method.

Answer 1

Answer on Question #73856 – Math – Linear Algebra

Question

Find the inverse of the matrix $A = \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & 4 \\ 1 & 2 & 2 \end{bmatrix}$ by Gauss-Jordan method.

Solution

Take two matrices: $A$ and the identity $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ . Reduce the matrix $A$ to

the identity matrix by the Gauss-Jordan method. After applying each operation to the first matrix, we apply the same operation to the second one. When the reduction of the first matrix to a single form is completed, the second matrix will be equal to $A^{-1}$ .

Step 1. Subtract the first line from the second.

$$A = \left[ \begin{array}{c c c} 1 & - 1 & 1 \\ 0 & - 1 & 3 \\ 1 & 2 & 2 \end{array} \right]; I = \left[ \begin{array}{c c c} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right].$$

Step 2. Subtract the first line from the third.

$$A = \left[ \begin{array}{c c c} 1 & - 1 & 1 \\ 0 & - 1 & 3 \\ 0 & 3 & 1 \end{array} \right]; I = \left[ \begin{array}{c c c} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{array} \right].$$

Step 3. Multiply the second line by $-1$ and third line by $\frac{1}{3}$ .

$$A = \left[ \begin{array}{c c c} 1 & - 1 & 1 \\ 0 & 1 & - 3 \\ 0 & 1 & \frac {1}{3} \end{array} \right]; I = \left[ \begin{array}{c c c} 1 & 0 & 0 \\ 1 & - 1 & 0 \\ - \frac {1}{3} & 0 & \frac {1}{3} \end{array} \right].$$

Step 4. Subtract the second line from the third.

$$A = \left[ \begin{array}{c c c} 1 & - 1 & 1 \\ 0 & 1 & - 3 \\ 0 & 0 & 3 \frac {1}{3} \end{array} \right]; I = \left[ \begin{array}{c c c} 1 & 0 & 0 \\ 1 & - 1 & 0 \\ - \frac {4}{3} & 1 & \frac {1}{3} \end{array} \right].$$

Step 5. Multiply the third line by $\frac{3}{10}$ .

$$A = \left[ \begin{array}{ccc} 1 & -1 & 1 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{array} \right]; I = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 1 & -1 & 0 \\ -\frac{4}{10} & \frac{3}{10} & \frac{1}{10} \end{array} \right].$$

Step 6. Subtract the third line, multiplied by $-3$, from the second.

$$A = \left[ \begin{array}{ccc} 1 & -1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]; I = \left[ \begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ -\frac{2}{10} & -\frac{1}{10} & \frac{3}{10} \\ -\frac{4}{10} & \frac{3}{10} & \frac{1}{10} \end{array} \right].$$

Step 7. Subtract the second line, multiplied by $-1$ and the third line from the first.

$$A = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]; I = \left[ \begin{array}{ccc} \frac{12}{10} & -\frac{4}{10} & \frac{2}{10} \\ -\frac{2}{10} & -\frac{1}{10} & \frac{3}{10} \\ -\frac{4}{10} & \frac{3}{10} & \frac{1}{10} \end{array} \right] = A^{-1}.$$

Answer: $A^{-1} = \begin{bmatrix} 1.2 & -0.4 & 0.2 \\ -0.2 & -0.1 & 0.3 \\ -0.4 & 0.3 & 0.1 \end{bmatrix}$.

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