Answer on Question #73856 – Math – Linear Algebra
Question
Find the inverse of the matrix A = [ 1 − 1 1 1 − 2 4 1 2 2 ] A = \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & 4 \\ 1 & 2 & 2 \end{bmatrix} A = ⎣ ⎡ 1 1 1 − 1 − 2 2 1 4 2 ⎦ ⎤
Solution
Take two matrices: A A A I = [ 1 0 0 0 1 0 0 0 1 ] I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} I = ⎣ ⎡ 1 0 0 0 1 0 0 0 1 ⎦ ⎤ A A A
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 A^{-1} A − 1
Step 1. Subtract the first line from the second.
A = [ 1 − 1 1 0 − 1 3 1 2 2 ] ; I = [ 1 0 0 − 1 1 0 0 0 1 ] . 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]. A = ⎣ ⎡ 1 0 1 − 1 − 1 2 1 3 2 ⎦ ⎤ ; I = ⎣ ⎡ 1 − 1 0 0 1 0 0 0 1 ⎦ ⎤ .
Step 2. Subtract the first line from the third.
A = [ 1 − 1 1 0 − 1 3 0 3 1 ] ; I = [ 1 0 0 − 1 1 0 − 1 0 1 ] . 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]. A = ⎣ ⎡ 1 0 0 − 1 − 1 3 1 3 1 ⎦ ⎤ ; I = ⎣ ⎡ 1 − 1 − 1 0 1 0 0 0 1 ⎦ ⎤ .
Step 3. Multiply the second line by − 1 -1 − 1 1 3 \frac{1}{3} 3 1
A = [ 1 − 1 1 0 1 − 3 0 1 1 3 ] ; I = [ 1 0 0 1 − 1 0 − 1 3 0 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]. A = ⎣ ⎡ 1 0 0 − 1 1 1 1 − 3 3 1 ⎦ ⎤ ; I = ⎣ ⎡ 1 1 − 3 1 0 − 1 0 0 0 3 1 ⎦ ⎤ .
Step 4. Subtract the second line from the third.
A = [ 1 − 1 1 0 1 − 3 0 0 3 1 3 ] ; I = [ 1 0 0 1 − 1 0 − 4 3 1 1 3 ] . 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]. A = ⎣ ⎡ 1 0 0 − 1 1 0 1 − 3 3 3 1 ⎦ ⎤ ; I = ⎣ ⎡ 1 1 − 3 4 0 − 1 1 0 0 3 1 ⎦ ⎤ .
Step 5. Multiply the third line by 3 10 \frac{3}{10} 10 3
A = [ 1 − 1 1 0 1 − 3 0 0 1 ] ; I = [ 1 0 0 1 − 1 0 − 4 10 3 10 1 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]. A = ⎣ ⎡ 1 0 0 − 1 1 0 1 − 3 1 ⎦ ⎤ ; I = ⎣ ⎡ 1 1 − 10 4 0 − 1 10 3 0 0 10 1 ⎦ ⎤ .
Step 6. Subtract the third line, multiplied by − 3 -3 − 3
A = [ 1 − 1 1 0 1 0 0 0 1 ] ; I = [ 1 2 0 0 − 2 10 − 1 10 3 10 − 4 10 3 10 1 10 ] . 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]. A = ⎣ ⎡ 1 0 0 − 1 1 0 1 0 1 ⎦ ⎤ ; I = ⎣ ⎡ 2 1 − 10 2 − 10 4 0 − 10 1 10 3 0 10 3 10 1 ⎦ ⎤ .
Step 7. Subtract the second line, multiplied by − 1 -1 − 1
A = [ 1 0 0 0 1 0 0 0 1 ] ; I = [ 12 10 − 4 10 2 10 − 2 10 − 1 10 3 10 − 4 10 3 10 1 10 ] = A − 1 . 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}. A = ⎣ ⎡ 1 0 0 0 1 0 0 0 1 ⎦ ⎤ ; I = ⎣ ⎡ 10 12 − 10 2 − 10 4 − 10 4 − 10 1 10 3 10 2 10 3 10 1 ⎦ ⎤ = A − 1 .
Answer: A − 1 = [ 1.2 − 0.4 0.2 − 0.2 − 0.1 0.3 − 0.4 0.3 0.1 ] A^{-1} = \begin{bmatrix} 1.2 & -0.4 & 0.2 \\ -0.2 & -0.1 & 0.3 \\ -0.4 & 0.3 & 0.1 \end{bmatrix} A − 1 = ⎣ ⎡ 1.2 − 0.2 − 0.4 − 0.4 − 0.1 0.3 0.2 0.3 0.1 ⎦ ⎤
Answer provided by https://www.AssignmentExpert.com
