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Question: Find the inverse of the matrix whose rows are 1,-1,2 and 3,0,-2 and 1,0,3...

Find the inverse of the matrix whose rows are 1,-1,2 and 3,0,-2 and 1,0,3

Answer

The inverse of the matrix is:

[0311211−11118110−111311]\begin{bmatrix} 0 & \frac{3}{11} & \frac{2}{11} \\ -1 & \frac{1}{11} & \frac{8}{11} \\ 0 & -\frac{1}{11} & \frac{3}{11} \end{bmatrix}

Explanation

Solution

The given matrix is A=[1−1230−2103]A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}.

To find the inverse of AA, we use the elementary row operations method by augmenting the matrix AA with the identity matrix II:

[A∣I]=[1−12∣10030−2∣010103∣001][A | I] = \begin{bmatrix} 1 & -1 & 2 & | & 1 & 0 & 0 \\ 3 & 0 & -2 & | & 0 & 1 & 0 \\ 1 & 0 & 3 & | & 0 & 0 & 1 \end{bmatrix}

Perform row operations to transform the left side into the identity matrix:

  1. R2→R2−3R1R_2 \rightarrow R_2 - 3R_1:

[1−12∣10003−8∣−310103∣001]\begin{bmatrix} 1 & -1 & 2 & | & 1 & 0 & 0 \\ 0 & 3 & -8 & | & -3 & 1 & 0 \\ 1 & 0 & 3 & | & 0 & 0 & 1 \end{bmatrix}

  1. R3→R3−R1R_3 \rightarrow R_3 - R_1:

[1−12∣10003−8∣−310011∣−101]\begin{bmatrix} 1 & -1 & 2 & | & 1 & 0 & 0 \\ 0 & 3 & -8 & | & -3 & 1 & 0 \\ 0 & 1 & 1 & | & -1 & 0 & 1 \end{bmatrix}

  1. R2↔R3R_2 \leftrightarrow R_3:

[1−12∣100011∣−10103−8∣−310]\begin{bmatrix} 1 & -1 & 2 & | & 1 & 0 & 0 \\ 0 & 1 & 1 & | & -1 & 0 & 1 \\ 0 & 3 & -8 & | & -3 & 1 & 0 \end{bmatrix}

  1. R1→R1+R2R_1 \rightarrow R_1 + R_2:

[103∣001011∣−10103−8∣−310]\begin{bmatrix} 1 & 0 & 3 & | & 0 & 0 & 1 \\ 0 & 1 & 1 & | & -1 & 0 & 1 \\ 0 & 3 & -8 & | & -3 & 1 & 0 \end{bmatrix}

  1. R3→R3−3R2R_3 \rightarrow R_3 - 3R_2:

[103∣001011∣−10100−11∣01−3]\begin{bmatrix} 1 & 0 & 3 & | & 0 & 0 & 1 \\ 0 & 1 & 1 & | & -1 & 0 & 1 \\ 0 & 0 & -11 & | & 0 & 1 & -3 \end{bmatrix}

  1. R3→−111R3R_3 \rightarrow -\frac{1}{11}R_3:

[103∣001011∣−101001∣0−111311]\begin{bmatrix} 1 & 0 & 3 & | & 0 & 0 & 1 \\ 0 & 1 & 1 & | & -1 & 0 & 1 \\ 0 & 0 & 1 & | & 0 & -\frac{1}{11} & \frac{3}{11} \end{bmatrix}

  1. R1→R1−3R3R_1 \rightarrow R_1 - 3R_3:

[100∣0311211011∣−101001∣0−111311]\begin{bmatrix} 1 & 0 & 0 & | & 0 & \frac{3}{11} & \frac{2}{11} \\ 0 & 1 & 1 & | & -1 & 0 & 1 \\ 0 & 0 & 1 & | & 0 & -\frac{1}{11} & \frac{3}{11} \end{bmatrix}

  1. R2→R2−R3R_2 \rightarrow R_2 - R_3:

[100∣0311211010∣−1111811001∣0−111311]\begin{bmatrix} 1 & 0 & 0 & | & 0 & \frac{3}{11} & \frac{2}{11} \\ 0 & 1 & 0 & | & -1 & \frac{1}{11} & \frac{8}{11} \\ 0 & 0 & 1 & | & 0 & -\frac{1}{11} & \frac{3}{11} \end{bmatrix}

The left side is the identity matrix, so the right side is the inverse matrix A−1A^{-1}.

A−1=[0311211−11118110−111311]=111[032−11180−13]A^{-1} = \begin{bmatrix} 0 & \frac{3}{11} & \frac{2}{11} \\ -1 & \frac{1}{11} & \frac{8}{11} \\ 0 & -\frac{1}{11} & \frac{3}{11} \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}