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Question: Find the inverse of matrix whose row 1 is 1,0,-1 row 2 is 2,3,2 and row 3 is 2,1,4...

Find the inverse of matrix whose row 1 is 1,0,-1 row 2 is 2,3,2 and row 3 is 2,1,4

Answer

[57−114314−2737−27−27−114314]\begin{bmatrix} \frac{5}{7} & -\frac{1}{14} & \frac{3}{14} \\ -\frac{2}{7} & \frac{3}{7} & -\frac{2}{7} \\ -\frac{2}{7} & -\frac{1}{14} & \frac{3}{14} \end{bmatrix}

Explanation

Solution

Let the given matrix be AA.

A=[10−1232214]A = \begin{bmatrix} 1 & 0 & -1 \\ 2 & 3 & 2 \\ 2 & 1 & 4 \end{bmatrix}

To find the inverse of matrix AA using elementary row operations, we augment the matrix AA with the identity matrix II of the same size, forming the augmented matrix [A∣I][A | I].

[A∣I]=[10−1∣100232∣010214∣001][A | I] = \begin{bmatrix} 1 & 0 & -1 & | & 1 & 0 & 0 \\ 2 & 3 & 2 & | & 0 & 1 & 0 \\ 2 & 1 & 4 & | & 0 & 0 & 1 \end{bmatrix}

Now, we apply elementary row operations to transform the left side of the augmented matrix into the identity matrix. The same operations applied to the right side will transform the identity matrix into the inverse of AA.

  1. R2→R2−2R1R_2 \rightarrow R_2 - 2R_1

    R3→R3−2R1R_3 \rightarrow R_3 - 2R_1

    [10−1∣100034∣−210016∣−201]\begin{bmatrix} 1 & 0 & -1 & | & 1 & 0 & 0 \\ 0 & 3 & 4 & | & -2 & 1 & 0 \\ 0 & 1 & 6 & | & -2 & 0 & 1 \end{bmatrix}

  2. R2↔R3R_2 \leftrightarrow R_3 (Swap R2R_2 and R3R_3 to get a 1 in the (2,2) position easily)

    [10−1∣100016∣−201034∣−210]\begin{bmatrix} 1 & 0 & -1 & | & 1 & 0 & 0 \\ 0 & 1 & 6 & | & -2 & 0 & 1 \\ 0 & 3 & 4 & | & -2 & 1 & 0 \end{bmatrix}

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

    [10−1∣100016∣−20100−14∣41−3]\begin{bmatrix} 1 & 0 & -1 & | & 1 & 0 & 0 \\ 0 & 1 & 6 & | & -2 & 0 & 1 \\ 0 & 0 & -14 & | & 4 & 1 & -3 \end{bmatrix}

  4. R3→−114R3R_3 \rightarrow -\frac{1}{14} R_3

    [10−1∣100016∣−201001∣−414−114314]=[10−1∣100016∣−201001∣−27−114314]\begin{bmatrix} 1 & 0 & -1 & | & 1 & 0 & 0 \\ 0 & 1 & 6 & | & -2 & 0 & 1 \\ 0 & 0 & 1 & | & -\frac{4}{14} & -\frac{1}{14} & \frac{3}{14} \end{bmatrix} = \begin{bmatrix} 1 & 0 & -1 & | & 1 & 0 & 0 \\ 0 & 1 & 6 & | & -2 & 0 & 1 \\ 0 & 0 & 1 & | & -\frac{2}{7} & -\frac{1}{14} & \frac{3}{14} \end{bmatrix}

  5. R1→R1+R3R_1 \rightarrow R_1 + R_3

    R2→R2−6R3R_2 \rightarrow R_2 - 6R_3

    R1R_1: (1,0,−1∣1,0,0)+(0,0,1∣−27,−114,314)=(1,0,0∣1−27,0−114,0+314)=(1,0,0∣57,−114,314)(1, 0, -1 | 1, 0, 0) + (0, 0, 1 | -\frac{2}{7}, -\frac{1}{14}, \frac{3}{14}) = (1, 0, 0 | 1-\frac{2}{7}, 0-\frac{1}{14}, 0+\frac{3}{14}) = (1, 0, 0 | \frac{5}{7}, -\frac{1}{14}, \frac{3}{14})

    R2R_2: (0,1,6∣−2,0,1)−6×(0,0,1∣−27,−114,314)=(0,1,0∣−2−6(−27),0−6(−114),1−6(314))=(0,1,0∣−2+127,614,1−1814)=(0,1,0∣−14+127,37,14−1814)=(0,1,0∣−27,37,−27)(0, 1, 6 | -2, 0, 1) - 6 \times (0, 0, 1 | -\frac{2}{7}, -\frac{1}{14}, \frac{3}{14}) = (0, 1, 0 | -2 - 6(-\frac{2}{7}), 0 - 6(-\frac{1}{14}), 1 - 6(\frac{3}{14})) = (0, 1, 0 | -2 + \frac{12}{7}, \frac{6}{14}, 1 - \frac{18}{14}) = (0, 1, 0 | \frac{-14+12}{7}, \frac{3}{7}, \frac{14-18}{14}) = (0, 1, 0 | -\frac{2}{7}, \frac{3}{7}, -\frac{2}{7})

    The augmented matrix is now:

    [100∣57−114314010∣−2737−27001∣−27−114314]\begin{bmatrix} 1 & 0 & 0 & | & \frac{5}{7} & -\frac{1}{14} & \frac{3}{14} \\ 0 & 1 & 0 & | & -\frac{2}{7} & \frac{3}{7} & -\frac{2}{7} \\ 0 & 0 & 1 & | & -\frac{2}{7} & -\frac{1}{14} & \frac{3}{14} \end{bmatrix}

The left side is the identity matrix, so the matrix on the right side is the inverse of AA.

The inverse of the matrix is [57−114314−2737−27−27−114314]\begin{bmatrix} \frac{5}{7} & -\frac{1}{14} & \frac{3}{14} \\ -\frac{2}{7} & \frac{3}{7} & -\frac{2}{7} \\ -\frac{2}{7} & -\frac{1}{14} & \frac{3}{14} \end{bmatrix}.