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Question: EX. Find the inverse of the matrix $A$, using elementary row operations, where $A = \begin{bmatrix} ...

EX. Find the inverse of the matrix AA, using elementary row operations, where A=[12−2−1300−21]A = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}.

Answer

[326112225]\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}

Explanation

Solution

To find the inverse of the matrix A=[12−2−1300−21]A = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} 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]=[12−2∣100−130∣0100−21∣001][A | I] = \begin{bmatrix} 1 & 2 & -2 & | & 1 & 0 & 0 \\ -1 & 3 & 0 & | & 0 & 1 & 0 \\ 0 & -2 & 1 & | & 0 & 0 & 1 \end{bmatrix}

Our goal is to transform the left side of the augmented matrix into the identity matrix by applying elementary row operations to the entire augmented matrix. The matrix on the right side will then be the inverse of AA.

Step 1: Make the element in the first row, first column (a11) equal to 1. It is already 1.

Step 2: Make the elements below a11 zero. Apply R2→R2+R1R_2 \rightarrow R_2 + R_1: [12−2∣100−1+13+20+(−2)∣0+11+00+00−21∣001]=[12−2∣10005−2∣1100−21∣001]\begin{bmatrix} 1 & 2 & -2 & | & 1 & 0 & 0 \\ -1+1 & 3+2 & 0+(-2) & | & 0+1 & 1+0 & 0+0 \\ 0 & -2 & 1 & | & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & -2 & | & 1 & 0 & 0 \\ 0 & 5 & -2 & | & 1 & 1 & 0 \\ 0 & -2 & 1 & | & 0 & 0 & 1 \end{bmatrix}

Step 3: Make the element in the second row, second column (a22) equal to 1. Apply R2→15R2R_2 \rightarrow \frac{1}{5}R_2: [12−2∣10001−2/5∣1/51/500−21∣001]\begin{bmatrix} 1 & 2 & -2 & | & 1 & 0 & 0 \\ 0 & 1 & -2/5 & | & 1/5 & 1/5 & 0 \\ 0 & -2 & 1 & | & 0 & 0 & 1 \end{bmatrix}

Step 4: Make the elements above and below a22 zero. Apply R1→R1−2R2R_1 \rightarrow R_1 - 2R_2: [1−2(0)2−2(1)−2−2(−2/5)∣1−2(1/5)0−2(1/5)0−2(0)01−2/5∣1/51/500−21∣001]=[10−6/5∣3/5−2/5001−2/5∣1/51/500−21∣001]\begin{bmatrix} 1 - 2(0) & 2 - 2(1) & -2 - 2(-2/5) & | & 1 - 2(1/5) & 0 - 2(1/5) & 0 - 2(0) \\ 0 & 1 & -2/5 & | & 1/5 & 1/5 & 0 \\ 0 & -2 & 1 & | & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & -6/5 & | & 3/5 & -2/5 & 0 \\ 0 & 1 & -2/5 & | & 1/5 & 1/5 & 0 \\ 0 & -2 & 1 & | & 0 & 0 & 1 \end{bmatrix} Apply R3→R3+2R2R_3 \rightarrow R_3 + 2R_2: [10−6/5∣3/5−2/5001−2/5∣1/51/500+2(0)−2+2(1)1+2(−2/5)∣0+2(1/5)0+2(1/5)1+2(0)]=[10−6/5∣3/5−2/5001−2/5∣1/51/50001/5∣2/52/51]\begin{bmatrix} 1 & 0 & -6/5 & | & 3/5 & -2/5 & 0 \\ 0 & 1 & -2/5 & | & 1/5 & 1/5 & 0 \\ 0 + 2(0) & -2 + 2(1) & 1 + 2(-2/5) & | & 0 + 2(1/5) & 0 + 2(1/5) & 1 + 2(0) \end{bmatrix} = \begin{bmatrix} 1 & 0 & -6/5 & | & 3/5 & -2/5 & 0 \\ 0 & 1 & -2/5 & | & 1/5 & 1/5 & 0 \\ 0 & 0 & 1/5 & | & 2/5 & 2/5 & 1 \end{bmatrix}

Step 5: Make the element in the third row, third column (a33) equal to 1. Apply R3→5R3R_3 \rightarrow 5R_3: [10−6/5∣3/5−2/5001−2/5∣1/51/50001∣225]\begin{bmatrix} 1 & 0 & -6/5 & | & 3/5 & -2/5 & 0 \\ 0 & 1 & -2/5 & | & 1/5 & 1/5 & 0 \\ 0 & 0 & 1 & | & 2 & 2 & 5 \end{bmatrix}

Step 6: Make the elements above a33 zero. Apply R1→R1+65R3R_1 \rightarrow R_1 + \frac{6}{5}R_3: [1+65(0)0+65(0)−6/5+65(1)∣3/5+65(2)−2/5+65(2)0+65(5)01−2/5∣1/51/50001∣225]=[100∣15/510/5601−2/5∣1/51/50001∣225]=[100∣32601−2/5∣1/51/50001∣225]\begin{bmatrix} 1 + \frac{6}{5}(0) & 0 + \frac{6}{5}(0) & -6/5 + \frac{6}{5}(1) & | & 3/5 + \frac{6}{5}(2) & -2/5 + \frac{6}{5}(2) & 0 + \frac{6}{5}(5) \\ 0 & 1 & -2/5 & | & 1/5 & 1/5 & 0 \\ 0 & 0 & 1 & | & 2 & 2 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & | & 15/5 & 10/5 & 6 \\ 0 & 1 & -2/5 & | & 1/5 & 1/5 & 0 \\ 0 & 0 & 1 & | & 2 & 2 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & | & 3 & 2 & 6 \\ 0 & 1 & -2/5 & | & 1/5 & 1/5 & 0 \\ 0 & 0 & 1 & | & 2 & 2 & 5 \end{bmatrix} Apply R2→R2+25R3R_2 \rightarrow R_2 + \frac{2}{5}R_3: [100∣3260+25(0)1+25(0)−2/5+25(1)∣1/5+25(2)1/5+25(2)0+25(5)001∣225]=[100∣326010∣5/55/52001∣225]=[100∣326010∣112001∣225]\begin{bmatrix} 1 & 0 & 0 & | & 3 & 2 & 6 \\ 0 + \frac{2}{5}(0) & 1 + \frac{2}{5}(0) & -2/5 + \frac{2}{5}(1) & | & 1/5 + \frac{2}{5}(2) & 1/5 + \frac{2}{5}(2) & 0 + \frac{2}{5}(5) \\ 0 & 0 & 1 & | & 2 & 2 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & | & 3 & 2 & 6 \\ 0 & 1 & 0 & | & 5/5 & 5/5 & 2 \\ 0 & 0 & 1 & | & 2 & 2 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & | & 3 & 2 & 6 \\ 0 & 1 & 0 & | & 1 & 1 & 2 \\ 0 & 0 & 1 & | & 2 & 2 & 5 \end{bmatrix}

The left side is now the identity matrix. The matrix on the right side is the inverse of AA. A−1=[326112225]A^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}.

The sequence of operations used is: R2→R2+R1R_2 \rightarrow R_2 + R_1, R2→15R2R_2 \rightarrow \frac{1}{5}R_2, R1→R1−2R2R_1 \rightarrow R_1 - 2R_2, R3→R3+2R2R_3 \rightarrow R_3 + 2R_2, R3→5R3R_3 \rightarrow 5R_3, R1→R1+65R3R_1 \rightarrow R_1 + \frac{6}{5}R_3, R2→R2+25R3R_2 \rightarrow R_2 + \frac{2}{5}R_3.

The final matrix on the right side is the inverse matrix A−1A^{-1}.