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Question: Find The inverse of matrix whose row 1 is 2,0,-1 row 2 is 5,1,0 and row 3 is 0,1,3...

Find The inverse of matrix whose row 1 is 2,0,-1 row 2 is 5,1,0 and row 3 is 0,1,3

Answer

[3−11−156−55−22]\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}

Explanation

Solution

Let the given matrix be AA.

A=[20−1510013]A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}

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

[A∣I]=[20−1∣100510∣010013∣001][A | I] = \begin{bmatrix} 2 & 0 & -1 & | & 1 & 0 & 0 \\ 5 & 1 & 0 & | & 0 & 1 & 0 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \end{bmatrix}

Apply elementary row operations to transform the left side into the identity matrix.

  1. R1↔R2R_1 \leftrightarrow R_2:

[510∣01020−1∣100013∣001]\begin{bmatrix} 5 & 1 & 0 & | & 0 & 1 & 0 \\ 2 & 0 & -1 & | & 1 & 0 & 0 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \end{bmatrix}

  1. R1→R1−2R2R_1 \rightarrow R_1 - 2R_2:

[112∣−21020−1∣100013∣001]\begin{bmatrix} 1 & 1 & 2 & | & -2 & 1 & 0 \\ 2 & 0 & -1 & | & 1 & 0 & 0 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \end{bmatrix}

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

[112∣−2100−2−5∣5−20013∣001]\begin{bmatrix} 1 & 1 & 2 & | & -2 & 1 & 0 \\ 0 & -2 & -5 & | & 5 & -2 & 0 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \end{bmatrix}

  1. R2↔R3R_2 \leftrightarrow R_3:

[112∣−210013∣0010−2−5∣5−20]\begin{bmatrix} 1 & 1 & 2 & | & -2 & 1 & 0 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \\ 0 & -2 & -5 & | & 5 & -2 & 0 \end{bmatrix}

  1. R1→R1−R2R_1 \rightarrow R_1 - R_2:

[10−1∣−21−1013∣0010−2−5∣5−20]\begin{bmatrix} 1 & 0 & -1 & | & -2 & 1 & -1 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \\ 0 & -2 & -5 & | & 5 & -2 & 0 \end{bmatrix}

  1. R3→R3+2R2R_3 \rightarrow R_3 + 2R_2:

[10−1∣−21−1013∣001001∣5−22]\begin{bmatrix} 1 & 0 & -1 & | & -2 & 1 & -1 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \\ 0 & 0 & 1 & | & 5 & -2 & 2 \end{bmatrix}

  1. R1→R1+R3R_1 \rightarrow R_1 + R_3:

[100∣3−11013∣001001∣5−22]\begin{bmatrix} 1 & 0 & 0 & | & 3 & -1 & 1 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \\ 0 & 0 & 1 & | & 5 & -2 & 2 \end{bmatrix}

  1. R2→R2−3R3R_2 \rightarrow R_2 - 3R_3:

[100∣3−11010∣0−3(5)0−3(−2)1−3(2)001∣5−22]\begin{bmatrix} 1 & 0 & 0 & | & 3 & -1 & 1 \\ 0 & 1 & 0 & | & 0 - 3(5) & 0 - 3(-2) & 1 - 3(2) \\ 0 & 0 & 1 & | & 5 & -2 & 2 \end{bmatrix}

[100∣3−11010∣−156−5001∣5−22]\begin{bmatrix} 1 & 0 & 0 & | & 3 & -1 & 1 \\ 0 & 1 & 0 & | & -15 & 6 & -5 \\ 0 & 0 & 1 & | & 5 & -2 & 2 \end{bmatrix}

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

A−1=[3−11−156−55−22]A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}

In summary, the inverse of the matrix A=[20−1510013]A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} is found by augmenting AA with the identity matrix II to form [A∣I][A|I] and applying elementary row operations to transform [A∣I][A|I] into [I∣A−1][I|A^{-1}].