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Mathematics Question on Matrices

Find the inverse of each of the matrices,if it exists.[13−2 −30−5 250]\begin{bmatrix} 1 & 3 & -2\\\ -3 & 0 & -5\\\ 2&5&0 \end{bmatrix}

Answer

Let A=[13−2 −30−5 250]\begin{bmatrix} 1 & 3 & -2\\\ -3 & 0 & -5\\\ 2&5&0 \end{bmatrix}

We know that A = IA

[13−2 −30−5 250]\begin{bmatrix} 1 & 3 & -2\\\ -3 & 0 & -5\\\ 2&5&0 \end{bmatrix}=[100 010 001]\begin{bmatrix} 1 & 0 & 0\\\ 0 & 1 & 0\\\ 0&0&1 \end{bmatrix}A

Applying R2→R2+3R1R_2 → R_2 + 3R_1 and R3→R3−2R1R_3 → R_3 − 2R_1, we have:

[13−2 09−11 0−14]\begin{bmatrix} 1 & 3 & -2\\\ 0 & 9 & -11\\\ 0&-1&4 \end{bmatrix}=[100 310 −201]\begin{bmatrix} 1 & 0 & 0\\\ 3 & 1 & 0\\\ -2&0&1 \end{bmatrix}A .

Applying R1→R1+3R3R_1 → R_1 + 3R_3 and R2→R2+8R3R_2 → R_2 +8R_3, we have:

[1010 0121 0−14]\begin{bmatrix} 1 & 0 & 10\\\ 0 & 1 & 21\\\ 0&-1&4 \end{bmatrix}=[−503 −1318 −201]\begin{bmatrix} -5 & 0 & 3\\\ -13 & 1 & 8\\\ -2&0&1 \end{bmatrix}A

Applying R3→R3+R2R_3 → R_3 + R_2 , we have

[1010 0121 0025]\begin{bmatrix} 1 & 0 & 10\\\ 0 & 1 & 21\\\ 0&0&25 \end{bmatrix}=[−503 −1318 −1519]\begin{bmatrix} -5 & 0 & 3\\\ -13 & 1 & 8\\\ -15&1&9 \end{bmatrix}A

Applying R3→125R3R_3 → \frac{1}{25}R_3 , we have

[1010 0121 001]\begin{bmatrix} 1 & 0 & 10\\\ 0 & 1 & 21\\\ 0&0&1 \end{bmatrix}=[−503 −1318 −35125925]\begin{bmatrix} -5 & 0 & 3\\\ -13 & 1 & 8\\\ -\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}

Applying R1→R1−10R3R_1 → R_1 -10R_3 and R2→R2−21R3R_2 → R_2 -21R_3, we have:

[100 010 001]\begin{bmatrix} 1 & 0 & 0\\\ 0 & 1 & 0\\\ 0&0&1 \end{bmatrix}=[1−25−35 −254251125 −35125925]\begin{bmatrix} 1 & -\frac25 & -\frac35\\\ -\frac25 & \frac{4}{25} & \frac{11}{25}\\\ -\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}A

therefore A-1=[1−25−35 −254251125 −35125925]\begin{bmatrix} 1 & -\frac25 & -\frac35\\\ -\frac25 & \frac{4}{25} & \frac{11}{25}\\\ -\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}