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Calculating the Inverse of a Matrix

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2025/12/02 03:43
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Calculating the Inverse of a Matrix

Calculating the inverse of a 4×44 \times 4 matrix needs at most 224224 steps. Wow.

Step 0

Let
(211−5−3−3020)\begin{pmatrix} 2 & 1 & 1 \\ -5 & -3 & -3 \\ 0 & 2 & 0 \end{pmatrix}
be the matrix you want to calculate.

Step 1: Create an Augmented Matrix

Add II to the right-hand side of the matrix. In this example, it becomes
(211100−5−3−3010020001)\begin{pmatrix} 2 & 1 & 1 & 1 & 0 & 0\\ -5 & -3 & -3 & 0 & 1 & 0\\ 0 & 2 & 0 & 0 & 0 & 1 \end{pmatrix}

Step 2: Calculating the Reduced Row Echelon Form

I hate this step.
Anyway, you want to make the left-hand side of the matrix to be II. You can do it through row operations:
  • Swap 2 rows.
  • Multiply each element in one row by a non-zero constant.
  • Add a multiple of another row to one row.
If you don't know how to do that, there are a lot of online resources about it.
Finally, we get
(1003100100012001−5−2−12)\begin{pmatrix} 1 & 0 & 0 & 3 & 1 & 0\\ 0 & 1 & 0 & 0 & 0 & \dfrac{1}{2}\\ 0 & 0 & 1 & -5 & -2 & -\dfrac{1}{2} \end{pmatrix}

Step 3: The right-hand side of this matrix is the inverse of the original matrix.

So the inverse of the matrix is
(3100012−5−2−12)\begin{pmatrix} 3 & 1 & 0 \\ 0 & 0 & \dfrac{1}{2} \\ -5 & -2 & -\dfrac{1}{2} \end{pmatrix}

Why does it work?

Doing a row operation is actually left-multiplying an elementary matrix. When the left-hand side becomes II, the combination of those elementary matrices is A−1A^{-1} because A−1A=IA^{-1}A=I. Since the same operations are done to the right-hand side, it becomes A−1I=A−1A^{-1}I=A^{-1}.

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