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Exercise 1768Linear algebra › Matrices and linear systemsL1Added 3 Nov 2026

Inverse of a 3×3 matrix by row reduction

#inverse#Gaussian elimination#3x3 matrices

Statement

Compute the inverse of
A=(123014560).A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}.

Hint

Apply Gaussian elimination to the augmented matrix [A∣I][A \mid I] until the left block becomes the identity.

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Solution

Augmented matrix.
(123100014010560001).\left(\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 4 & 0 & 1 & 0 \\ 5 & 6 & 0 & 0 & 0 & 1 \end{array}\right).

Elimination. L3←L3−5L1L_3 \leftarrow L_3 - 5L_1:
(1231000140100−4−15−501).\left(\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 4 & 0 & 1 & 0 \\ 0 & -4 & -15 & -5 & 0 & 1 \end{array}\right).
L3←L3+4L2L_3 \leftarrow L_3 + 4L_2:
(123100014010001−541).\left(\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 4 & 0 & 1 & 0 \\ 0 & 0 & 1 & -5 & 4 & 1 \end{array}\right).
L1←L1−3L3L_1 \leftarrow L_1 - 3L_3, L2←L2−4L3L_2 \leftarrow L_2 - 4L_3:
(12016−12−301020−15−4001−541).\left(\begin{array}{ccc|ccc} 1 & 2 & 0 & 16 & -12 & -3 \\ 0 & 1 & 0 & 20 & -15 & -4 \\ 0 & 0 & 1 & -5 & 4 & 1 \end{array}\right).
L1←L1−2L2L_1 \leftarrow L_1 - 2L_2:
(100−2418501020−15−4001−541).\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & -24 & 18 & 5 \\ 0 & 1 & 0 & 20 & -15 & -4 \\ 0 & 0 & 1 & -5 & 4 & 1 \end{array}\right).

Conclusion.
A−1=(−2418520−15−4−541).A^{-1} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}.

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