Exercise 134Linear algebra › Linear mapsL1Added 3 Nov 2026

Inverse of an endomorphism

#inverse#endomorphism#matrix inverse

Statement

Let $f:\mathbb{R}^3\to\mathbb{R}^3$ have matrix
$A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{pmatrix}.$
Show that $f$ is an isomorphism and compute $f^{-1}(x,y,z)$ .

Hint

$A$ is upper triangular with nonzero diagonal entries, hence invertible. Solve $f(x',y',z')=(x,y,z)$ by back-substitution.

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Solution

Invertibility. $\det A = 1\cdot1\cdot1=1\neq0$ , so $A$ is invertible and $f$ is an isomorphism.

Inverse. Solve $A(x',y',z')^T=(x,y,z)^T$ :
$x'+2y'+3z'=x, \qquad y'+4z'=y, \qquad z'=z.$
Back-substituting: $z'=z$ ; $y'=y-4z$ ; $x'=x-2y'-3z'=x-2(y-4z)-3z=x-2y+5z$ .

Conclusion.
$f^{-1}(x,y,z)=(x-2y+5z,\ y-4z,\ z), \qquad A^{-1}=\begin{pmatrix} 1 & -2 & 5 \\ 0 & 1 & -4 \\ 0 & 0 & 1 \end{pmatrix}.$

Check. Row by row, $AA^{-1}$ gives $\begin{pmatrix} 1 & -2+2 & 5-8+3 \\ 0 & 1 & -4+4 \\ 0 & 0 & 1 \end{pmatrix}=I$ .

Inverse of an endomorphism

Statement

Solution

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