The Inverse Matrix of an Upper Triangular Matrix with Variables

Linear algebra problems and solutions

Problem 275

Let $A$ be the following $3\times 3$ upper triangular matrix.
\[A=\begin{bmatrix}
1 & x & y \\
0 &1 &z \\
0 & 0 & 1
\end{bmatrix},\] where $x, y, z$ are some real numbers.

Determine whether the matrix $A$ is invertible or not. If it is invertible, then find the inverse matrix $A^{-1}$.

 
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Solution.

We form the augmented matrix
\[[A\mid I]= \left[\begin{array}{rrr|rrr}
1 & x & y & 1 &0 & 0 \\
0 & 1 & z & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 & 1 \\
\end{array} \right]\] and apply elementary row operations as follows.
\begin{align*}
\left[\begin{array}{rrr|rrr}
1 & x & y & 1 &0 & 0 \\
0 & 1 & z & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{R_1-xR_2}
\left[\begin{array}{rrr|rrr}
1 & 0 & y-xz & 1 & -x & 0 \\
0 & 1 & z & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 & 1 \\
\end{array} \right]\\[10pt] \xrightarrow{\substack{R_1-(y-xz)R_3\\ R_2-zR_3}}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & 1 & -x & xz-y \\
0 & 1 & 0 & 0 & 1 & -z \\
0 & 0 & 1 & 0 & 0 & 1 \\
\end{array} \right].
\end{align*}
We could reduce the matrix $A$ into the identity matrix $I$.
Thus, the matrix $A$ is invertible and the right $3\times 3$ matrix is the inverse matrix of $A^{-1}$.
Hence,
\[A^{-1}=\begin{bmatrix}
1 & -x & xz-y \\
0 & 1 & -z \\
0 & 0 & 1
\end{bmatrix}.\]


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1 Response

  1. 02/03/2017

    […] The inverse element of the matrix [begin{bmatrix} 1 & x & y \ 0 &1 &z \ 0 & 0 & 1 end{bmatrix}] is given by [begin{bmatrix} 1 & -x & xz-y \ 0 & 1 & -z \ 0 & 0 & 1 end{bmatrix}.] For a proof, see the post The inverse matrix of an upper triangular matrix with variables. […]

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