Question

In Exercises 21–23, use determinants to find out if the matrix is invertible.

22. \(\left( {\begin{aligned}{*{20}{c}}5&1&{ - 1}\\1&{ - 3}&{ - 2}\\0&5&3\end{aligned}} \right)\)

Short answer

The matrix is invertible.

Step-by-step solution

State the condition of invertibility of the matrix using determinant

Matrix A of the order\(n \times n\)(square matrix) isinvertible if thedeterminant of the matrix is not 0 (or \(\det \left( A \right) \ne 0\)).

Check the invertibility of the matrix using the determinant

Consider the matrix\(A = \left( {\begin{aligned}{*{20}{c}}5&1&{ - 1}\\1&{ - 3}&{ - 2}\\0&5&3\end{aligned}} \right)\).

Compute the determinate of matrix A, as shown below:

\(\begin{aligned}{c}\det \left( A \right) = \left| {\begin{aligned}{*{20}{c}}5&1&{ - 1}\\1&{ - 3}&{ - 2}\\0&5&3\end{aligned}} \right|\\ = 5 \cdot \left| {\begin{aligned}{*{20}{c}}{ - 3}&{ - 2}\\5&3\end{aligned}} \right| - 1 \cdot \left| {\begin{aligned}{*{20}{c}}1&{ - 2}\\0&3\end{aligned}} \right| - 1 \cdot \left| {\begin{aligned}{*{20}{c}}1&{ - 3}\\0&5\end{aligned}} \right|\\ = 5 \cdot \left( { - 3\left( 3 \right) - 5\left( { - 2} \right)} \right) - \left( {1\left( 3 \right) - 0\left( { - 2} \right)} \right) - \left( {1\left( 5 \right) - 0\left( { - 3} \right)} \right)\\ = 5\left( 1 \right) - 3 - 5\\ = 2 - 5\\ = - 3\end{aligned}\)

Since \(\det \left( A \right) \ne 0\), the matrix is invertible.