Question

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

23. \(\left( {\begin{aligned}{*{20}{c}}2&0&0&6\\1&{ - 7}&{ - 5}&0\\3&8&6&0\\0&7&5&4\end{aligned}} \right)\)

Short answer

The matrix is not 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 (\(\det \left( A \right) \ne 0\)).

Check the invertibility of the matrix using determinant

Consider the matrix\(A = \left( {\begin{aligned}{*{20}{c}}2&0&0&6\\1&{ - 7}&{ - 5}&0\\3&8&6&0\\0&7&5&4\end{aligned}} \right)\).

Compute the determinate of matrix A, as shown below:

\(\begin{aligned}{c}\det \left( A \right) = {\left( { - 1} \right)^{1 + 1}} \cdot \left( 2 \right)\left| {\begin{aligned}{*{20}{c}}{ - 7}&{ - 5}&0\\8&6&0\\7&5&4\end{aligned}} \right| + {\left( { - 1} \right)^{1 + 2}} \cdot \left( 0 \right)\left| {\begin{aligned}{*{20}{c}}1&{ - 5}&0\\3&6&0\\0&5&4\end{aligned}} \right| + {\left( { - 1} \right)^{1 + 3}} \cdot \left( 0 \right)\left| {\begin{aligned}{*{20}{c}}1&{ - 7}&0\\3&8&0\\0&7&4\end{aligned}} \right|\\ + {\left( { - 1} \right)^{1 + 4}} \cdot \left( 8 \right)\left| {\begin{aligned}{*{20}{c}}1&{ - 7}&{ - 5}\\3&8&6\\0&7&5\end{aligned}} \right|\\ = 2 \cdot \left| {\begin{aligned}{*{20}{c}}{ - 7}&{ - 5}&0\\8&6&0\\7&5&4\end{aligned}} \right| - 0 + 0 - 8 \cdot \left| {\begin{aligned}{*{20}{c}}1&{ - 7}&{ - 5}\\3&8&6\\0&7&5\end{aligned}} \right|\\ = 2 \cdot \left| {\begin{aligned}{*{20}{c}}{ - 7}&{ - 5}&0\\8&6&0\\7&5&4\end{aligned}} \right| - 8 \cdot \left| {\begin{aligned}{*{20}{c}}1&{ - 7}&{ - 5}\\3&8&6\\0&7&5\end{aligned}} \right|\end{aligned}\)

Evaluate the determinant\(\left| {\begin{aligned}{*{20}{c}}{ - 7}&{ - 5}&0\\8&6&0\\7&5&4\end{aligned}} \right|\).

\(\begin{aligned}{c}\left| {\begin{aligned}{*{20}{c}}{ - 7}&{ - 5}&0\\8&6&0\\7&5&4\end{aligned}} \right| = \left( { - 7} \right) \cdot \left| {\begin{aligned}{*{20}{c}}6&0\\5&4\end{aligned}} \right| + 5 \cdot \left| {\begin{aligned}{*{20}{c}}8&0\\7&4\end{aligned}} \right| + 0\left| {\begin{aligned}{*{20}{c}}8&6\\7&5\end{aligned}} \right|\\ = - 7\left( {6\left( 4 \right) - 0\left( 5 \right)} \right) + 5\left( {8\left( 4 \right) - 0} \right) + 0\\ = - 168 + 160\\ = - 8\end{aligned}\)

Evaluate the determinant\(\left| {\begin{aligned}{*{20}{c}}1&{ - 7}&{ - 5}\\3&8&6\\0&7&5\end{aligned}} \right|\).

\(\begin{aligned}{c}\left| {\begin{aligned}{*{20}{c}}1&{ - 7}&{ - 5}\\3&8&6\\0&7&5\end{aligned}} \right| = 1 \cdot \left| {\begin{aligned}{*{20}{c}}8&6\\7&5\end{aligned}} \right| + 7 \cdot \left| {\begin{aligned}{*{20}{c}}3&6\\0&5\end{aligned}} \right| - 5\left| {\begin{aligned}{*{20}{c}}3&8\\0&7\end{aligned}} \right|\\ = \left( {8\left( 5 \right) - 6\left( 7 \right)} \right) + 7\left( {3\left( 5 \right) - 0} \right) - 5\left( {3\left( 7 \right) - 0\left( 8 \right)} \right)\\ = 40 - 42 + 105 - 105\\ = - 2\end{aligned}\)

Obtain determinant A of the matrix.

\(\begin{aligned}{c}\det \left( A \right) = 2\left( { - 8} \right) - 8\left( { - 2} \right)\\ = - 16 + 16\\ = 0\end{aligned}\)

Since \(\det \left( A \right) = 0\), the matrix is not invertible.