Find the determinant of the matrix \[A=\left[ \begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 5 & 1 & 2 & 3 \\ 4 & 5 & 4 & 3 \\ 2 & 2 & -4 & 5 \end{array} \right]\nonumber \]
Solution
We will use the properties of determinants outlined above to find \(\det \left(A\right)\). First, add \(-5\) times the first row to the second row. Then add \(-4\) times the first row to the third row, and \(-2\) times the first row to the fourth row. This yields the matrix \[B=\left[ \begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 0 & -9 & -13 & -17 \\ 0 & -3 & -8 & -13 \\ 0 & -2 & -10 & -3 \end{array} \right]\nonumber \] Notice that the only row operation we have done so far is adding a multiple of a row to another row. Therefore, by Theorem 3.2.4, \(\det \left(B\right) = \det \left(A\right).\)
At this stage, you could use Laplace Expansion to find \(\det \left(B\right)\). However, we will continue with row operations to find an even simpler matrix to work with.
Add \(-3\) times the third row to the second row. By Theorem 3.2.4 this does not change the value of the determinant. Then, multiply the fourth row by \(-3\). This results in the matrix \[C=\left[ \begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 0 & 0 & 11 & 22 \\ 0 & -3 & -8 & -13 \\ 0 & 6 & 30 & 9 \end{array} \right]\nonumber \] Here, \(\det \left(C\right) = -3 \det \left(B\right)\), which means that \(\det \left( B\right) =\left(-\frac{1}{3}\right) \det \left( C\right)\)
Since \(\det \left(A\right) = \det \left(B\right)\), we now have that \(\det \left(A\right) = \left(-\frac{1}{3}\right) \det \left( C\right)\). Again, you could use Laplace Expansion here to find \(\det \left(C\right)\). However, we will continue with row operations.
Now replace the add \(2\) times the third row to the fourth row. This does not change the value of the determinant by Theorem 3.2.4. Finally switch the third and second rows. This causes the determinant to be multiplied by \(-1.\) Thus \(\det \left( C\right) = -\det \left( D\right)\) where \[D=\left[ \begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 0 & -3 & -8 & -13 \\ 0 & 0 & 11 & 22 \\ 0 & 0 & 14 & -17 \end{array} \right]\nonumber \]
Hence, \(\det \left(A\right) = \left(-\frac{1}{3}\right) \det \left( C\right) = \left(\frac{1}{3}\right) \det \left( D\right)\)
You could do more row operations or you could note that this can be easily expanded along the first column. Then, expand the resulting \(3 \times 3\) matrix also along the first column. This results in \[\det \left( D\right) =1\left( -3\right) \left\vert \begin{array}{cc} 11 & 22 \\ 14 & -17 \end{array} \right\vert = 1485\nonumber \] and so \(\det \left( A\right) =\left(\frac{1}{3}\right) \left( 1485\right) =495.\)