Calculation by expansion
Task number: 4548
Expand the determinants of the following matrices according to the second row and then calculate its value:
Hint
If we denote by \(\boldsymbol A^{i,j}\) the matrix obtained from the matrix \(\boldsymbol A\) by discarding the \(i\)-th row and \(j\)-th column, then the Laplace expansion of the determinant of a matrix of order \(n\) by the \(i\)-th row is:
\[\det\boldsymbol A=\sum_{j=1}^n a_{i,j} (-1)^{i+j} \det\boldsymbol A^{i,j}\]Laplace expansion is computationally inefficient already when the corresponding row contains at least two non-zero elements. Its recurrent application leads to an exponential computation similar to the direct application of the determinant definition.
Variant
Over \(\mathbb Z_5\): \(\begin{pmatrix} 1 & 2 & 3 \\ 4 & 4 & 1 \\ 2 & 3 & 3 \\ \end{pmatrix} \)Solution
\(\begin{vmatrix} 1 & 2 & 3 \\ 4 & 4 & 1 \\ 2 & 3 & 3 \\ \end{vmatrix} = -4 \begin{vmatrix} 2 & 3 \\ 3 & 3 \\ \end{vmatrix} + 4 \begin{vmatrix} 1 & 3 \\ 2 & 3 \\ \end{vmatrix} - \begin{vmatrix} 1 & 2 \\ 2 & 3 \\ \end{vmatrix} = 2+3+1=1\)Answer
The determinant is \(1\).Variant
Over \(\mathbb R\): \(\begin{pmatrix} 0 & 1 & 2 & 0 \\ 3 & 0 & 0 & 4 \\ 0 & 5 & 6 & 0 \\ 7 & 0 & 0 & 8 \\ \end{pmatrix} \)Solution
\(\begin{vmatrix} 0 & 1 & 2 & 0 \\ 3 & 0 & 0 & 4 \\ 0 & 5 & 6 & 0 \\ 7 & 0 & 0 & 8 \\ \end{vmatrix} = -3 \begin{vmatrix} 1 & 2 & 0 \\ 5 & 6 & 0 \\ 0 & 0 & 8 \\ \end{vmatrix} +0 \begin{vmatrix} 0 & 2 & 0 \\ 0 & 6 & 0 \\ 7 & 0 & 8 \\ \end{vmatrix} -0 \begin{vmatrix} 0 & 1 & 0 \\ 0 & 5 & 0 \\ 7 & 0 & 8 \\ \end{vmatrix} +4 \begin{vmatrix} 0 & 1 & 2 \\ 0 & 5 & 6 \\ 7 & 0 & 0 \\ \end{vmatrix} = 96-112=-16\)Answer
The determinant is \(-16\).Variant
Over \(\mathbb R\): \(\begin{pmatrix} 0 & 1 & 2 & 0 \\ 3 & 100 & 100 & 4 \\ 0 & 5 & 6 & 0 \\ 7 & 100 & 100 & 8 \\ \end{pmatrix} \)Solution
\(\begin{vmatrix} 0 & 1 & 2 & 0 \\ 3 & 100 & 100 & 4 \\ 0 & 5 & 6 & 0 \\ 7 & 100 & 100 & 8 \\ \end{vmatrix} = -3 \begin{vmatrix} 1 & 2 & 0 \\ 5 & 6 & 0 \\ 100 & 100 & 8 \\ \end{vmatrix} +100 \begin{vmatrix} 0 & 2 & 0 \\ 0 & 6 & 0 \\ 7 & 0 & 8 \\ \end{vmatrix} -100 \begin{vmatrix} 0 & 1 & 0 \\ 0 & 5 & 0 \\ 7 & 0 & 8 \\ \end{vmatrix} +4 \begin{vmatrix} 0 & 1 & 2 \\ 0 & 5 & 6 \\ 7 & 100 & 100 \\ \end{vmatrix} =96-112=-16\)The determinant has the same value as in the previous variant, because the coefficients of 100 have been multiplied by the determinants of the singular matrices and these have value 0.
Answer
The determinant is \(-16\).
