Calculation by transformations
Task number: 4547
Hint
The determinant of an upper triangular matrix is equal to the product of the elements on the diagonal. This product corresponds to the identity. The other permutations map at least one index on a smaller value \(i>p(i)\) and thus the corresponding product contains at least one zero and does not affect the value of the determinant.
Adding the \(t\)-multiple of the \(j\)-th row to the \(i\)-th row for \(i\ne j\) does not change the value of the determinant. Using modifications of this type, a matrix can be converted to a triangular matrix with the same value of the determinant.
Beware, swapping two rows will change the sign of the determinant. Multiplying a row by the scalar \(t\) changes the value of the determinant to \(t\)-multiple. These modifications should be avoided when calculating the determinant.
Because \(\det \boldsymbol A=\det \boldsymbol A^{\mathsf T}\), it is possible to use column transformations in addition to row transformations and possibly alternate them during the calculation.
Variant
Over \(\mathbb R\): \(\begin{pmatrix} 7 & -3 \\ 5 & -6 \end{pmatrix} \)Solution
\( \begin{vmatrix} 7 & -3 \\ 5 & -6 \end{vmatrix}= \begin{vmatrix} 7 & -3 \\ 5-\tfrac{5}{7} 7& -6-\tfrac{5}{7}(-3) \end{vmatrix}= \begin{vmatrix} 7 & -3 \\ 0 & -\tfrac{27}{7} \end{vmatrix}= 7\cdot{}\tfrac{-27}{7}= -27 \)To avoid fractions, we can reduce the elements in the first column sequentially, e.g. subtracting 5 from 7 to get 2, etc. It is also possible to eliminate \(-3\) by applying \(-6\) to the lower triangular matrix and transposing that.
Answer
The determinant is \(-27\).Variant
Over \(\mathbb Z_5\): \(\begin{pmatrix} 1 & 2 & 3 \\ 4 & 4 & 1 \\ 2 & 3 & 3 \\ \end{pmatrix} \)Solution
\( \begin{vmatrix} 1 & 2 & 3 \\ 4 & 4 & 1 \\ 2 & 3 & 3 \\ \end{vmatrix}= \begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 4 & 2 \\ \end{vmatrix}= \begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \\ \end{vmatrix}= 1 \cdot{}1 \cdot{} 1 =1 \)First, we added the first row to the second and three times the first to the third. Then we added the second to the third.
Answer
The determinant is \(1\).Variant
Over \(\mathbb Z_5\): \(\begin{pmatrix} 4 & 4 & 1 \\ 2 & 3 & 3 \\ 1 & 2 & 3 \\ \end{pmatrix} \)Solution
\( \begin{vmatrix} 4 & 4 & 1 \\ 2 & 3 & 3 \\ 1 & 2 & 3 \\ \end{vmatrix}= \begin{vmatrix} 4 & 4 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & 4 \\ \end{vmatrix}= \begin{vmatrix} 4 & 4 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \\ \end{vmatrix}= 4 \cdot{}1 \cdot{} 4 =1 \)Compared to the previous variant, the value of the determinant has not changed because the rows have been reordered according to the permutation with a positive sign.
Answer
The determinant is \(1\).
