Calculation by the definition
Task number: 4546
Hint
The definition of the determinant of a square matrix of order \(n\) is \[\det\boldsymbol A=\sum_{p\in S_n} \operatorname{sgn}(p)\prod_{i=1}^n a_{i,p(i)}\]
The group \(S_2\) contains two permutations: the identity \((1{,}2)\) with a positive sign and the permutation \((2{,}1)\) with a negative sign. Hence: \[ \begin{vmatrix} a_{1{,}1} & a_{1{,}2} \\ a_{2{,}1} & a_{2{,}2} \\ \end{vmatrix} = +a_{1{,}1}a_{2{,}2} -a_{1{,}2}a_{2{,}1} \]
Similarly, \(S_3\) contains six permutations, three of which are positive-signed \((1{,}2,3),(2{,}3,1),(3{,}1,2)\) and three with a negative sign: \((1{,}3,2),(3{,}2,1),(2{,}1,3)\). Hence: \[ \begin{vmatrix} a_{1{,}1} & a_{1{,}2} & a_{1{,}3} \\ a_{2{,}1} & a_{2{,}2} & a_{2{,}3} \\ a_{3{,}1} & a_{3{,}2} & a_{3{,}3} \\ \end{vmatrix} = +a_{1{,}1}a_{2{,}2}a_{3{,}3} +a_{1{,}2}a_{2{,}3}a_{3{,}1} +a_{1{,}3}a_{2{,}1}a_{3{,}2} -a_{1{,}1}a_{2{,}3}a_{3{,}2} -a_{1{,}3}a_{2{,}2}a_{3{,}1} -a_{1{,}2}a_{2{,}1}a_{3{,}3} \]
Variant
Over \(\mathbb R\): \(\begin{pmatrix} 7 & -1 \\ 5 & -2 \end{pmatrix} \)Solution
\(\begin{vmatrix} 7 & -1 \\ 5 & -2 \end{vmatrix}= 7\cdot(-2)-(-1)\cdot 5=-9 \)Answer
The determinant is \(-9\).Variant
Over \(\mathbb R\): \(\begin{pmatrix} 7 & -3 \\ 5 & -6 \end{pmatrix} \)Solution
\(\begin{vmatrix} 7 & -3 \\ 5 & -6 \end{vmatrix}= 7\cdot(-6)-(-3)\cdot 5=-27 \)Answer
The determinant is \(-27\). Notice that by multiplying the second column by three, the value of the determinant has also changed three times.Variant
Over \(\mathbb C\): \(\begin{pmatrix} 2-\mathrm i & \mathrm i+3 \\ \mathrm i-3 & 2+ \mathrm i \end{pmatrix} \)Solution
\(\begin{vmatrix} 2-\mathrm i & \mathrm i+3 \\ \mathrm i-3 & 2+ \mathrm i \end{vmatrix}= (2-\mathrm i)\cdot(2+ \mathrm i)-(\mathrm i+3)\cdot (\mathrm i-3)=15 \)Answer
The determinant is \(15\).Variant
Over \(\mathbb Z_5\): \(\begin{pmatrix} 1 & 2 & 3 \\ 4 & 4 & 1 \\ 2 & 3 & 3 \\ \end{pmatrix} \)Solution
\(\begin{vmatrix} 1 & 2 & 3 \\ 4 & 4 & 1 \\ 2 & 3 & 3 \\ \end{vmatrix}= 1 \cdot{} 4 \cdot{} 3 + 2 \cdot{} 1 \cdot{} 2 + 3 \cdot{} 4 \cdot{} 3 - 1 \cdot{} 1 \cdot{} 3 - 3 \cdot{} 4 \cdot{} 2 - 2 \cdot{} 4 \cdot{} 3 = 2 + 4 + 1 + 2 + 1 + 1 = 1 \)Answer
The determinant is \(1\).
