Respuesta :
[tex]\huge{\underline{\underline{\boxed{\sf{\purple{Answer}}}}}}[/tex]
- Since column 3 of the given matrix consists of numbers which are all constant multiples of each other, there is a theorem which states that the determinant of this matrix is 0 .
- However, I will prove it from calculation just to verify and show how you would calculate such determinants should you not be aware of the theorems of linear matrix algebra.
- Use co-factor expansion along any row or column of your choice.
- This involves adding the products of the entries in each row with their co-factors[tex] { - 1}^({i + j} )[/tex]for an entry in row I and column j, multiplied by the minor of the entry, formed by evaluating the resulting determinant when you delete row I and column j.
In this case, since the given matrix is 3x3 indimension, it will result in three 2×2 determinants which we can find from definition[tex] \binom{a \: b}{c \: d} [/tex] = ad-bc
I will take co-factor expansion along row 1 as an example to find the determinant of the given matrix as :
∆=[tex] 1( - 1) {}^{1 + 1} {} \binom{56}{8 \: 9} + 2( - 1) {}^{1 + 2} \binom{4 \: 6}{7 \: 9} + 3( - 1) {}^{1 + 3} \binom{4 \: 5}{7 \: 8} \\ = 1(45 - 48) +( - 2)(36 - 42) + 3(32 - 35) \\ = - 3 + 12 - 9 \\ = 0 [/tex]
