Let {Delta _1} = begin{vmatrix} {a_1} & {b_1} | vedclass

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(C) The product of two determinants of order $3 	imes 3$ can be expressed as a single determinant of order $3 	imes 3$ where each element is the sum

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(C) The product of two determinants of order $3 	imes 3$ can be expressed as a single determinant of order $3 	imes 3$ where each element is the sum of products of corresponding elements of the rows or columns of the two determinants.Specifically,if we multiply the rows of ${\Delta _1}$ with the rows of ${\Delta _2}$,each entry in the resulting determinant is of the form $(a_i \alpha_j + b_i \beta_j + c_i \gamma_j)$.Since each of the $9$ entries in the resulting $3 	imes 3$ determinant is a sum of $3$ terms,we can expand the determinant using the linearity property of determinants.For a $3 	imes 3$ determinant,if each of the $9$ entries is a sum of $3$ terms,the total number of determinants formed by this expansion is $3^3 = 27$.

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