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4 votes
4 votes
Let
$$
A=\left[\begin{array}{llll}
1 & 2 & 3 & 4 \\
4 & 1 & 2 & 3 \\
3 & 4 & 1 & 2 \\
2 & 3 & 4 & 1
\end{array}\right]
$$
And
$$
B=\left[\begin{array}{llll}
3 & 4 & 1 & 2 \\
4 & 1 & 2 & 3 \\
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1
\end{array}\right]
$$
Let $\operatorname{det}(\mathrm{A})$ and $\operatorname{det}(\mathrm{B})$ denote the determinants of the matrices $\mathrm{A}$ and $\mathrm{B}$, respectively.
  1. $\operatorname{det} \mathrm{AB}=(\operatorname{det} A) \times(\operatorname{det} B)$
  2. $\operatorname{det}(\mathrm{A})=0$
  3. $\operatorname{det} \mathrm{B}=-\operatorname{det} \mathrm{A}$
  4. $\operatorname{det} \mathrm{A}=\operatorname{det} \mathrm{B}$
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2 Answers

2 votes
2 votes

the elements of both the matrix $A,B$ are the same. in matrix $B$ $R_1,R_3$ are interchanged.

as we know the important property of any determinants: 

  • if two rows (or 2 columns) of a determinant are interchanged the sign of the value of the determinant is changed.

here only one time rows is changed so the determinant should be multiply by $(-1)$.

 $\therefore det (B)= -det(A)$

For Option A, 

If $A, B$ are arbitrary $n × n$ matrices, then $det(AB) = det(A)* det(B)$

here $\Delta(A)=-160$, $\Delta(B)=-160 \rightarrow \Delta(AB)=-25600$

So Option $(A, C)$ is correct.

edited by
Answer:

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