Maths · Evaluation of determinants
and suppose that det. =2 \) then the det.(B) equals, where
\( \operatorname{Let} A=\left|\begin{array}{lll}\boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} \\ \boldsymbol{p} & \boldsymbol{q} & \boldsymbol{r} \\ \boldsymbol{x} & \boldsymbol{y} & \boldsymbol{z}\end{array}\right| \) and suppose that det. \( (A)=2 \) then the det.(B) equals, where \( \boldsymbol{B}=\left|\begin{array}{ccc}\mathbf{4} \boldsymbol{x} & \mathbf{2} \boldsymbol{a} & -\boldsymbol{p} \\ \mathbf{4} \boldsymbol{y} & \mathbf{2} \boldsymbol{b} & -\boldsymbol{q} \\ \boldsymbol{4} \boldsymbol{z} & \boldsymbol{2} \boldsymbol{c} & -\boldsymbol{t}\end{array}\right| \)
- A. \( \operatorname{det}(B)=-2 \)
- B. \( \operatorname{det}(B)=-8 \)
- C. \( \operatorname{det}(B)=-16 \)
- D. \( \operatorname{det}(B)=8 \)
Step-by-step solution
We have det(A) = 2. B can be obtained from A^T by permuting columns (cyclic shift: col1 ← col3, col2 ← col1, col3 ← col2) which does not change the determinant sign (even permutation), then scaling columns by 4, 2, and -1 respectively, multiplying the determinant by 4*2*(-1) = -8. Since det(A^T) = det(A) = 2, det(B) = -8 * 2 = -16.
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