Đáp án:
Câu 7: $C.\ \det(A^T.A^{-1})= 1$
Câu 8: $C.\ I =\dfrac{\pi}{3}$
Giải thích các bước giải:
Câu 7:
Ta có: $\det A = 2$
$+)\quad \det(-A)= (-1)^4\det(A)= 2$
$+)\quad \det(2A^2)= 2^4\det(A\times A)= 2^4\det^2(A)= 64$
$+)\quad \det(A^T.A^{-1})= \det(A^T).\det(A^{-1}) = \det(A).\det(A^{-1})= 1$
Câu 8:
$D =\{(x,y)\Big|\ -1\leqslant x \leqslant 1,\ -1\leqslant y \leqslant 1\}$
$I = \displaystyle\iint\limits_D \dfrac{x^2}{y^2+1}dxdy$
$\Leftrightarrow I = \displaystyle\int\limits_{-1}^1dx\displaystyle\int\limits_{-1}^1\dfrac{x^2}{y^2+1}dy$
$\Leftrightarrow I = \displaystyle\int\limits_{-1}^1\left(x^2\arctan y\Bigg|_{-1}^1\right)dx$
$\Leftrightarrow I = \displaystyle\int\limits_{-1}^1\dfrac{\pi}{2}x^2dx$
$\Leftrightarrow I = \dfrac{\pi}{6}x^3\Bigg|_{-1}^1$
$\Leftrightarrow I = \dfrac{\pi}{3}$
