ここで勉強すれば数学検定１級の壁は超えられるか。

$\begin{vmatrix} \dfrac {5}{36} & \dfrac {2}{9}-\dfrac {\sqrt {15}}{15} & \dfrac {5}{36}-\dfrac {\sqrt {15}}{30} \\ \dfrac {5}{36}+\dfrac {\sqrt {15}}{24} & \dfrac {2}{9} & \dfrac {5}{36}-\dfrac {\sqrt {15}}{24} \\ \dfrac {5}{36}+\dfrac {\sqrt {15}}{30} & \dfrac {2}{9}+\dfrac {\sqrt {15}}{15} & \dfrac {5}{36} \end{vmatrix}$

の行列式を求める。

２行－１行　と　３行－１行をする。

$=\begin{vmatrix} \dfrac {5}{36}&\dfrac {2}{9}-\dfrac {\sqrt {15}}{15}&\dfrac {5}{36}-\dfrac {\sqrt {15}}{30}\\ \dfrac {\sqrt {15}}{24}&\dfrac {\sqrt {15}}{15}&\dfrac {\sqrt {15}}{30}-\dfrac {\sqrt {15}}{24}\\ \dfrac {\sqrt {15}}{30}&\dfrac {2\sqrt {15}}{15}&\dfrac {\sqrt {15}}{30}\end{vmatrix}$

２行と３行をそれぞれ$\sqrt {15}$でくくると

$=\left( \sqrt {15}\right) ^{2}\begin{vmatrix} \dfrac {5}{36} & \dfrac {2}{9}-\dfrac {\sqrt {15}}{15} & \dfrac {5}{36}-\dfrac {\sqrt {15}}{30} \\ \dfrac {1}{24} & \dfrac {1}{15} & -\dfrac {1}{120} \\ \dfrac {1}{30} & \dfrac {2}{15} & \dfrac {1}{30} \end{vmatrix}$

２行目の分母から１２０　３行目の分母から３０をくくって

$=\dfrac {15}{120.30}\begin{vmatrix} \dfrac {5}{36} & \dfrac {2}{9}-\dfrac {\sqrt {15}}{15} & \dfrac {5}{36}-\dfrac {\sqrt {15}}{30} \\ 5 & 8 & -1 \\ 1 & 4 & 1 \end{vmatrix}$

１行－３行×５／３６　　　と　　２行－３行×５　をすると

$=\dfrac {1}{240}\begin{pmatrix} 0 & -\dfrac {1}{3}-\dfrac {\sqrt {15}}{15} & -\dfrac {\sqrt {15}}{30} \\ 0 & -12 & -6 \\ 1 & 4 & 1 \end{pmatrix}$

$=\left( \left( \dfrac {1}{3}+\dfrac {\sqrt {15}}{15}\right) \times 6-\dfrac {\sqrt {15}}{30}\times 12\right) \dfrac {1}{240}=\dfrac {1}{120}$・・・答え