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Determinant of $$$\left[\begin{array}{ccc}5 - \lambda & 8 & 16\\4 & 1 - \lambda & 8\\-4 & -4 & - \lambda - 11\end{array}\right]$$$

The calculator will find the determinant of the square $$$3$$$x$$$3$$$ matrix $$$\left[\begin{array}{ccc}5 - \lambda & 8 & 16\\4 & 1 - \lambda & 8\\-4 & -4 & - \lambda - 11\end{array}\right]$$$, with steps shown.

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Your Input

Calculate $$$\left|\begin{array}{ccc}5 - \lambda & 8 & 16\\4 & 1 - \lambda & 8\\-4 & -4 & - \lambda - 11\end{array}\right|$$$.

Solution

Subtract column $$$2$$$ multiplied by $$$\frac{5}{8} - \frac{\lambda}{8}$$$ from column $$$1$$$: $$$C_{1} = C_{1} - \left(\frac{5}{8} - \frac{\lambda}{8}\right) C_{2}$$$.

$$$\left|\begin{array}{ccc}5 - \lambda & 8 & 16\\4 & 1 - \lambda & 8\\-4 & -4 & - \lambda - 11\end{array}\right| = \left|\begin{array}{ccc}0 & 8 & 16\\- \frac{\left(\lambda - 9\right) \left(\lambda + 3\right)}{8} & 1 - \lambda & 8\\- \frac{\lambda}{2} - \frac{3}{2} & -4 & - \lambda - 11\end{array}\right|$$$

Subtract column $$$2$$$ multiplied by $$$2$$$ from column $$$3$$$: $$$C_{3} = C_{3} - 2 C_{2}$$$.

$$$\left|\begin{array}{ccc}0 & 8 & 16\\- \frac{\left(\lambda - 9\right) \left(\lambda + 3\right)}{8} & 1 - \lambda & 8\\- \frac{\lambda}{2} - \frac{3}{2} & -4 & - \lambda - 11\end{array}\right| = \left|\begin{array}{ccc}0 & 8 & 0\\- \frac{\left(\lambda - 9\right) \left(\lambda + 3\right)}{8} & 1 - \lambda & 2 \lambda + 6\\- \frac{\lambda}{2} - \frac{3}{2} & -4 & - \lambda - 3\end{array}\right|$$$

Expand along row $$$1$$$:

$$$\left|\begin{array}{ccc}0 & 8 & 0\\- \frac{\left(\lambda - 9\right) \left(\lambda + 3\right)}{8} & 1 - \lambda & 2 \lambda + 6\\- \frac{\lambda}{2} - \frac{3}{2} & -4 & - \lambda - 3\end{array}\right| = \left(0\right) \left(-1\right)^{1 + 1} \left|\begin{array}{cc}1 - \lambda & 2 \lambda + 6\\-4 & - \lambda - 3\end{array}\right| + \left(8\right) \left(-1\right)^{1 + 2} \left|\begin{array}{cc}- \frac{\left(\lambda - 9\right) \left(\lambda + 3\right)}{8} & 2 \lambda + 6\\- \frac{\lambda}{2} - \frac{3}{2} & - \lambda - 3\end{array}\right| + \left(0\right) \left(-1\right)^{1 + 3} \left|\begin{array}{cc}- \frac{\left(\lambda - 9\right) \left(\lambda + 3\right)}{8} & 1 - \lambda\\- \frac{\lambda}{2} - \frac{3}{2} & -4\end{array}\right| = - 8 \left|\begin{array}{cc}- \frac{\left(\lambda - 9\right) \left(\lambda + 3\right)}{8} & 2 \lambda + 6\\- \frac{\lambda}{2} - \frac{3}{2} & - \lambda - 3\end{array}\right|$$$

The determinant of a 2x2 matrix is $$$\left|\begin{array}{cc}a & b\\c & d\end{array}\right| = a d - b c$$$.

$$$\left|\begin{array}{cc}- \frac{\left(\lambda - 9\right) \left(\lambda + 3\right)}{8} & 2 \lambda + 6\\- \frac{\lambda}{2} - \frac{3}{2} & - \lambda - 3\end{array}\right| = \left(- \frac{\left(\lambda - 9\right) \left(\lambda + 3\right)}{8}\right)\cdot \left(- \lambda - 3\right) - \left(2 \lambda + 6\right)\cdot \left(- \frac{\lambda}{2} - \frac{3}{2}\right) = \frac{\lambda^{3}}{8} + \frac{5 \lambda^{2}}{8} + \frac{3 \lambda}{8} - \frac{9}{8}$$$

Finally, $$$\left(-8\right)\cdot \left(\frac{\lambda^{3}}{8} + \frac{5 \lambda^{2}}{8} + \frac{3 \lambda}{8} - \frac{9}{8}\right) = - \left(\lambda - 1\right) \left(\lambda + 3\right)^{2}.$$$

Answer

$$$\left|\begin{array}{ccc}5 - \lambda & 8 & 16\\4 & 1 - \lambda & 8\\-4 & -4 & - \lambda - 11\end{array}\right| = - \left(\lambda - 1\right) \left(\lambda + 3\right)^{2}\approx - 9 \left(0.333333333333333 \lambda + 1\right)^{2} \left(\lambda - 1\right)$$$A