Question on determinant

Question

Given $x \in \Bbb R$ and

$$P = \begin {bmatrix}1&1&1\\0&2&2\\0&0&3\end {bmatrix}, \qquad Q=\begin {bmatrix}2&x&x\\0&4&0\\x&x&6\end {bmatrix}, \qquad R=PQP^{-1}$$

show that

$$\det R = \det \begin {bmatrix}2&x&x\\0&4&0\\x&x&5\end {bmatrix}+8$$

for all $x \in \Bbb R$.

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My attempt: $|R|=\frac{|P||Q|}{|P|}=|Q|$ $$\left|\begin {array}&2&x&x\\0&4&0\\x&x&6\end {array}\right|=\left|\begin {array}&2&x&x\\0&4&0\\x&x&5\end {array}\right|+\left|\begin {array}&2&x&x\\0&4&0\\x&x&1\end {array}\right|=\left|\begin {array}&2&x&x\\0&4&0\\x&x&6\end {array}\right|+8-4x^2$$

What's my mistake?

Answer 1

You used multilinearity incorrectly. It should be

$$\left|\begin {array}&2&x&x\\0&4&0\\x&x&6\end {array}\right|=\left|\begin {array}&2&x&x\\0&4&0\\x&x&5\end {array}\right|+\left|\begin {array}&2&x&x\\0&4&0\\0&0&1\end {array}\right|=\left|\begin {array}&2&x&x\\0&4&0\\x&x&5\end {array}\right|+8.$$

Answer 2

Hint.

$$ \det\left|\begin {array}&2&x&x\\0&4&0\\x&x&6\end {array}\right|=\det\left|\begin {array}&0&0&4\\2&x&x\\x&6&x\end {array}\right| = \det\left|\begin {array}&0&0&4\\2&x&x\\x&5&x\end {array}\right|+8 $$