Consider the matrices A=pmatrix2&-24&-2pmatrix and B=pmatrix3&91&3pmatrix.

Question

Consider the matrices $A=\begin{pmatrix}2&-2\4&-2\end{pmatrix}$ and $B=\begin{pmatrix}3&9\1&3\end{pmatrix}$. If matrices $P$ and $Q$ are such that $PA=B$ and $AQ=B$, then the absolute value of the sum of the diagonal elements of $2(P+Q)$ is \_\_\_\_\_\_\_\_.

Answer: 34

Accepted Answer

$A^{-1}=\dfrac{1}{4}\begin{pmatrix}-2&2\-4&2\end{pmatrix}=\dfrac12\begin{pmatrix}-1&1\-2&1\end{pmatrix}$.

$P=BA^{-1}=\dfrac12\begin{pmatrix}3&9\1&3\end{pmatrix}\begin{pmatrix}-1&1\-2&1\end{pmatrix}=\dfrac12\begin{pmatrix}-21&12\-7&4\end{pmatrix}$.

$Q=A^{-1}B=\dfrac12\begin{pmatrix}-1&1\-2&1\end{pmatrix}\begin{pmatrix}3&9\1&3\end{pmatrix}=\dfrac12\begin{pmatrix}-2&-6\-5&-15\end{pmatrix}$.

$\operatorname{tr}(2(P+Q)) = \operatorname{tr}(2P)+\operatorname{tr}(2Q) = (-21+4)+(-2-15) = -17-17=-34$.

So $|\operatorname{tr}(2(P+Q))|=34$.

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