Let A+2B=pmatrix1&2&06&-3&3-5&3&1pmatrix and 2A-B=pmatrix2&-1&52&-1&60&1&2pmatrix.

Question

Let $A+2B=\begin{pmatrix}1&2&0\6&-3&3\-5&3&1\end{pmatrix}$ and $2A-B=\begin{pmatrix}2&-1&5\2&-1&6\0&1&2\end{pmatrix}$. If $\operatorname{Tr}(A)$ denotes the sum of all diagonal elements of $A$, then $\operatorname{Tr}(A)-\operatorname{Tr}(B)$ has value equal to

Accepted Answer

Multiplying the second equation by 2: $4A-2B=\begin{pmatrix}4&-2&10\4&-2&12\0&2&4\end{pmatrix}$. Adding to the first: $5A=\begin{pmatrix}5&0&10\10&-5&15\-5&5&5\end{pmatrix}$, so $A=\begin{pmatrix}1&0&2\2&-1&3\-1&1&1\end{pmatrix}$, giving $\operatorname{Tr}(A)=1-1+1=1$.

From $A+2B=\begin{pmatrix}1&2&0\6&-3&3\-5&3&1\end{pmatrix}$: $B=\dfrac12\left[\begin{pmatrix}1&2&0\6&-3&3\-5&3&1\end{pmatrix}-\begin{pmatrix}1&0&2\2&-1&3\-1&1&1\end{pmatrix}ight]=\begin{pmatrix}0&1&-1\2&-1&0\-2&1&0\end{pmatrix}$, giving $\operatorname{Tr}(B)=0-1+0=-1$.

$\operatorname{Tr}(A)-\operatorname{Tr}(B)=1-(-1)=2$.

Solutions · 1