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અહી $A+2 B=\left[\begin{array}{ccc}1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1\end{array}\right]$ અને $2 A - B =\left[\begin{array}{ccc}2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2\end{array}\right] $ આપેલ છે જો If $\operatorname{Tr}( A )$ એ શ્રેણિક $A $ ના વિકર્ણો ઘટકોનો સરવાળો દર્શાવે છે તો $\operatorname{Tr}( A )-\operatorname{Tr}( B )$ ની કિમંત મેળવો.
$1$
$2$
$0$
$3$
Solution
$A +2 B =\left(\begin{array}{ccc}1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1\end{array}\right) \quad \ldots(1)$
$2 A – B =\left(\begin{array}{ccc}2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2\end{array}\right)$
$\Rightarrow 4 A -2 B =\left(\begin{array}{ccc}4 & -2 & 10 \\ 4 & -2 & 12 \\ 0 & 2 & 4\end{array}\right)$ $……(2)$
$(1)+(2) \Rightarrow 5 A =\left(\begin{array}{ccc}5 & 0 & 10 \\ 10 & -5 & 15 \\ -5 & 5 & 5\end{array}\right)$
$A=\left(\begin{array}{ccc}1 & 0 & 2 \\ 2 & -1 & 3 \\ -1 & 1 & 1\end{array}\right)$ and $2 A=\left(\begin{array}{ccc}2 & 0 & 4 \\ 4 & -2 & 6 \\ -2 & 2 & 2\end{array}\right)$
$\therefore B=\left(\begin{array}{ccc}2 & 0 & 4 \\ 4 & -2 & 6 \\ -2 & 2 & 2\end{array}\right)-\left(\begin{array}{ccc}2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2\end{array}\right)$
$B =\left(\begin{array}{ccc}0 & 1 & -1 \\ 2 & -1 & 0 \\ -2 & 1 & 0\end{array}\right)$
$\operatorname{tr}(A)=1-1+1=1$
$\operatorname{tr}(B)=-1$
$\operatorname{tr}(A)=1$ and $\operatorname{tr}(B)=-1$
$\therefore \operatorname{tr}(A)-\operatorname{tr}(B)=2$