Hello guys Good afternoon 10th Std you have ekam kasoti paper on 1st Feb. Those who now or don't know?
Answers
Answer:
Solution:-
Given equation is
\begin{lgathered}\implies\sf{\left[\begin{array}{c c c}4 & -2 \\ 4 & 0\end{array}\right] + 3A = \left[\begin{array}{c c c}-2 & -2 \\ 1 & -3\end{array}\right]}\end{lgathered}⟹[44−20]+3A=[−21−2−3]
We have to find the value of A
\begin{lgathered}\sf \implies3A =\sf{\left[\begin{array}{c c c} - 2 & -2 \\ 1 & - 3\end{array}\right]} - {\left[\begin{array}{c c c}4 & -2 \\ 4 & 0\end{array}\right]}\end{lgathered}⟹3A=[−21−2−3]−[44−20]
Now we can write as
\begin{lgathered}\sf \implies3A =\sf{\left[\begin{array}{c c c} - 2 & -2 \\ 1 & - 3\end{array}\right]} + {\left[\begin{array}{c c c} - 4 & + 2 \\ - 4 & 0\end{array}\right]}\end{lgathered}⟹3A=[−21−2−3]+[−4−4+20]
Now
\begin{lgathered}\sf \implies3A =\sf{\left[\begin{array}{c c c} - 6 & 0 \\ - 3 & - 3\end{array}\right]}\end{lgathered}⟹3A=[−6−30−3]
\begin{lgathered}\sf \implies \: A =\sf{\left[\begin{array}{c c c} \dfrac{ - 6}{3} & & 0 \\ \\ \dfrac{ - 3}{3} & & \dfrac{ - 3}{3} \end{array}\right]}\end{lgathered}⟹A=⎣⎢⎢⎡3−63−303−3⎦⎥⎥⎤
\begin{lgathered}\sf \implies A =\sf{\left[\begin{array}{c c c} - 2 & 0 \\ - 1 & - 1\end{array}\right]}\end{lgathered}⟹A=[−2−10−1]
Answer
\begin{lgathered}\sf \implies A =\sf{\left[\begin{array}{c c c} - 2 & 0 \\ - 1 & - 1\end{array}\right]}\end{lgathered}⟹A=[−2−10−1]
Solution:-
Given equation is
\begin{lgathered}\implies\sf{\left[\begin{array}{c c c}4 & -2 \\ 4 & 0\end{array}\right] + 3A = \left[\begin{array}{c c c}-2 & -2 \\ 1 & -3\end{array}\right]}\end{lgathered}⟹[44−20]+3A=[−21−2−3]
We have to find the value of A
\begin{lgathered}\sf \implies3A =\sf{\left[\begin{array}{c c c} - 2 & -2 \\ 1 & - 3\end{array}\right]} - {\left[\begin{array}{c c c}4 & -2 \\ 4 & 0\end{array}\right]}\end{lgathered}⟹3A=[−21−2−3]−[44−20]
Now we can write as
\begin{lgathered}\sf \implies3A =\sf{\left[\begin{array}{c c c} - 2 & -2 \\ 1 & - 3\end{array}\right]} + {\left[\begin{array}{c c c} - 4 & + 2 \\ - 4 & 0\end{array}\right]}\end{lgathered}⟹3A=[−21−2−3]+[−4−4+20]
Now
\begin{lgathered}\sf \implies3A =\sf{\left[\begin{array}{c c c} - 6 & 0 \\ - 3 & - 3\end{array}\right]}\end{lgathered}⟹3A=[−6−30−3]
\begin{lgathered}\sf \implies \: A =\sf{\left[\begin{array}{c c c} \dfrac{ - 6}{3} & & 0 \\ \\ \dfrac{ - 3}{3} & & \dfrac{ - 3}{3} \end{array}\right]}\end{lgathered}⟹A=⎣⎢⎢⎡3−63−303−3⎦⎥⎥⎤
\begin{lgathered}\sf \implies A =\sf{\left[\begin{array}{c c c} - 2 & 0 \\ - 1 & - 1\end{array}\right]}\end{lgathered}⟹A=[−2−10−1]
Answer
\begin{lgathered}\sf \implies A =\sf{\left[\begin{array}{c c c} - 2 & 0 \\ - 1 & - 1\end{array}\right]}\end{lgathered}⟹A=[−2−10−1]