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Question Number 217164 by Tawa11 last updated on 03/Mar/25

Answered by issac last updated on 03/Mar/25

$$xi+2yi+azi+bxj-3yj-zj+4xk+cky+2kz$$$$(i+bj+4)x+(2i-3j+c)y+(ai-j+2k)z$$$$equal=0$$$$solve$$$$i+bj=-4$$$$2i-3j=-c$$$$ai-j=-2k$$$$\left(\begin{array}{ccc}1& b& 0\\ 2& -3& 0\\ a& -1& 2\end{array}\right)\left(\begin{array}{c}i\\ j\\ k\end{array}\right)=\left(\begin{array}{c}4\\ -c\\ 0\end{array}\right)$$$$\left(\begin{array}{c}i\\ j\\ k\end{array}\right)={\left(\begin{array}{ccc}1& b& 0\\ 2& -3& 0\\ a& -1& 2\end{array}\right)}^{-1}\left(\begin{array}{c}4\\ -c\\ 0\end{array}\right)$$$$\left(\begin{array}{c}i\\ j\\ k\end{array}\right)=\frac{\left(\begin{array}{c}4\\ -c\\ 0\end{array}\right)}{{\left(\begin{array}{ccc}1& b& 0\\ 2& -3& 0\\ a& -1& 2\end{array}\right)}^1}=\frac{\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{c}4\\ -\mathrm{c}\\ 0\end{array}\right)}{{\left(\begin{array}{ccc}1& b& 0\\ 2& -3& 0\\ a& -1& 2\end{array}\right)}^1}$$$$\left(\begin{array}{c}i\\ j\\ k\end{array}\right)=\left(\begin{array}{ccc}1& 0& 1\\ 0& -\frac{1}{3}& 1\\ 0& 0& \frac{1}{2}\end{array}\right)\left(\begin{array}{c}4\\ -c\\ 0\end{array}\right)=\left(\begin{array}{c}4\\ \frac{1}{3}\\ 0\end{array}\right)$$$$4+\frac{1}{3}b=-4$$$$b=-24$$$$8-1=-c$$$$c=-7$$$$4a-\frac{1}{3}=14\to 12\mathrm{a}=43$$$$a=\frac{43}{12}$$

Commented by Tawa11 last updated on 06/Mar/25

$$\mathrm{Thanks}\mathrm{sir}.$$

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