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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 ((1,( b),( 0)),(2,(−3),( 0)),(a,(−1),( 2)) ) ((i),(j),(k) )= ((( 4)),((−c)),(( 0)) ) ((i),(j),(k) )= ((1,( b),0),(2,(−3),0),(a,(−1),2) )^(−1) ((( 4)),((−c)),(( 0)) ) ((i),(j),(k) )=( ((( 4)),((−c)),(( 0)) )/ ((1,( b),0),(2,(−3),0),(a,(−1),2) )^1 )=(( ((1,0,0),(0,1,0),(0,0,1) ) ((( 4)),((−c)),(( 0)) ))/ ((1,b,0),(2,(−3),0),(a,(−1),2) )^1 ) ((i),(j),(k) )= ((1,( 0),1),(0,(−(1/3)),1),(0,( 0),(1/2)) ) ((( 4)),((−c)),(( 0)) )= ((4),((1/3)),(0) ) 4+(1/3)b=−4 b=−24 8−1=−c c=−7 4a−(1/3)=14 → 12a=43 a=((43)/(12))

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

Commented by Tawa11 last updated on 06/Mar/25

Thanks sir.

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

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