A) \[x=-5,\text{ }y=3,\text{ }z=2\]
B) \[x=5,y=-3,z=2\]
C) \[x=5,y=3,\text{ }z=-2\]
D) \[x=5,y=-3,z=-2\]
Correct Answer: B
Solution :
Given, \[\left[ \begin{matrix} 4 & x-z \\ 2+y & xz \\ \end{matrix} \right]=\left[ \begin{matrix} 4 & 3 \\ -1 & 10 \\ \end{matrix} \right]\] On comparing the corresponding elements, we get \[x-z=3\] ?.(i) \[2+y=-1\] \[\Rightarrow \] \[y=-1-2=-3\] ?.(ii) and \[xz=10\] ?.(iii) Now, \[{{(x+z)}^{2}}={{(x-z)}^{2}}+4xz\] \[={{(3)}^{2}}+4\times 10\] \[=9+40\] \[\Rightarrow \] \[{{(x+z)}^{2}}=49\] \[\Rightarrow \] \[x+z=7\] ?.(iv) On adding Eq. (i) and (iv), we get \[2x=10\] \[\Rightarrow \] \[x=5\] From Eq. (i), we get \[5-z=3\] \[\Rightarrow \] \[-z=3-5\] \[\Rightarrow \] \[z=2\] Hence, \[x=5,\,y=-3,z=2\]
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