A) 0
B) 679
C) 779
D) 1000
Correct Answer: A
Solution :
\[\left| \,\begin{matrix} 265 & 240 & 219 \\ 240 & 225 & 198 \\ 219 & 198 & 181 \\ \end{matrix}\, \right|=\left| \,\begin{matrix} 25 & 21 & 219 \\ 15 & 27 & 198 \\ 21 & 17 & 181 \\ \end{matrix}\, \right|\] {Applying \[{{C}_{1}}\to {{C}_{1}}-{{C}_{2}};\,{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}\}\] =\[\left| \,\begin{matrix} 4 & 21 & 9 \\ -12 & 27 & -72 \\ 4 & 17 & 11 \\ \end{matrix}\, \right|\] {Applying \[{{C}_{1}}\to {{C}_{1}}-{{C}_{2}};\,{{C}_{3}}\to {{C}_{3}}-10{{C}_{2}}\}\] = \[\,\left| \,\begin{matrix} 4 & 21 & 9 \\ -12 & 27 & -72 \\ 0 & -4 & 2 \\ \end{matrix}\, \right|\] {Applying \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\]} =\[4\,\left| \,\begin{matrix} 1 & 21 & 9 \\ -3 & 27 & -72 \\ 0 & -4 & 2 \\ \end{matrix}\, \right|=4\,\left| \,\begin{matrix} 1 & 21 & 9 \\ 0 & 90 & -45 \\ 0 & -4 & 2 \\ \end{matrix}\, \right|\]by \[{{R}_{2}}\to 3{{R}_{1}}+{{R}_{2}}\] = \[2X=\left[ \begin{matrix} 3 & 2 \\ 0 & -2 \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 2 \\ 7 & 4 \\ \end{matrix} \right]\,\Rightarrow \,2X=\left[ \begin{matrix} 4 & 4 \\ 7 & 2 \\ \end{matrix} \right]\]= 0.
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