3 and 4 .Determinants and Matrices
medium

ધારો કે $A$ એ  $3 \times 3$ વાસ્તવિક શ્રેણિક છે કે જેથી  $A \left(\begin{array}{l}1 \\ 1 \\ 0\end{array}\right)=\left(\begin{array}{l}1 \\ 1 \\ 0\end{array}\right) ; A \left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right)=\left(\begin{array}{c}-1 \\ 0 \\ 1\end{array}\right)$ અને $A \left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right)=\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)$. જો $X =\left( x _{1}, x _{2}, x _{3}\right)^{ T }$ અને $I$ એ કક્ષા $3$ વાળો એકમ  શ્રેણિક હોય, તો સંહતિ  $( A -2 I ) X =\left(\begin{array}{l}4 \\ 1 \\ 1\end{array}\right)$ ને .............  

A

એક પણ ઉકેલ નથી

B

આનંત ઉકેલ છે 

C

અનન્ય ઉકેલ છે 

D

બરાબર બે  ઉકેલ છે 

(JEE MAIN-2022)

Solution

$A =\left[\begin{array}{lll} a _{1} & b _{1} & c _{1} \\ a _{2} & b _{2} & c _{2} \\ a _{3} & b _{3} & c _{3}\end{array}\right]$

$A \left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right]=\left[\begin{array}{l} c _{1} \\ c _{2} \\ c _{3}\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$

$\Rightarrow c _{1}=1, c _{2}=1, c _{3}=2$

$A \left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right]=\left[\begin{array}{l} c _{1}+ a _{1} \\ c _{2}+ a _{2} \\ c _{3}+ a _{3}\end{array}\right]=\left[\begin{array}{c}-1 \\ 0 \\ 1\end{array}\right]$

$\Rightarrow a _{1}=-2, a _{2}=-1, a _{3}=-1$

$A \left[\begin{array}{l}1 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{l} a _{1}+ b _{1} \\ a _{2}+ b _{2} \\ a _{3}+ b _{3}\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 0\end{array}\right]$

$\Rightarrow b _{1}=3, b _{2}=2, b _{3}=1$

$\Rightarrow \quad A =\left[\begin{array}{lll}-2 & 3 & 1 \\ -1 & 2 & 1 \\ -1 & 1 & 2\end{array}\right]$

$\Rightarrow A -2 I =\left[\begin{array}{ccc}-4 & 3 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 0\end{array}\right]$

$| A -2 I |=0$

Now, $\left[\begin{array}{lll}-4 & 3 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 0\end{array}\right]\left[\begin{array}{l} x _{1} \\ x _{2} \\ x _{3}\end{array}\right]=\left[\begin{array}{l}4 \\ 1 \\ 1\end{array}\right]$

$-4 x_{1}+3 x_{2}+x_{3}=4 \quad \ldots .$ ($1$)

$- x _{1}+ x _{3}=1 \ldots .$ ($2$)

$- x _{1}+ x _{2}=1 \ldots .$ ($2$)

(1) $-[(2)+3(3)]$

$0=0 \Rightarrow$ infinite solutions

Standard 12
Mathematics

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