JEE Main 2022 — Mathematics Algebra

Let the matrix be $A=[\begin{matrix}{a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2} \\ {a}_{3} & {b}_{3} & {c}_{3}\end{matrix}]$A=[a1​a2​a3​​b1​b2​b3​​c1​c2​c3​​]

Now $A[\begin{matrix}0 \\ 0 \\ 1\end{matrix}]=[\begin{matrix}{c}_{1} \\ {c}_{2} \\ {c}_{3}\end{matrix}]=[\begin{matrix}1 \\ 1 \\ 2\end{matrix}]$A[001​]=[c1​c2​c3​​]=[112​]

$$\Rightarrow {c}_{1}=1,{c}_{2}=1,{c}_{3}=2$$⇒c1​=1,c2​=1,c3​=2

$$A[\begin{matrix}1 \\ 0 \\ 1\end{matrix}]=[\begin{matrix}{c}_{1}+{a}_{1} \\ {c}_{2}+{a}_{2} \\ {c}_{3}+{a}_{3}\end{matrix}]=[\begin{matrix}-1 \\ 0 \\ 1\end{matrix}]$$A[101​]=[c1​+a1​c2​+a2​c3​+a3​​]=[−101​]

$$\Rightarrow {a}_{1}=-2,{a}_{2}=-1,{a}_{3}=-1$$⇒a1​=−2,a2​=−1,a3​=−1

$$A[\begin{matrix}1 \\ 1 \\ 0\end{matrix}]=[\begin{matrix}{a}_{1}+{b}_{1} \\ {a}_{2}+{b}_{2} \\ {a}_{3}+{b}_{3}\end{matrix}]=[\begin{matrix}1 \\ 1 \\ 0\end{matrix}]$$A[110​]=[a1​+b1​a2​+b2​a3​+b3​​]=[110​]

$$\Rightarrow {b}_{1}=3,{b}_{2}=2,{b}_{3}=1$$⇒b1​=3,b2​=2,b3​=1

So matrix $A=[\begin{matrix}-2 & 3 & 1 \\ -1 & 2 & 1 \\ -1 & 1 & 2\end{matrix}]$A=[−2−1−1​321​112​]

$$\Rightarrow A-2I=[\begin{matrix}-4 & 3 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 0\end{matrix}]$$⇒A−2I=[−4−1−1​301​110​]

i.e. $|A-2I|=0$∣A−2I∣=0

Now $X=[\begin{matrix}{x}_{1} \\ {x}_{2} \\ {x}_{3}\end{matrix}]$X=[x1​x2​x3​​]

So, $[\begin{matrix}-4 & 3 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 0\end{matrix}][\begin{matrix}{x}_{1} \\ {x}_{2} \\ {x}_{3}\end{matrix}]=[\begin{matrix}4 \\ 1 \\ 1\end{matrix}]$[−4−1−1​301​110​][x1​x2​x3​​]=[411​]

$$-4{x}_{1}+3{x}_{2}+{x}_{3}=4...(1)$$−4x1​+3x2​+x3​=4...(1)

$$-{x}_{1}+{x}_{3}=1...(2)$$−x1​+x3​=1...(2)

$$-{x}_{1}+{x}_{2}=1...(3)$$−x1​+x2​=1...(3)

Solving the above system of equations we get infinite solutions.