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Consider the following $3 \times 3$ matrices.

$M_{1}=\begin{pmatrix} 0&1&1 \\ 1&0&1 \\ 1&1&0 \end{pmatrix}$

$M_{2}=\begin{pmatrix} 1&0&1 \\ 0&0&0 \\ 1&0&1 \end{pmatrix}$

How may $0-1$ column vectors of the form

$X$= $\begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix}$

are there such that $M_{1}X = M_{2}X$ (modulo $2$)? (modulo $2$ means all operations are done modulo $2$, i.e, $3 = 1$ (modulo $2$), $4 = 0$ (modulo $2$)).

1. None
2. Two
3. Three
4. Four
5. Eight
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0
is the ans D? @Arjun Sir
0
Option b

X'=[0 0 0]

X'=[1 1 1]
0
Did you use trial and error method?
+1
Solve for $(M_1-M_2)X = 2k$. You'll get $x_1 + x_3 = even, x_3 = (1/0), x_2+x_3 = even$. Only $[1\ 1\ 1]^T$ and $[0 \ 0\ 0]^T$ satisfies.

$M_1X =\begin{pmatrix} 0 &1 &1 \\ 1&0 &1 \\ 1&1 &0 \end{pmatrix}$ $\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3} \end{pmatrix}$  =$\begin{pmatrix} x_{2}+x_{3}\\ x_{1}+x_{3}\\ x_{1}+x_{2} \end{pmatrix}$

$M_2X=\begin{pmatrix} 1 &0 &1 \\ 0&0 &0 \\ 1& 0 & 1 \end{pmatrix}$ $\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3} \end{pmatrix}$   = $\begin{pmatrix} x_{1}+x_{3}\\ 0\\ x_{1}+x_{3} \end{pmatrix}$

Given $M_{1}X=M_{2}X \mod 2$

So, by comparing both matrices, we can say

$(x_2+x_3) \mod 2 = (x_1+x_3) \mod 2$

$(x_1+x_3) \mod 2=0$      // so either $x_1=x_3=0$ OR $x_1=x_3=1$

$(x_1+x_2) \mod 2 =(x_1+x_3)\mod 2$

So, we can see when $x_1=x_3=0$ then  $x_{2}=0$ and when $x_1=x_3=1$ then $x_{2}=1$

Eventually there can be two $\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$ and $\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}$ $0-1$ column vectors for $X.$

Correct Answer: $B$
by Boss (14.7k points)
edited
0

X1+X3 = 2
why we are taking them equal

@arvin