# ISI2014-DCG-8

1 vote
217 views

If $M$ is a $3 \times 3$ matrix such that $\begin{bmatrix} 0 & 1 & 2 \end{bmatrix}M=\begin{bmatrix}1 & 0 & 0 \end{bmatrix}$ and $\begin{bmatrix}3 & 4 & 5 \end{bmatrix} M = \begin{bmatrix}0 & 1 & 0 \end{bmatrix}$ then $\begin{bmatrix}6 & 7 & 8 \end{bmatrix}M$ is equal to

1. $\begin{bmatrix}2 & 1 & -2 \end{bmatrix}$
2. $\begin{bmatrix}0 & 0 & 1 \end{bmatrix}$
3. $\begin{bmatrix} -1 & 2 & 0 \end{bmatrix}$
4. $\begin{bmatrix} 9 & 10 & 8 \end{bmatrix}$

recategorized

Here it's given that $\begin{bmatrix} 0 & 1 & 2 \end{bmatrix}M=\begin{bmatrix}1 & 0 & 0 \end{bmatrix} \tag{i}$

and

$\begin{bmatrix} 3 & 4 & 5 \end{bmatrix}M=\begin{bmatrix}0 & 1 & 0 \end{bmatrix} \tag{ii}$

Now $2\times\mathrm{no(ii)-\mathrm{no(i)}}\Rightarrow$

\begin{align} 2\begin{bmatrix} 3 & 4 & 5 \end{bmatrix}M-\begin{bmatrix} 0 & 1 & 2 \end{bmatrix}M &=2\begin{bmatrix}0 & 1 & 0 \end{bmatrix}-\begin{bmatrix}1 & 0 & 0 \end{bmatrix}\\\Rightarrow \left( 2\begin{bmatrix} 3 & 4 & 5 \end{bmatrix}-\begin{bmatrix} 0 & 1 & 2 \end{bmatrix} \right)M &=2\begin{bmatrix}0 & 1 & 0 \end{bmatrix}-\begin{bmatrix}1 & 0 & 0 \end{bmatrix}\\\Rightarrow \left( \begin{bmatrix} 6 & 8 & 10 \end{bmatrix}-\begin{bmatrix} 0 & 1 & 2 \end{bmatrix} \right)M &=\begin{bmatrix}0 & 2 & 0 \end{bmatrix}-\begin{bmatrix}1 & 0 & 0 \end{bmatrix}\\\Rightarrow \begin{bmatrix} (6-0) & (8-1) & (10-2) \end{bmatrix}M &=\begin{bmatrix}(0-1) & (2-0) & (0-0) \end{bmatrix}\\\therefore \begin{bmatrix} 6 & 7 & 8 \end{bmatrix}M &=\begin{bmatrix}-1 & 2 & 0 \end{bmatrix}\end{align}

So the correct answer is C.

1 vote

Check the ans

0
it's correct :)

# c is correct answer .

HINT: just assume M as ' a to i' row wise , setup the equations . you can solve last column of answer easily and the only option that matches with it is c only i.e 0

0
Why not B)?

If I take like  0,1,2 as 0th element, 1st element .... of an array, then is it nt like identity matrix?
0
it is looking like identity will come but it is false.

solve it properly , you will get  C only
0
Can u plz. solve it?

## Related questions

1 vote
1
131 views
Suppose that $A$ is a $3 \times 3$ real matrix such that for each $u=(u_1, u_2, u_3)’ \in \mathbb{R}^3, \: u’Au=0$ where $u’$ stands for the transpose of $u$. Then which one of the following is true? $A’=-A$ $A’=A$ $AA’=I$ None of these
The value of $\lambda$ such that the system of equation $\begin{array}{} 2x & – & y & + & 2z & = & 2 \\ x & – & 2y & + & z & = & -4 \\ x & + & y & + & \lambda z & = & 4 \end{array}$ has no solution is $3$ $1$ $0$ $-3$
For the matrices $A = \begin{pmatrix} a & a \\ 0 & a \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$, $(B^{-1}AB)^3$ is equal to $\begin{pmatrix} a^3 & a^3 \\ 0 & a^3 \end{pmatrix}$ $\begin{pmatrix} a^3 & 3a^3 \\ 0 & a^3 \end{pmatrix}$ $\begin{pmatrix} a^3 & 0 \\ 3a^3 & a^3 \end{pmatrix}$ $\begin{pmatrix} a^3 & 0 \\ -3a^3 & a^3 \end{pmatrix}$
The values of $\eta$ for which the following system of equations $\begin{array} {} x & + & y & + & z & = & 1 \\ x & + & 2y & + & 4z & = & \eta \\ x & + & 4y & + & 10z & = & \eta ^2 \end{array}$ has a solution are $\eta=1, -2$ $\eta=-1, -2$ $\eta=3, -3$ $\eta=1, 2$