edited by
2,664 views

8 Answers

2 votes
2 votes
$Rank = 2$ means two $non-zero$ rows. Therefore, the last row must be all $zero$. Now, let $A, B, C$ be first, second, and third row respectively. To get row $C$ all zero, the operation we need to perform is $C = 3C - B$. Therefore if $a=b=\frac{1}{3}$, the operation will cause row $C$  to be all zero. Hence option $D$ is correct.
0 votes
0 votes
It's a elementary question of matrix. There are numerous ways to solve this one but I like the below approach most

 

   $\begin{pmatrix} 1 &1 &2 &2 \\ 1& 1 &1 &3 \\ a& b & b & 1 \end{pmatrix}$ $C_{2}'=C_{2}-C_{1}$ , $C_{3}'=C_{3}-2*C_{1}$ , $C_{4}'=C_{3}-2*C_{1}$

= $\begin{pmatrix} 1 &0  &0  &0 \\  1& 0 &-1  &1 \\  a&  b-a & b-2a  & 1-2a  \end{pmatrix}$ $R_{2}'=R_{2}-R_{1}$ , $R_{3}'=R_{3}-a*R_{1}$

= $\begin{pmatrix} 1 &0  &0  &0 \\  0& 0 &-1  &1 \\  0&  b-a & b-2a  & 1-2a  \end{pmatrix}$ $C_{3}'=C_{3}+C_{2}$

= $\begin{pmatrix} 1 &0  &0  &0 \\  0& 0 &0  &1 \\  0&  b-a & 1+b-4a  & 1-2a  \end{pmatrix}$ Now interchange $C_{4}$ and $C_{2}$

=$\begin{pmatrix} 1 &0  &0  &0 \\  0& 1 &0  &0 \\  0&  1-2a & 1+b-4a  & b-a  \end{pmatrix}$ $R_{3}'=R_{3}-(1-2a)*R_{2}$

=$\begin{pmatrix} 1 &0  &0  &0 \\  0& 1 &0  &0 \\  0&  0 & 1+b-4a  & b-a  \end{pmatrix}$

 

Now as it's mentioned that rank of the matrix is 2, all elements of the third row must be zero

So, 1+b-4a=0 & b-a=0 $\Rightarrow$ a=b=$\frac{1}{3}$
0 votes
0 votes

Dimension of the Matrix is 3 x 4, Let’s consider m = 3 and n = 4

As, Rank of the Matrix = 2 which is less than m and less than n i.e Rank < m & Rank < n. So we can say the last row will be a zero row in Echelon Form.

$\begin{pmatrix}1&1&2&2\\1&1&1&3\\a&b&b&1\end{pmatrix},R_{2}=R_{2}-R_{1}\longrightarrow \begin{pmatrix}1&1&2&2\\0&0&-1&1\\a&b&b&1\end{pmatrix},R_{3}=(R_{3}/b)-1\longrightarrow \begin{pmatrix}1&1&2&2\\0&0&-1&1\\(a/b)-1&0&-1&(1/b)-2\end{pmatrix},R_{3}=R_{3}-R_{2}\longrightarrow \begin{pmatrix}1&1&2&2\\0&0&-1&1\\(a/b)-1&0&0&(1/b)-3\end{pmatrix}$

Now as 3rd row should be a zero row.

Therefore :

$\\(a/b) = 1 \\ \Rightarrow a=b$   ,and $\\(1/b)-3=0\\\Rightarrow b=1/3$

So, Option D is correct.

Answer:

Related questions

0 votes
0 votes
0 answers
1
go_editor asked Sep 15, 2018
1,095 views
Suppose the rank of the matrix $\begin{pmatrix} 1 & 1 & 2 & 2 \\ 1 & 1 & 1 & 3 \\ a & b & b & 1 \end{pmatrix}$ is 2 for some real numbers $a$ and $b$. Then the $b$ equals...
8 votes
8 votes
3 answers
2
jjayantamahata asked Mar 26, 2018
1,569 views
If $A$ is a $2 \times 2$ matrix such that trace $A = det \ A = 3,$ then what is the trace of $A^{-1}$?$1$$\left(\dfrac{1}{3}\right)$$\left(\dfrac{1}{6}\right)$$\left(\dfr...
0 votes
0 votes
0 answers
3
go_editor asked Sep 15, 2018
596 views
If $A$ is a $2 \times 2$ matrix such that $trace \: A = det \: A =3$, then what is the trace of $A^{-1}$?$1$$1/3$$1/6$$1/2$
0 votes
0 votes
0 answers
4
go_editor asked Sep 15, 2018
364 views
The diagonal elements of a square matrix $M$ are odd integers while the off-diagonals are even integers. Then$M$ must be singular$M$ must be nonsingularthere is not enoug...