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Best answer
23 votes
23 votes
The easiest solution

the smallest square submatrix is of 3*3

and rank =2 is given, so determinant of any submatrix of order 3*3 must be 0

let us take

$\begin{pmatrix} 1 &2 &2 \\ 1&1 &3 \\ b&b &1 \end{pmatrix}$

determinant = 1(1-3b)-2(1-3b)+2(b-b)=0

-1(1-3b)=0

-1+3b=0

b=$\frac{1}{3}$

Correct Answer: $D$
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4 votes
4 votes
try to reduce the matrix in row echolen form $\begin{bmatrix} 1 & 1& 2 &2 \\ 0 & 0 & -1 & 1\\ 0 & b-a & b-2a& 1-2a \end{bmatrix}$

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

since rank is 2 third row must be zero

implies b = a and 1 + b-4a=0

which gives a = b = 1/3
3 votes
3 votes
$\begin{bmatrix} 1 &1 & 2 & 2\\ 1 & 1 &1 & 3\\ a& b& b& 1 \end{bmatrix}$

First observe the matrix, this matrix will have rank 2 meaning only two independent row and columns are there. We use row approach as it is easier for this matrix.

Linear dependent rows means we can write it as combination of other rows, therefore we can write $R_{3}$ as some linear combination of $R_{1}$ and $R_{2}$. this also consequently means that i can make $R_{3}$ zero using row operation

Another hint in question is presence of 1 in $R_{3}$, to get $R_{3}$ as zero we need to make 1 zero, so what can be used to make it zero?

 

Using $R_{1}$ and $R_{2}$, say we subtract  x$R_{1}$+y$R_{2}$from $R_{3}$, we end up with

$\begin{bmatrix} 1 &1 & 2 & 2\\ 1 & 1 &1 & 3\\ a-(x+y)& b-(x+y)& b-(2x+y)& 1-(2x+3y) \end{bmatrix}$.

Its obvious that  b-(x+y) and b-(2x+y) can only be zero together when x is zero, also 1-(2x+3y) is zero when y = 1/3.

subtract $R_{2}$/3 from $R_{3}$, we end up with $\begin{bmatrix} 1 &1 & 2 & 2\\ 1 & 1 &1 & 3\\ a-1/3& b-1/3& b-1/3& 0 \end{bmatrix}$

this gives us a=1/3 and b = 1/3.
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2 votes
2 votes
$\color{blue}{\text{When a=1,b=1}}$
$\begin{pmatrix}1 & 1 &2&2\\1&1 &1&3\\1&1&1&1 \end{pmatrix}$ $R_3\leftarrow R_3-R_1$
= $\begin{pmatrix}1 & 1 &2&2\\1&1 &1&3\\0&0&-1&-1 \end{pmatrix}$ $R_2\leftarrow R_2-R_1$

=$\begin{pmatrix}1 & 1 &2&2\\0&0 &-1&1\\0&0&-1&-1 \end{pmatrix}$ $R_3\leftarrow R_3+R_2$

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

$∴\color{maroon}{\text{Rank of the matrix will be 3}}$

$\color{blue}{\text{When a=3,b=3}}$
$\begin{pmatrix}1 & 1 &2&2\\1&1 &1&3\\3&3&3&1 \end{pmatrix}$ $R_2\leftarrow R_2-R_1$
= $\begin{pmatrix}1 & 1 &2&2\\0&0 &-1&1\\3&3&3&1 \end{pmatrix}$ $R_1\leftarrow 3R_1$

=$\begin{pmatrix}3 & 3 &6&6\\0&0 &-1&1\\3&3&3&1 \end{pmatrix}$ $R_3\leftarrow R_3-R_1$

=$\begin{pmatrix}3 & 3 &6&6\\0&0 &-1&1\\0&0&-3&-5 \end{pmatrix}$ $R_2\leftarrow 3R_2$

=$\begin{pmatrix}3 & 3 &6&6\\0&0 & -3&3\\0&0&-3&-5\end{pmatrix}$$R_3\leftarrow R_3-R_2$

=$\begin{pmatrix}3&3&6&6\\0&0 &-3&3\\0&0&0&-8\end{pmatrix}$

$∴\color{maroon}{\text{Rank of the matrix will be 3}}$

$\color{blue}{\text{When a=1/2 ,b=1/2}}$

$\begin{pmatrix}1 & 1 &2&2\\1&1 &1&3\\1/2&1/2&1/2&1 \end{pmatrix}$ $R_2\leftarrow R_2-R_1$
= $\begin{pmatrix}1 & 1 &2&2\\0&0 &-1&1\\1/2&1/2&1/2&1 \end{pmatrix}$ $R_1\leftarrow \dfrac{1}{2}R_1$

=$\begin{pmatrix}1/2 & 1/2 &1&1\\0&0 &-1&1\\1/2&1/2&1/2&1 \end{pmatrix}$ $R_3\leftarrow R_3-R_1$

=$\begin{pmatrix}1/2 & 1/2 &1&1\\0&0 &-1&1\\0&0&-1/2&0 \end{pmatrix}$ $R_2\leftarrow \dfrac{1}{2}R_2$

=$\begin{pmatrix}1/2 & 1/2 &1&1\\0&0 & -1/2&1/2\\0&0&-1/2&0\end{pmatrix}$$R_3\leftarrow R_3-R_2$

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

$∴\color{maroon}{\text{Rank of the matrix will be 3}}$

$\color{blue}{\text{When a=1/3, b=1/3}}$

$\begin{pmatrix}1 & 1 &2&2\\1&1 &1&3\\1/3&1/3&1/3&1 \end{pmatrix}$ $R_2\leftarrow R_2-R_1$
= $\begin{pmatrix}1 & 1 &2&2\\0&0 &-1&1\\1/3&1/3&1/3&1\end{pmatrix}$ $R_1\leftarrow \dfrac{1}{3}R_1$

=$\begin{pmatrix}1/3 & 1/3 &2/3&2/3\\0&0 &-1&1\\1/3&1/3&1/3&1 \end{pmatrix}$ $R_3\leftarrow R_3-R_1$

=$\begin{pmatrix}1/3 & 1/3 &2/3&2/3\\0&0 &-1&1\\0&0&-1/3&1/3 \end{pmatrix}$ $R_3\leftarrow R_3-\dfrac{1}{3}R_2$

=$\begin{pmatrix}1/3 & 1/3 &2/3&2/3\\0&0 & -1&1\\0&0&0&0\end{pmatrix}$

$∴\color{maroon}{\text{Rank of the matrix will be 2}}$

$∴\color{gold}{ \text{When b =1/3 then only Rank of the matrix will be 2}}$
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