ISI2017-MMA-29

Question

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 $b$ equals

• A. $1$
• B. $3$
• C. $1/2$
• D. $1/3$

Answer 1

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: $D$

Comments

Rank of matrix is maximum number of linearly independent rows. 

Let rows of the given matrix be $R_1, R_2, R_3$.

Now, since the rank of the matrix is given as $2$, 

and as we can see $R_1$ and $R_2$ are linearly independent, $R_3$ must be a linear combination of $R_1$ or $R_2$. 

So, now check the last element in $R_3$. It's $1$. 

which could come by doing one of the following transformations - 

    1. $R_3 = R_3 - \frac{1}{2}R_1$  ---> Generates $R_3$ as $(\begin{matrix} a & a & b&  1 \end{matrix})$  

    2. $R_3 = R_3 - \frac{1}{3} R_2$ ---> Generates $R_3$ as $(\begin{matrix} a & b & b &  1 \end{matrix})$ 

So, second way is to go. hence option (D).

Note that $a = b$ in the second case above, but that's not relevant to the question.

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Generates $R_3$ as $(\begin{matrix} a & a & b&  1 \end{matrix})$

How did you get this?

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@zeeshanmohnavi, submatrix is informal way of saying minor of a matrix,both are same, it is like captital of India( aka Delhi )

Answer 2

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

Comments

Didn't understand the second step

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Please show the operations performed for row reductions?

Answer 3

$\begin{bmatrix} 1 &1 & 2 & 2\\ 1 & 1 &1 & 3\\ a& b& b& 1 \end{bmatrix}$

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.

Comments

Please correct me if I’m wrong, but is there a typo here:

Its obvious that  b-(x+y) and b-(x+2y) can only be zero together when x is zero

where instead of b-(x+2y) it should be b-(2x+y) according to the matrix.

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you are right @amichopra. thanks for pointing it out, it has been edited.

Answer 4

$\color{blue}{\text{When a=1,b=1}}$
$\begin{pmatrix}1 & 1 &2&2\\1&1 &1&3\\1&1&1&1 \end{pmatrix}$ $R_3\leftarrow R_3-R_1$
= $\begin{pmatrix}1 & 1 &2&2\\1&1 &1&3\\0&0&-1&-1 \end{pmatrix}$ $R_2\leftarrow R_2-R_1$

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

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

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

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

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

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

=$\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$

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

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

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

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

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

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

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

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

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

Comments

why you have taken a & b same??

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To eliminate the rows(or columns), $a$ and $b$ should be equal. Because the $1^{st}$ col. and $2^{nd}$ col of the matrix is the same except $a$ and $b$. If we take both of them same, it would be easy to eliminate the columns(or rows) and will be easy to find the rank.