# ISI2017-MMA-29

1.1k 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 $b$ equals

1. $1$
2. $3$
3. $1/2$
4. $1/3$

edited

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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can you explain how to arrive to the logic....according to grehwal book if the rank is n then there will be atleat 1 minor of order n which is non zero, it doesnt talk about submatrix.
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My professor told me about submatrix
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Yes it is a concept of submatrix
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I studied the MIT 18.06 lecture series by Gilbert Strang but this property wasn't discussed there. Please share some links where I can study more about this "submatrix being 0" property?
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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.

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
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Didn't understand the second step
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Please show the operations performed for row reductions?
$\color{blue}{\text{When a=1,b=1}}$
$\begin{pmatrix}1 & 1 &2&2\\1&1 &1&3\\1&1&1&1 \end{pmatrix}$ $R_3\leftarrow R_3-R_1$
= $\begin{pmatrix}1 & 1 &2&2\\1&1 &1&3\\0&0&-1&-1 \end{pmatrix}$ $R_2\leftarrow R_2-R_1$

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

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

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

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

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

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

=$\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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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.
1 vote Correct me if I am wrong.

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=$\frac{1}{3}$
$Rank = 2$ 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.

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