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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 | 540 views

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: $D$
by Boss (18.4k points)
edited
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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?
+1

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.

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

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

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

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

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

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

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

=$\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$

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

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

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

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

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

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

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

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

$∴\color{gold}{ \text{When b =1/3 then only Rank of the matrix will be 2}}$
by Boss (15.4k points)
edited
0
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.
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
by Loyal (5.7k points)
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Didn't understand the second step
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Please show the operations performed for row reductions?
+1 vote Correct me if I am wrong.

by Active (1.5k points)