GATE2017 EC

Question

The rank of the matrix

$\begin{bmatrix} 1 & -1 & 0 &0 & 0\\ 0 & 0 & 1 &-1 &0 \\ 0 &1 &-1 &0 &0 \\ -1 & 0 &0 & 0 &1 \\ 0&0 & 0 & 1 & -1 \end{bmatrix}$

is ________.

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Ans 5?

Comments

Yes because determinant value changes..

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@Verma Ashish

What u mean by "determinant value changes."??

Do u mean less than $4$ determinant not possible??

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Rank is the largest matrix or sub-matrix for which determinant is not 0.

Here the largest submatrix is of size 4*4 whose deteminant is not 0. So rank =4.

Answer 1

$\begin{bmatrix} 1 & -1 & 0 &0 & 0\\ 0 & 0 & 1 &-1 &0 \\ 0 &1 &-1 &0 &0 \\ -1 & 0 &0 & 0 &1 \\ 0&0 & 0 & 1 & -1 \end{bmatrix}$

Applying $C_{1}\leftarrow C_{1}+C_{2}+C_{3}+C_{4}+C_{5}$

$\begin{bmatrix} {\color{Red} 0} & -1 & 0 &0 & 0\\ {\color{Red} 0} & 0 & 1 &-1 &0 \\ {\color{Red} 0} &1 &-1 &0 &0 \\ {\color{Red} 0} & 0 &0 & 0 &1 \\ {\color{Red} 0}&0 & 0 & 1 & -1 \end{bmatrix}$

Applying $R_{1}\leftarrow R_{1}+R_{2}+R_{3}+R_{4}+R_{5}$

$\begin{bmatrix} {\color{Red}0} & {\color{Red} 0}& {\color{Red} 0}&{\color{Red} 0}& {\color{Red} 0}\\ 0 & 0 & 1 &-1 &0 \\ 0 &1 &-1 &0 &0 \\ 0 & 0 &0 & 0 &1 \\ 0&0 & 0 & 1 & -1 \end{bmatrix}$

Applying $C_{3}\leftarrow C_{3}+C_{2}$

$\begin{bmatrix} 0 & 0 & {\color{Red} 0} &0 & 0\\ 0 & 0 & {\color{Red} 1} &-1 &0 \\ 0 &1 &{\color{Red} 0} &0 &0 \\ 0 & 0 &{\color{Red} 0} & 0 &1 \\ 0&0 & {\color{Red} 0} & 1 & -1 \end{bmatrix}$

Applying $C_{4}\leftarrow C_{4}+C_{3}$

$\begin{bmatrix} 0 & 0 & 0 &{\color{Red} 0} & 0\\ 0 & 0 & 1 &{\color{Red} 0} &0 \\ 0 &1 &0 &{\color{Red} 0} &0 \\ 0 & 0 &0 & {\color{Red} 0} &1 \\ 0&0 & 0 & {\color{Red} 1} & -1 \end{bmatrix}$

Applying $C_{5}\leftarrow C_{5}+C_{4}$

$\begin{bmatrix} 0 & 0 & 0 &0 & {\color{Red} 0}\\ 0 & 0 & 1 &0 &{\color{Red} 0} \\ 0 &1 &0 &0 &{\color{Red} 0} \\ 0 & 0 &0 & 0 &{\color{Red} 1} \\ 0&0 & 0 & 1 & {\color{Red} 0} \end{bmatrix}$

$\because$ there are $4$ Linearly independent rows

$\therefore$ the rank of the given matrix is $4$

Comments

what r u doing?? Plz rechk.

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I tried to eliminate all -1.

please correct me if i am wrong.

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Chk ur 2nd matrix

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what is wrong in it ?

Answer 2

After Row-Reduce you will get :

\begin{pmatrix}
 1&-1&0&0&0 &\\0&0&1&-1&0 &\\0&1&-1&0&0 &\\-1&0&0&0&1 &\\0&0&0&1&-1
\end{pmatrix}
Applying $R_4\longleftarrow R_4+R_1$

\begin{pmatrix}
1&-1&0&0&0 &\\0&0&1&-1&0 &\\0&1&-1&0&0 &\\0&-1&0&0&1 &\\0&0&0&1&-1
\end{pmatrix}

Interchange $R_2\longleftrightarrow R_3$

\begin{pmatrix}
1&-1&0&0&0 &\\0&1&-1&0&0 &\\0&0&1&-1&0 &\\0&-1&0&0&1 &\\0&0&0&1&-1
\end{pmatrix}

Applying $R_4\longleftarrow R_4+R_2$

\begin{pmatrix}
 1&-1&0&0&0 &\\0&1&-1&0&0 &\\0&0&1&-1&0 &\\0&0&-1&0&1 &\\0&0&0&1&-1
\end{pmatrix}

Applying $R_4\longleftarrow R_4+R_3$

\begin{pmatrix}
1&-1&0&0&0 &\\0&1&-1&0&0 &\\0&0&1&-1&0 &\\0&0&0&-1&1 &\\0&0&0&1&-1
\end{pmatrix}

Applying $R_4\longleftarrow -(R_4)$

\begin{pmatrix}
1&-1&0&0&0 &\\0&1&-1&0&0 &\\0&0&1&-1&0 &\\0&0&0&1&-1 &\\0&0&0&1&-1
\end{pmatrix}

Applying $R_5\longleftarrow R_5-R_4$

\begin{pmatrix}
1&-1&0&0&0 &\\0&1&-1&0&0 &\\0&0&1&-1&0 &\\0&0&0&1&-1 &\\0&0&0&0&0
\end{pmatrix}

Applying $R_3\longleftarrow R_3+R_4$

\begin{pmatrix}
1&-1&0&0&0 &\\0&1&-1&0&0 &\\0&0&1&0&-1 &\\0&0&0&1&-1 &\\0&0&0&0&0
\end{pmatrix}

Applying $R_2\longleftarrow R_2+R_3$

\begin{pmatrix}
1&-1&0&0&0 &\\0&1&0&0&-1 &\\0&0&1&0&-1 &\\0&0&0&1&-1 &\\0&0&0&0&0
\end{pmatrix}

Applying $R_1\longleftarrow R_1+R_2$

\begin{pmatrix}
1&0&0&0&-1 &\\0&1&0&0&-1 &\\0&0&1&0&-1 &\\0&0&0&1&-1 &\\0&0&0&0&0
\end{pmatrix}
The total leading coefficient in the reduced row echelon form is $4$. Hence Rank$=4$

Comments

please show the steps for converting it in echleon form.

Answer 3

Convert matrix in the Echelon form and the number of non-zero rows is the answer. Hence the answer is 4.

Answer 4

$\begin{bmatrix} {\color{Red} 1} & -1& 0& 0& 0\\ 0& 0& 1& -1& 0\\ 0& 1& -1& 0& 0\\ -1& 0& 0& 0& 1\\ 0& 0& 0& 1& -1 \end{bmatrix} \xrightarrow[]{R4 \ \leftarrow \ R4 \ + \ R1} \begin{bmatrix} 1 & -1& 0& 0& 0\\ 0& 0& 1& -1& 0\\ 0& 1& -1& 0& 0\\ 0& -1& 0& 0& 1\\ 0& 0& 0& 1& -1 \end{bmatrix} \xrightarrow[]{R2\ \Leftrightarrow \ R3} \begin{bmatrix} {\color{Red} 1} & -1& 0& 0& 0\\ 0& {\color{Red} 1}& -1& 0& 0\\ 0& 0& 1& -1& 0\\ 0& -1& 0& 0& 1\\ 0& 0& 0& 1& -1 \end{bmatrix} \xrightarrow[]{R4 \ \leftarrow \ R4 \ + \ R2} \begin{bmatrix} {\color{Red} 1} & -1& 0& 0& 0\\ 0& {\color{Red} 1}& -1& 0& 0\\ 0& 0& {\color{Red} 1}& -1& 0\\ 0& 0& -1& 0& 1\\ 0& 0& 0& 1& -1 \end{bmatrix} \xrightarrow[]{R4 \ \leftarrow \ R4 \ + \ R3} \begin{bmatrix} {\color{Red} 1} & -1& 0& 0& 0\\ 0& {\color{Red} 1}& -1& 0& 0\\ 0& 0& {\color{Red} 1}& -1& 0\\ 0& 0& 0& {\color{Red} {-1}}& 1\\ 0& 0& 0& 1& -1 \end{bmatrix} \xrightarrow[]{R5 \ \leftarrow \ R5 \ + \ R4} \begin{bmatrix} {\color{Red} 1} & -1& 0& 0& 0\\ 0& {\color{Red} 1}& -1& 0& 0\\ 0& 0& {\color{Red} 1}& -1& 0\\ 0& 0& 0& {\color{Red} {-1}}& 1\\ 0& 0& 0& 0& 0 \end{bmatrix}$

 

Here, we can see that there are 4 columns with pivot elements (red colored elements). Therefore, the rank of the matrix = 4.