# If the given matrix $A = \left[ {\begin{array}{*{20}{c}}

1&3&1 \\

2&1&{ - 1} \\

3&0&1

\end{array}} \right]$, then rank(A) is equal to

$

\left( a \right)4 \\

\left( b \right)1 \\

\left( c \right)2 \\

\left( d \right)3 \\

$

Last updated date: 20th Mar 2023

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Answer

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Hint: In this question, the rank of the matrix is equal to the number of non-zero rows in the matrix after reducing it to the echelon form. In echelon form we only apply row operation.

Complete step-by-step answer:

Given, $A = \left[ {\begin{array}{*{20}{c}}

1&3&1 \\

2&1&{ - 1} \\

3&0&1

\end{array}} \right]$

Now, we have to convert the above matrix into echelon form. Echelon forms the same upper triangular matrix. In echelon form we only apply row operation.

$A = \left[ {\begin{array}{*{20}{c}}

1&3&1 \\

2&1&{ - 1} \\

3&0&1

\end{array}} \right]$

Apply row operation, ${R_2} \to {R_2} - 2{R_1}$

$A = \left[ {\begin{array}{*{20}{c}}

1&3&1 \\

0&{ - 5}&{ - 3} \\

3&0&1

\end{array}} \right]$

Now apply row operation, ${R_3} \to {R_3} - 3{R_1}$

$A = \left[ {\begin{array}{*{20}{c}}

1&3&1 \\

0&{ - 5}&{ - 3} \\

0&{ - 9}&{ - 2}

\end{array}} \right]$

Again, apply row operation, ${R_3} \to {R_3} - \dfrac{{9{R_2}}}{5}$

$A = \left[ {\begin{array}{*{20}{c}}

1&3&1 \\

0&{ - 5}&{ - 3} \\

0&0&{\dfrac{{17}}{5}}

\end{array}} \right]$

We can see the above matrix is an upper triangular matrix. Now it is converted into echelon form so the rank of the matrix is equal to the number of non-zero rows.

$A = \left[ {\begin{array}{*{20}{c}}

1&3&1 \\

0&{ - 5}&{ - 3} \\

0&0&{\dfrac{{17}}{5}}

\end{array}} \right]$

In this matrix there are no non zero rows so the rank of this matrix is 3.

Hence, Rank(A)=3

So, the correct option is (d).

Note: Whenever we face such types of problems we use some important points. First we convert matrix into echelon form by using some row operations then observe how many non- zero rows in echelon form matrix and we know the number of non-zero rows in echelon form is equal to rank of matrix.

Complete step-by-step answer:

Given, $A = \left[ {\begin{array}{*{20}{c}}

1&3&1 \\

2&1&{ - 1} \\

3&0&1

\end{array}} \right]$

Now, we have to convert the above matrix into echelon form. Echelon forms the same upper triangular matrix. In echelon form we only apply row operation.

$A = \left[ {\begin{array}{*{20}{c}}

1&3&1 \\

2&1&{ - 1} \\

3&0&1

\end{array}} \right]$

Apply row operation, ${R_2} \to {R_2} - 2{R_1}$

$A = \left[ {\begin{array}{*{20}{c}}

1&3&1 \\

0&{ - 5}&{ - 3} \\

3&0&1

\end{array}} \right]$

Now apply row operation, ${R_3} \to {R_3} - 3{R_1}$

$A = \left[ {\begin{array}{*{20}{c}}

1&3&1 \\

0&{ - 5}&{ - 3} \\

0&{ - 9}&{ - 2}

\end{array}} \right]$

Again, apply row operation, ${R_3} \to {R_3} - \dfrac{{9{R_2}}}{5}$

$A = \left[ {\begin{array}{*{20}{c}}

1&3&1 \\

0&{ - 5}&{ - 3} \\

0&0&{\dfrac{{17}}{5}}

\end{array}} \right]$

We can see the above matrix is an upper triangular matrix. Now it is converted into echelon form so the rank of the matrix is equal to the number of non-zero rows.

$A = \left[ {\begin{array}{*{20}{c}}

1&3&1 \\

0&{ - 5}&{ - 3} \\

0&0&{\dfrac{{17}}{5}}

\end{array}} \right]$

In this matrix there are no non zero rows so the rank of this matrix is 3.

Hence, Rank(A)=3

So, the correct option is (d).

Note: Whenever we face such types of problems we use some important points. First we convert matrix into echelon form by using some row operations then observe how many non- zero rows in echelon form matrix and we know the number of non-zero rows in echelon form is equal to rank of matrix.

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