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# 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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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.

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.