Question

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 \\
 $

Answer 1

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.