Question

Find the rank of matrix: \[{\text{A = }}\left[ {{\text{ }}\begin{array}{*{20}{c}}
  1&1&1 \\
  1&1&1 \\
  1&1&1
\end{array}} \right]\]

Answer 1

Hint: Proceed the solution by applying elementary row operation and find row echelon form, through which we can get minimum non zero rows. And as we know that Rank is equal to the number of non-zero rows, hence we can reach our answer.

Complete step-by-step answer:

We know that,
The rank of a matrix is the maximum number of linearly independent column vectors in the matrix or the maximum number of linearly independent row vectors in the matrix. Both definitions are equivalent.

Linearly independent row vectors can be found by elementary row operation

\[{\text{A = }}\left[ {{\text{ }}\begin{array}{*{20}{c}}
  1&1&1 \\
  1&1&1 \\
  1&1&1
\end{array}} \right]\]

On applying R2​→R2​−R1​ ; (on subtracting row 1 from row 2 we get)

\[{\text{A = }}\left[ {{\text{ }}\begin{array}{*{20}{c}}
  1&1&1 \\
  0&0&0 \\
  1&1&1
\end{array}} \right]\]
Here row2 has become completely zero, it shows that, row2 is depending on row 1 hence row vector2 is not linearly independent row vectors

Now On applying R3​→R3​−R1​ ; (on subtracting row 1 from row 3 we get)

\[{\text{A = }}\left[ {{\text{ }}\begin{array}{*{20}{c}}
  1&1&1 \\
  0&0&0 \\
  0&0&0
\end{array}} \right]\]

Here row3 has become completely zero, it shows that, row3 is depending on row 1 hence row vector3 is not linearly independent row vectors

Now in matrix A, only row 1 becomes non-zero and any row operation can’t make it completely zero. Hence we can say that row vector 1 is linearly independent of row vectors. This thing can also be said there is only one non-zero row remaining.
So Rank = number of non-zero rows =1

Note: The rank of a matrix can also be calculated using determinants. The rank of a matrix is the order of the largest non-zero square submatrix. In this particular question, \[\left| {\text{A}} \right|{\text{ = }}\left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&1&1 \\
  1&1&1
\end{array}} \right|{\text{ = 0}}\]
Every Submatrix of order 2, \[\left| {{{\text{A}}_{{\text{sub}}}}} \right|{\text{ = }}\left| {\begin{array}{*{20}{c}}
  1&1 \\
  1&1
\end{array}} \right|{\text{ = 0}}\] will also be zero

Submatrix of order 1, \[\left| {{{\text{A}}_{{\text{sub}}}}} \right|{\text{ = }}\left| 1 \right| \ne {\text{0}}\]
Order of the largest non-zero square submatrix=1
Hence rank =1.