
If I is a (9 x 9) unit matrix, then rank (I) =
(a) 0
(b) 3
(c) 6
(d) 9
Answer
563.1k+ views
Hint: To solve this question, we will, first of all, define a unit matrix. Then we see which order of the matrix unit is taken in our question. The rank of a matrix is the maximum number of linearly independent rows in a matrix. So, observing the definition of the order of the matrix and rank of the matrix, we will get our answer.
Complete step-by-step solution
The unit matrix is every \[n\times n\] square matrix made up of all zeroes except for the elements of the main diagonal that are all ones. So, basically, a \[2\times 2\] order unit matrix is given by
\[\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]\]
And similarly, \[3\times 3\] the order unit matrix is given by
\[\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right]\]
So an \[n\times n\] order unit matrix is given by
\[\left[ \begin{matrix}
1 & 0 & ..... & 0 \\
0 & 1 & .... & 0 \\
0 & ..... & 1 & 0 \\
0 & 0 & ..... & 1 \\
\end{matrix} \right]\]
If I is a \[9\times 9\] unit matrix, then I is of the form
\[I=\left[ \begin{matrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{matrix} \right]_{9\times 9}^{{}}\]
Let us define the Rank of a matrix. The Rank of a matrix is defined as
(i) The maximum number of linearly independent column vectors in the matrix or
(ii) The maximum number of linearly independent rows of vectors in the matrix.
We see that that matrix is given as I has all lows linearly independent rows. Therefore, the number of linearly independent rows of the matrix I = 9 = order of the matrix.
\[\Rightarrow Rank\left( I \right)=9\]
Hence, option (d) is the right answer.
Note: There is no difference between a unit matrix and an identity matrix. In fact, the unit matrix is another name of identity. It is named as the unit matrix because the determinant of a unit or identity matrix is given by “I”.
Complete step-by-step solution
The unit matrix is every \[n\times n\] square matrix made up of all zeroes except for the elements of the main diagonal that are all ones. So, basically, a \[2\times 2\] order unit matrix is given by
\[\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]\]
And similarly, \[3\times 3\] the order unit matrix is given by
\[\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right]\]
So an \[n\times n\] order unit matrix is given by
\[\left[ \begin{matrix}
1 & 0 & ..... & 0 \\
0 & 1 & .... & 0 \\
0 & ..... & 1 & 0 \\
0 & 0 & ..... & 1 \\
\end{matrix} \right]\]
If I is a \[9\times 9\] unit matrix, then I is of the form
\[I=\left[ \begin{matrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{matrix} \right]_{9\times 9}^{{}}\]
Let us define the Rank of a matrix. The Rank of a matrix is defined as
(i) The maximum number of linearly independent column vectors in the matrix or
(ii) The maximum number of linearly independent rows of vectors in the matrix.
We see that that matrix is given as I has all lows linearly independent rows. Therefore, the number of linearly independent rows of the matrix I = 9 = order of the matrix.
\[\Rightarrow Rank\left( I \right)=9\]
Hence, option (d) is the right answer.
Note: There is no difference between a unit matrix and an identity matrix. In fact, the unit matrix is another name of identity. It is named as the unit matrix because the determinant of a unit or identity matrix is given by “I”.
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