Let A be a non zero matrix of order m × n. The Rank of A is defined to be the greatest positive integer r such that A has at least one non-zero minor of order r
For a non-zero m × n matrix A
0 < rank of A ≤ min {m, n}
For a non-zero matrix A of order n,
rank of A < , or = n according as A is singular or non-singular
SOLUTION
GIVEN
[tex]A= \displaystyle\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} [/tex]
TO DETERMINE
The rank of the matrix
[tex]\sf{A + {A}^{2} }[/tex]
CONCEPT TO BE IMPLEMENTED
Let A be a non zero matrix of order m × n. The Rank of A is defined to be the greatest positive integer r such that A has at least one non-zero minor of order r
For a non-zero m × n matrix A
0 < rank of A ≤ min {m, n}
For a non-zero matrix A of order n,
rank of A < , or = n according as A is singular or non-singular
EVALUATION
Here it is given that
[tex]A= \displaystyle\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} [/tex]
So
[tex]{A}^{2} = \displaystyle\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} .\displaystyle\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} = \displaystyle\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} [/tex]
Now
[tex]\sf{A + {A}^{2} }[/tex]
[tex] = \displaystyle\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} + \displaystyle\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} [/tex]
[tex] = \displaystyle\begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix} [/tex]
Now
[tex]\sf{det(A + {A}^{2} ) = 4 – 0 = 4 \ne0}[/tex]
[tex]\sf{ \therefore \: \: A + {A}^{2} \: \: is \:a \: 2 \times 2 \: non \: singular \: matrix }[/tex]
So rank of the matrix [tex]\sf{A + {A}^{2} }[/tex] is 2
━━━━━━━━━━━━━━━━
Learn more from Brainly :-
1. A is a square matrix of order 3 having a row of zeros, then the determinant of A is
https://brainly.in/question/28455441
2. If [1 2 3] B = [34], then the order of the matrix B is
https://brainly.in/question/9030091