Question

m 1
Let A = 0 0 1 then the rank of the matrix A+ A2 is
0 0​

Answer 1

SOLUTION

GIVEN

$A= \displaystyle\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$

TO DETERMINE

The rank of the matrix

$\sf{A + {A}^{2} }$

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

$A= \displaystyle\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$

So

${A}^{2} = \displaystyle\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} .\displaystyle\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} = \displaystyle\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$

Now

$\sf{A + {A}^{2} }$

$= \displaystyle\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} + \displaystyle\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$

$= \displaystyle\begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}$

Now

$\sf{det(A + {A}^{2} ) = 4 - 0 = 4 \ne0}$

$\sf{ \therefore \: \: A + {A}^{2} \: \: is \:a \: 2 \times 2 \: non \: singular \: matrix }$

So rank of the matrix $\sf{A + {A}^{2} }$ is 2

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