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

Question

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

in progress 0
Jade 5 months 2021-07-03T07:15:19+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-03T07:16:32+00:00

    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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