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S seema garhwal
      X has of order   .   7X also  has of order  .       Z has of order   .   5Z also  has of order   . Mtarices 7X and 5Z can only be subtracted  if they both have same order  i.e  =    and it is given that  p=n. We can say that both matrices have order of . Thus, order of   is . Option (B) is correct.

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S seema garhwal
P and Y are of order   and  respectivly.   PY will be defined only if k=3, i.e. order of PY is . W and Y are of order   and  respectivly.   WY is  defined because the number of columns of W is equal to the number of rows of Y which is 3, i.e. the order of WY is  Matrices PY and WY can only be added if they both have same order i.e =    implies p=n. Thus, are restrictions on n, k and p so that...

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S seema garhwal
The bookshop has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. The total amount the bookshop will receive from selling all the books:   The total amount the bookshop will receive from selling all the books is 20160.        

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S seema garhwal
Let Rs. x be invested in the first bond. Money invested in second bond = Rs (3000-x) The first bond pays 5% interest per year and the second bond pays 7% interest per year. To obtain an annual total interest of  Rs. 1800,  we have                           (simple interest for 1 year  )   Thus, to obtain an annual total interest of  Rs. 2000, the trust fund should invest Rs 5000 in the...

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S seema garhwal
Let Rs. x be invested in the first bond. Money invested in second bond = Rs (3000-x) The first bond pays 5% interest per year and the second bond pays 7% interest per year. To obtain an annual total interest of  Rs. 1800,  we have                           (simple interest for 1 year  )   Thus, to obtain an annual total interest of  Rs. 1800, the trust fund should invest Rs 15000 in the...

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S seema garhwal
To prove :  L.H.S :    R.H.S :                                                                                                                                                                                                                                                                            Hence, we can see L.H.S = R.H.S i.e. .                

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S seema garhwal
                     We have,                Hence, the value of k is 1.            

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S seema garhwal
First, find the square of matrix A and then multiply it with A to get the cube of matrix A                        L.H.S :                                            Hence, L.H.S = R.H.S    i.e..            

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S seema garhwal
First, we will find ou the value of the square of matrix A                                                

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S seema garhwal
To prove the following multiplication of three by three matrices are not equal                                                                                  Hence,    i.e. .    

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S seema garhwal
To prove:                               Hence, the right-hand side not equal to the left-hand side, that is  .          

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S seema garhwal
To prove :      Hence, we have L.H.S. = R.H.S i.e. .                

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S seema garhwal
If two matrices are equal than corresponding elements are also equal. Thus, we have                                            Put the value of x Hence, we have       

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S seema garhwal
         Adding both the matrix in LHS and rewriting                                Adding equation 1 and 2, we get                      Put the value of x in equation 2, we have                              

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S seema garhwal
Multiplying with constant terms and rearranging we can rewrite the matrix as Dividing by 2 on both sides  

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S seema garhwal
             Now equating LHS and RHS we can write the following equations                                                                                                                                                                                                                                              

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S seema garhwal
Substituting the value of Y in the above equation

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S seema garhwal
        (ii)  and                                    Multiply equation 1 by 3 and equation 2 by 2 and subtract them,                                                          Putting value of Y in equation 1 , we get                                                           

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S seema garhwal
      (i) The given matrices are       and                                  Adding equation 1 and 2, we get                                                Putting the value of X in equation 1, we get                                          

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S seema garhwal
The simplification is explained in the following step     the final answer is an identity matrix of order 2
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