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Math Help - matrice problem (Urgent! ASAP please)

  1. #1
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    Question matrice problem (Urgent! ASAP please)

    Hello ppl. I need to solve these 2 matrices. please help me solving these and explain for me the solution. Thanks in advance
    Attached Thumbnails Attached Thumbnails matrice problem (Urgent! ASAP please)-image033.jpg   matrice problem (Urgent! ASAP please)-image035.jpg  
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  2. #2
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    1) "matrices" is the plural form of the singular word "matrix".
    2) What is to be done? Are they determinants?
    3) Have you considered expansion by minors or row reductions under which the value of the determinant is unchanged?
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    1) "matrices" is the plural form of the singular word "matrix".
    2) What is to be done? Are they determinants?
    3) Have you considered expansion by minors or row reductions under which the value of the determinant is unchanged?
    1) ok! Im sorry and thank you for correcting my mistake.

    2) the question says : Find the determinan of this matrix

    3) i don't understand what do you mean?
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  4. #4
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    Quote Originally Posted by Narek View Post
    1) ok! Im sorry and thank you for correcting my mistake.

    2) the question says : Find the determinan of this matrix

    3) i don't understand what do you mean?
    For the second,

     = \left| {\begin{array}{*{20}c}<br />
   1 & 2 & {2^2 } & {2^3 }  \\<br />
   1 & 3 & {3^2 } & {3^3 }  \\<br />
   1 & 5 & {5^2 } & {5^3 }  \\<br />
   1 & 6 & {6^2 } & {6^3 }  \\<br /> <br />
 \end{array} } \right| \hfill \\

     {\text{Apply  }}R_2  \to R_2  - R_1 ,{\text{  }}R_3  \to R_3  - R_1 ,{\text{  }}R_4  \to R_4  - R_1  \hfill \\

      = \left| {\begin{array}{*{20}c}<br />
   1 & 2 & {2^2 } & {2^3 }  \\<br />
   0 & {3 - 2} & {3^2  - 2^2 } & {3^3  - 2^3 }  \\<br />
   0 & {5 - 2} & {5^2  - 2^2 } & {5^3  - 2^3 }  \\<br />
   0 & {6 - 2} & {6^2  - 2^2 } & {6^3  - 2^3 }  \\<br /> <br />
 \end{array} } \right| \hfill \\

     = \left| {\begin{array}{*{20}c}<br />
   {3 - 2} & {3^2  - 2^2 } & {3^3  - 2^3 }  \\<br />
   {5 - 2} & {5^2  - 2^2 } & {5^3  - 2^3 }  \\<br />
   {6 - 2} & {6^2  - 2^2 } & {6^3  - 2^3 }  \\<br /> <br />
 \end{array} } \right| \hfill \\

     {\text{Take common }}\left( {3 - 2} \right){\text{ from }}R_1 {\text{, }}\left( {5 - 2} \right){\text{ from }}R_2 {\text{ and }}\left( {6 - 2} \right){\text{ from }}R_3  \hfill \\

     = \left( {3 - 2} \right)\left( {5 - 2} \right)\left( {6 - 2} \right)\left| {\begin{array}{*{20}c}<br />
   1 & {3 + 2} & {3^2  + 3.2 + 2^2 }  \\<br />
   1 & {5 + 2} & {5^2  + 5.2 + 2^2 }  \\<br />
   1 & {6 + 2} & {6^2  + 6.2 + 2^2 }  \\<br /> <br />
 \end{array} } \right| \hfill \\

     \left[ {{\text{since,  }}a^2  - b^2  = \left( {a - b} \right)\left( {a + b} \right){\text{  and  }}a^3  - b^3  = \left( {a - b} \right)\left( {a^2  + ab + b^2 } \right)} \right] \hfill \\

     = 1 \times 3 \times 4\left| {\begin{array}{*{20}c}<br />
   1 & 5 & {19}  \\<br />
   1 & 7 & {39}  \\<br />
   1 & 8 & {52}  \\<br /> <br />
 \end{array} } \right| \hfill \\

      {\text{Now, apply  }}R_2  \to R_2  - R_1 ,{\text{  }}R_3  \to R_3  - R_1 , \hfill \\


    Now, finish up. its value is 72
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  5. #5
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    Quote Originally Posted by Shyam View Post
    For the second,

     = \left| {\begin{array}{*{20}c}<br />
1 & 2 & {2^2 } & {2^3 } \\<br />
1 & 3 & {3^2 } & {3^3 } \\<br />
1 & 5 & {5^2 } & {5^3 } \\<br />
1 & 6 & {6^2 } & {6^3 } \\<br /> <br />
\end{array} } \right| \hfill \\

     {\text{Apply }}R_2 \to R_2 - R_1 ,{\text{ }}R_3 \to R_3 - R_1 ,{\text{ }}R_4 \to R_4 - R_1 \hfill \\

     = \left| {\begin{array}{*{20}c}<br />
1 & 2 & {2^2 } & {2^3 } \\<br />
0 & {3 - 2} & {3^2 - 2^2 } & {3^3 - 2^3 } \\<br />
0 & {5 - 2} & {5^2 - 2^2 } & {5^3 - 2^3 } \\<br />
0 & {6 - 2} & {6^2 - 2^2 } & {6^3 - 2^3 } \\<br /> <br />
\end{array} } \right| \hfill \\

     = \left| {\begin{array}{*{20}c}<br />
{3 - 2} & {3^2 - 2^2 } & {3^3 - 2^3 } \\<br />
{5 - 2} & {5^2 - 2^2 } & {5^3 - 2^3 } \\<br />
{6 - 2} & {6^2 - 2^2 } & {6^3 - 2^3 } \\<br /> <br />
\end{array} } \right| \hfill \\

     {\text{Take common }}\left( {3 - 2} \right){\text{ from }}R_1 {\text{, }}\left( {5 - 2} \right){\text{ from }}R_2 {\text{ and }}\left( {6 - 2} \right){\text{ from }}R_3 \hfill \\

     = \left( {3 - 2} \right)\left( {5 - 2} \right)\left( {6 - 2} \right)\left| {\begin{array}{*{20}c}<br />
1 & {3 + 2} & {3^2 + 3.2 + 2^2 } \\<br />
1 & {5 + 2} & {5^2 + 5.2 + 2^2 } \\<br />
1 & {6 + 2} & {6^2 + 6.2 + 2^2 } \\<br /> <br />
\end{array} } \right| \hfill \\

     \left[ {{\text{since, }}a^2 - b^2 = \left( {a - b} \right)\left( {a + b} \right){\text{ and }}a^3 - b^3 = \left( {a - b} \right)\left( {a^2 + ab + b^2 } \right)} \right] \hfill \\

     = 1 \times 3 \times 4\left| {\begin{array}{*{20}c}<br />
1 & 5 & {19} \\<br />
1 & 7 & {39} \\<br />
1 & 8 & {52} \\<br /> <br />
\end{array} } \right| \hfill \\

     {\text{Now, apply }}R_2 \to R_2 - R_1 ,{\text{ }}R_3 \to R_3 - R_1 , \hfill \\


    Now, finish up. its value is 72
    my friend, note that 4th column, 2th raw is 3^2 not 3^3
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  6. #6
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    Quote Originally Posted by Narek View Post
    my friend, note that 4th column, 2th raw is 3^2 not 3^3
    That does not matter so much. Now, you try with the same steps as I did, taking 3^2 instead of 3^3 in 4th column, 2nd row.

    If still not able to solve, then please ask.
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  7. #7
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    hmmmmmm

    yeah, its almost clear, except the part that:

    according to which rule, u said :
    R2 --> R2 - R1
    R3 --> R3 - R1
    R4 --> R4 - R1
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  8. #8
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    Quote Originally Posted by Narek View Post
    yeah, its almost clear, except the part that:

    according to which rule, u said :
    R2 --> R2 - R1
    R3 --> R3 - R1
    R4 --> R4 - R1
    R2 --> R2 - R1 mean, in second Row do this --> subtract first row from second row.
    this is called elementary row transformation

    did you get it now??? If not, please ask.
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