matrice problem (Urgent! ASAP please)

**Narek** · September 20th 2008, 08:07 AM

Hello ppl. I need to solve these 2 matrices. please help me solving these and explain for me the solution. Thanks in advance

**TKHunny** · September 20th 2008, 10:25 AM

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?

**Narek** · September 21st 2008, 06:08 PM

Originally Posted by TKHunny

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?

**Shyam** · September 21st 2008, 08:15 PM

Originally Posted by Narek

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,

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

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

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

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

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

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

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

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

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

Now, finish up. its value is 72

**Narek** · September 22nd 2008, 07:17 AM

Originally Posted by Shyam

For the second,

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

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

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

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

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

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

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

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

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

Now, finish up. its value is 72

my friend, note that 4th column, 2th raw is 3^2

not 3^3

**Shyam** · September 22nd 2008, 07:30 AM

Originally Posted by Narek

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.

**Narek** · September 22nd 2008, 07:40 AM

yeah, its almost clear, except the part that:

according to which rule, u said :
R2 --> R2 - R1
R3 --> R3 - R1
R4 --> R4 - R1

**Shyam** · September 22nd 2008, 08:08 AM

Originally Posted by Narek

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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