# Math Help - matrice problem (Urgent! ASAP please)

1. ## 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

2. 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?

3. 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?

4. 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

5. 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

6. 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.

7. ## hmmmmmm

yeah, its almost clear, except the part that:

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

8. 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