Which two elementary operations transform the following matrix:
Into this one:Code:2 -6 3 -3 -6 4 -2 -6 -6 0 -1 -3 -9 6 4 7 3 3 -9 2
Thanks for any help!Code:2 -6 3 -3 -6 4 -2 -6 -6 0 -27 99 -18 27 78 7 3 3 -9 2
What exactly your difficulty? I see that only the third row has changed so whatever elementary operations (row operations) are done must be done to it. There are three types of row operations: swap two rows (obviously not here), multiply one row by number, add a multiple of one row to another.
You might try this: if we were to multiply the row by a, then add b times the first row, the first number would be -a+ 2b= -27 and the second number would be -3a- 6b= 99. That gives two equations to solve for a and b. Now try the third number: For those values of a and b, is -9a+ 3b= -18? If yes, check the other numbers but you are probably done. If not try b times the second row: -a+ 4b= -27, -3a-2b= 99, etc.
If none of those work, then it must be that "add a multiple of one row to the third row" was used twice: Try a times the first row pluls b times the second row added to the third: 2a+ 4b- 1= -27, 2a- 2b- 3= 99, again two equations in two unknowns that can be checked by the other numbers in the third row.
Tedious, but just algebra.
Hello, Incanus!
Never seen a guessing game with matrices.
Wonder what the purpose of the problem is?
Which two elementary operations transform
this matrix .$\displaystyle \begin{pmatrix}2 &\text{-}6&3&\text{-}3&\text{-}6\\ 4&\text{-}2&\text{-}6&\text{-}6&0\\ \text{-}1&\text{-}3&\text{-}9&6&4\\ 7&3&3&\text{-}9&2\end{pmatrix}$ . into this one: . $\displaystyle \begin{pmatrix}2&\text{-}6&3&\text{-}3&\text{-}6\\ 4&\text{-}2&\text{-}6&\text{-}6&0\\ \text{-}27 &99&\text{-}18&27&78\\ 7&3&3&\text{-}9&2\end{pmatrix}$
. . $\displaystyle \begin{array}{c}\text{Multiply }R_3\text{ by -3.} \\ \\[-4mm]
\text{Multiply }R_1\text{ by -15.} \\ \\[-4mm]
\text{Add and replace }R_3. \end{array}$