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Posted on Jan 02, 2017
SOURCE: Solving matrix
Go to your matrix button and enter a "3x4" matrix.
Then enter it as follows:
-3 4 5 7
4 3 2 9
-5 5 3 -10
Then exit out and go to "2nd->matrix->math->rref(". Then press enter.
Your screen should look like this:
Then go to matrix and select your 3x4 matrix, press enter and close it with a parathesis. Your screen should look like this:
Press enter and the screen should say this:
1 0 0 2
0 1 0 -3
0 0 1 5
Hope this cleared up the confusion!
Posted on Mar 14, 2009
2y - x = 3
x = 3y - 5
Add the two equations side by side,
2y - x + x = 3 + 3y - 5
2y = 3y - 2
y = 2
Plug this in the second equation to get x,
x = 3(2) - 5
x = 1
So the solution is x = 1, y = 2
or in ordered pair notation (1, 2)
Posted on Jun 29, 2009
One not way would be to define a 4x4 matrix Mat A to hold the coefficients of the linear system. Then define a 4x1 column vector Mat V to hold the constants on the right.
Define a third 4x4 matrix Mat B you may leave filled with 0.
On command line, in Run Mat screen enter (Mat A) ^(-1) and store it in the zero-filled matrix Mat B. this is the inverse of Mat A.
If the inverse of Mat A exists, and it does in this case, the solution of the system is obtained as the column vector, resulting from the multiplication of Mat B by column vector Mat V
You can even shorten the procedure by just calculating ((Mat A)^-1)X (Mat V) [EXE]
Multiplication operator is the regular [times] key.
Posted on Nov 16, 2010
Tips for a great answer:
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(a+b)2 = (a+b)(a+b) = ... ?
(a+b)2 = a2 + 2ab + b2
You can easily see why it works, in this diagram:
2. Subtract Times Subtract And what happens if you square a binomial with a minus inside?
(a-b)2 = (a-b)(a-b) = ... ?
(a-b)2 = a2 - 2ab + b2
3. Add Times Subtract And then there is one more special case... what if you multiply (a+b) by (a-b) ?
(a+b)(a-b) = ... ?
(a+b)(a-b) = a2 - b2
That was interesting! It ended up very simple.
And it is called the "difference of two squares" (the two squares are a2 and b2).
This illustration may help you see why it works:
a2 - b2 is equal to (a+b)(a-b) Note: it does not matter if (a-b) comes first:
(a-b)(a+b) = a2 - b2
The Three Cases Here are the three results we just got:
(a+b)2 = a2 + 2ab + b2 } (the "perfect square trinomials") (a-b)2 = a2 - 2ab + b2 (a+b)(a-b) = a2 - b2 (the "difference of squares") Remember those patterns, they will save you time and help you solve many algebra puzzles.
Using Them So far we have just used "a" and "b", but they could be anything.
We can use the (a+b)2 case where "a" is y, and "b" is 1:
(y+1)2 = (y)2 + 2(y)(1) + (1)2 = y2 + 2y + 1
We can use the (a-b)2 case where "a" is 3x, and "b" is 4:
(3x-4)2 = (3x)2 - 2(3x)(4) + (4)2 = 9x2 - 24x + 16
We know that the result will be the difference of two squares, because:
(a+b)(a-b) = a2 - b2
(4y+2)(4y-2) = (4y)2 - (2)2 = 16y2 - 4
Sometimes you can recognize the pattern of the answer:
Example: can you work out which binomials to multiply to get 4x2 - 9
Hmmm... is that the difference of two squares?
Yes! 4x2 is (2x)2, and 9 is (3)2, so we have:
4x2 - 9 = (2x)2 - (3)2
And that can be produced by the difference of squares formula:
(a+b)(a-b) = a2 - b2
Like this ("a" is 2x, and "b" is 3):
(2x+3)(2x-3) = (2x)2 - (3)2 = 4x2 - 9
So the answer is that you can multiply (2x+3) and (2x-3) to get 4x2 - 9
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