Need help on how to solve matrices?
Hi, If you have a graphing calculator, make a 4 x 5 matrix and enter these numbers with 0 for anything missing: [1 . 1 . 1 . 1 . 0] [8 .-1 . 8 .-1 . 0] [3 . 3 .-1 . 3 .-4] = [A] [1 .-2 . 3 .-1 . 0] Use the matrix command under math rref([A]) This will solve the system for you. [1 . 0 . 0 . 0 . -1] [0 . 1 . 0 . 0 . .2] [0 . 0 . 1 . 0 . .1] [0 . 0 . 0 . 1 . -2] The answer is x = -1, y = 2, z = 1, and w = -2. Without a calculator, it is a much longer process. [1 . 1 . 1 . 1 . 0] <=Multiply by -8 and add this to row 2. [8 .-1 . 8 .-1 . 0] [3 . 3 .-1 . 3 .-4] [1 .-2 . 3 .-1 . 0] [1 . 1 . 1 . 1 . 0] <=Multiply by -3 and add this to row 3. [0 .-9 . 0 .-9 . 0] [3 . 3 .-1 . 3 .-4] [1 .-2 . 3 .-1 . 0] [1 . 1 . 1 . 1 . 0] <=Multiply by -1 and add this to row 4. [0 .-9 . 0 .-9 . 0] [0 . 0 .-4 . 0 .-4] [1 .-2 . 3 .-1 . 0] [1 . 1 . 1 . 1 . 0] [0 .-9 . 0 .-9 . 0] [0 . 0 .-4 . 0 .-4] [0 .-3 . 2 .-2 . 0] Multiply row 2 by -1/9 to get a 1 in the 2nd row, 2nd column. [1 . 1 . 1 . 1 . 0] [0 . 1 . 0 . 1 . 0] <=Multiply by -1 and add this to row 1. [0 . 0 .-4 . 0 .-4] [0 .-3 . 2 .-2 . 0] [1 . 0 . 1 . 0 . 0] [0 . 1 . 0 . 1 . 0] <=Multiply by 3 and add this to row 4. [0 . 0 .-4 . 0 .-4] [0 .-3 . 2 .-2 . 0] [1 . 0 . 1 . 0 . 0] [0 . 1 . 0 . 1 . 0] [0 . 0 .-4 . 0 .-4] [0 . 0 . 2 . 1 . 0] Divide row 3 by -4 to get a 1 in the 3rd row, 3rd column. [1 . 0 . 1 . 0 . 0] [0 . 1 . 0 . 1 . 0] [0 . 0 . 1 . 0 . 1] <=Multiply by -1 and add this to row 1. [0 . 0 . 2 . 1 . 0] [1 . 0 . 0 . 0 .-1] [0 . 1 . 0 . 1 . 0] [0 . 0 . 1 . 0 . 1] <=Multiply by -2 and add this to row 4. [0 . 0 . 2 . 1 . 0] [1 . 0 . 0 . 0 .-1] [0 . 1 . 0 . 1 . 0] [0 . 0 . 1 . 0 . 1]. [0 . 0 . 0 . 1 .-2] <=Multiply by -1 and add this to row 2. [1 . 0 . 0 . 0 .-1] [0 . 1 . 0 . 0 . 2] [0 . 0 . 1 . 0 . 1]. [0 . 0 . 0 . 1 . -2] The answer is x = -1, y = 2, z = 1, and w = -2 I hope that helps!! :-)
Can someone help me with math matrices?
You have 3 equations in F, C and R (for the number of pallets of face, common, and refractory bricks). We want to come up with a matrix equation Mv = b where v = f c r -- that's a vector with the variables you have to solve for -- and where b (for budget) = 12000 14500 6000 You want Labor costs = 12000 = 50f + 50c + 75r Material costs = 14500 = 75f + 60c + 100r Utility costs = 6000 = 35f + 30c + 45r The info is given by type of brick, but each row of M needs to have the factors for each of the costs, so M looks like: 50 50 35 75 60 100 35 100 45 Check that over and you should see that Mv = b. It's very important that you see that. Multiply out Mv and you should see that solving the system of equations gives you the number of pallets of each that you are looking for. Now you invert M, and calculate v = M^-1 times b Using Excel, I get M^-1 = -0.12 0 0.2 0.05 -0.15 0.25 0.06 0.1 -0.3 and M^-1 b = -240 -75 370 LIke cidyah, I get negative values in the solution. It looks like something in your question may have been typed wrong.
Can someone please help me with expanding brackets in Algebra?
You start with the first two brackets, a rule I have been taught is F.O.I.L which is first outsides insides lasts. You do this with only two brackets at a time like so: (X+3)(4X-7) Multiply the firsts so X * 4X = 4X^2 Multiply insides so 3 * 4X = 12X Multiply outsides so X * -7 = -7X Multiply lasts so -7 * 3 = -21 add these all together to get 4X^2 + 5X - 21 then we put this in brackets and multiply it by the last bracket (don't use the rule for this one) (4X^2 + 5X - 21)(2X+5) Now we just multiply each value by the ones in the second bracket like so: 4X^2 * 2X = 8X^3 4X * 5 = 20X 5X * 2X = 10X^2 5X * 5 = 25X -21 * 2X = -42X -21 * 5 = -105 now we add this all together to get: 8X^3 + 10X^2 + 3X - 105 I hope I explained that well enough!
Can someone help me with my algebra assignment? (solving equations through matrices)?
Standard form: 2x+3y-4z=4; x+2y-5z=6; 4x+5y-2z=0; 2, 3, -4, 4 1, 2, -5, 6 4, 5, -2, 0 ----------------------- substitute/eliminate x = -3/2y+2z+2 2, 3, -4, 4 0, 1/2, -3, 4 0, -1, 6, -8 ----------------------- substitute/eliminate y = +6z+8 2, 0, 14, -20 0, 1/2, -3, 4 0, 0, 0, 0 ----------------------- dependent system - infinitely many solutions. Standard form: 2x+y-z=6; 4x+2y-2z=12; -x-1/2y+1/2z=-3; 2, 1, -1, 6 4, 2, -2, 12 -1, -1/2, 1/2, -3 ----------------------- substitute/eliminate x = -1/2y+1/2z+3 2, 1, -1, 6 0, 0, 0, 0 0, 0, 0, 0 ----------------------- dependent system - infinitely many solutions.
Can someone help me with this math problem?
Taking this exactly as it is written, √(1+4√(x))=√(x)+1 . . . . . . . . Square both sides 1 + 4√x = x + 2√x + 1 Let y = √x, and write the equation as 1 + 4y = y² + 2y + 1 y² + 2y - 4y = 0 y² - 2y = 0 y (y - 2) = 0, therefore y = 0 or y = -2, and hence √x = 0 or ��x = -2 so x = 0 or x = 4 √x = 0 is a trivial solution; however if you substitute √x = -2 into the original equation you get 1 + 4(-2) = -2 + 1 -7 = -1 which is obviously unacceptable. And I am therefore doubtful about the equation as written. Do you mean it to be √(1+4√(x))=√(x+1) ? Squaring this equation gives 1 + 4√x = x + 1 Again, let y = √x, then write it as 1 + 4y = y² + 1 y² - 4y = 0 y (y - 4) = 0, hence y = 0 or y = 4, therefore √x = 0 or √x = 4, so x = 0 or x = 16 and again, substituting 4 for √x, √(1+4√(x))=√(x+1) becomes √(1+4(4))=√(16+1) √17 = √17 which seems OK, and confirms that the equation ought to be √(1+4√(x))=√(x+1) If you are going to use brackets, it is rather essential to use them correctly !)
A Matrix is the representation of a transformation on a linear space. Linear space geometrically, could be a 1D surface (line), 2D surface (plane)....... So a matrix can tell us how a particular linear surface will morph under a transformation represented by it.There are several properties associated with a matrix which will help us in characterizing the resulting linear surface. Some basic properties are as follows:Size : M X N matrix tells us that it will map a N dimensional vector to a M dimensional one. The exact details of the of the linear surfaces depend on the Rank of the matrix.Rank: This basically tells us about the dimensions of the morphed surface. So a Matrix of Rank 1 will morph any linear surface it is acting on to a 1 Dimensional surface i.e a line. Similarly a Rank 2 matrix will morph any linear surface to a plane.There are several other properties of matrices which can be exploited while solving equations. Matrices can be used to solve linear equations. Linear equations generally have linear surfaces for solutions (not necessarily sub-spaces). Matrices are a tool to deal with linear surfaces. In fact many properties of solutions to linear equations can be tied to Null space , Column space ,Row space of a transformation.
For parentheses and brackets, you can write[math]1^\text{st}[/math] Bracket:[math]\LaTeX[/math] code: \left( \frac{x}{y} \right) Output: [math]\left( \frac{x}{y} \right)[/math][math]2^\text{nd}[/math] Bracket:[math]\LaTeX[/math] code: \left\{ \frac{x}{y} \right\} Output:[math]\left\{ \frac{x}{y} \right\}[/math][math]3^\text{rd}[/math] Bracket:[math]\LaTeX[/math] code: \left[ \frac{x}{y} \right] Output: [math]\left[ \frac{x}{y} \right][/math]Besides,[math]\LaTeX[/math] code: \left< \frac{x}{y} \right> Output: [math]\left< \frac{x}{y} \right>[/math]** this < and > can be replaced by \langle and \rangle respectively.Those \left and \right commands are used to adjust the sizes automatically. But if you want to adjust their size explicitly, you follow this codes in this table below:[math]\begin{array}{|c|c} \LaTeX \,\text{Code} & \text{Output} \\ \hline \text{\big( \Big( \bigg( \Bigg(}\, - \text{\Bigg) \bigg) \Big) \big)} & \big( \Big( \bigg( \Bigg( - \Bigg) \bigg) \Big) \big) & \\ \text{\big\{ \Big\{ \bigg\{ \Bigg\{}\, - \text{\Bigg\} \bigg\} \Big\} \big\}} & \big\{ \Big\{ \bigg\{ \Bigg\{ - \Bigg\} \bigg\} \Big\} \big\} & \\ \text{\big[ \Big[ \bigg[ \Bigg[}\, - \text{\Bigg] \bigg] \Big] \big]} & \big[ \Big[ \bigg[ \Bigg[ - \Bigg] \bigg] \Big] \big] & \\ \text{\big< \Big< \bigg< \Bigg<}\, - \text{\Bigg> \bigg> \Big> \big>} & \big< \Big< \bigg< \Bigg< - \Bigg> \bigg> \Big> \big> & \\ \end{array}[/math]Moreover,\tiny ) \Tiny \}\scriptsize ]\small )\normalsize \} \large ]\Large )\LARGE \}\huge ]\Huge )For more [math]\LaTeX[/math] you can visit this links:An introduction to beautiful math on QuoraShareLaTeX, Online LaTeX EditorGood luck.
ATA:The printf line executes because it doesn't form part of the if statement. Notice the semi-colon ";" following the if? That means the entire if structure stops at that point, then the rest of the program continues. If you omit the semi-colon, then the printf call becomes the "then" portion of the if statement. I would advise you always follow if statements with open and close brackets for the then portion { ...the code you want to execute ... }, even if only one line of code. It makes this much more clear, and avoids issues if you later change something.x gets overwritten in the statement y=x=10;. What that does is place the value 10 inside x, then duplicate the same value which x now has inside of y. Note, the order of this may differ from compiler to compiler. In this case it doesn't make a difference, but sometimes doing these multiple assignments at once can have strange effects. Try to avoid having more than one assign action inside a code sequence, especially if not simply initializing with a literal / constant.The programming language C doesn't "actually" have a boolean type. An integer value of 0 is considered to be "false", while any other value is considered "true". So what you see there is the "false" being translated into a value of 0, and then being stored inside of the variable z (which is typed as an integer).As sample for 1:if (x == y){ // Ensure the THEN portion is surrounded by brackets printf("%d %d\n", x, y); // Indent to make it even more clear} // This is where the THEN portion stops, so to add another action, just place it inside the brackets.// If you wanted an ELSE portion, follow with the else keyword.// And as a suggestion, also do the surround-with-bracketselse{ // Do something else instead of the printf}Notice blank lines are simply ignored. In C/C++ a blank line is the same as a space. It's called whitespace, and can be multiple whitespaces following each other - still just means one spacer for C/C++. So any line of code following another directly, or 500 blank lines further on with 1000 spaces in between and a 100 tabs thrown in just for fun is still the same to the C compiler. So use these whitespaces to organize your code so "you" can read it easier - the compiler ignores them.
Please help me solve this matrix?
The bottom row says 0 = 0, which is always true. This system has infinitely many solutions. For instance, you can pick x and y to be any real numbers, then use x - z = 2/3 to find the value of z and y + w = 5/3 to find the value of w. If the bottom row said something contradictory like 0 = 5, then there would be no solutions. You shouldn't be surprised to get an infinite number of solutions since you started with 4 unknowns but only 3 equations.
I can think of one possible reason. [ ] is used to denote intervals. eg. [math]\displaystyle [0,1][/math] means "the interval from 0 to 1, inclusive."( ) is also used to denote intervals. eg [math]\displaystyle (0,1)[/math] means "the interval from 0 to 1, exclusive."That kind of leaves { } as the only choice that looks bracket-like. It's just a matter of convention. It's just that most people are accustomed to using { } that it's universally understood that { } is used for set notation.