If you have fewer equations than the number of variables, you will be able to learn some limiting information about the variables (such as x = 3y and y = 2z), but you cannot get a precise solution. For this article, we will be working toward getting a unique solution only.
If you have more variables, you will just continue the line as long as necessary. For example, if you are trying to solve a system with six variables, your standard form would look like Au+Bv+Cw+Dx+Ey+Fz =G. For this article, we will focus on systems with only three variables. Solving a larger system is exactly the same, but just takes more time and more steps. Note that in standard form, the operations between the terms is always addition. If your equation has subtraction instead of addition, you will need to work with this later my making your coefficient negative. If it helps you remember, you can rewrite the equation and make the operation addition and the coefficient negative. For example, you can rewrite the equation 3x-2y+4z=1 as 3x+(-2y)+4z=1.
For example, suppose you have a system that consists of the three equations 3x+y-z=9, 2x-2y+z=-3, and x+y+z=7. The top row of your matrix will contain the numbers 3,1,-1,9, since these are the coefficients and solution of the first equation. Note that any variable that has no coefficient showing is assumed to have a coefficient of 1. The second row of the matrix will be 2,-2,1,-3, and the third row will be 1,1,1,7. Be sure to align the x-coefficients in the first column, the y-coefficients in the second, the z-coefficients in the third, and the solution terms in the fourth. When you finish working with the matrix, these columns will be important in writing your solution.
You can indicate any specific position in a matrix by using a combination of R and C. For example, to pinpoint the term in the second row, third column, you could call it R2C3.
3 1 -1 9 2 -2 1 -3 1 1 1 7 You will be working with some basic operations to create the “solution matrix. ” The solution matrix will look like this:[6] X Research source 1 0 0 x 0 1 0 y 0 0 1 z Notice that the matrix consists of 1’s in a diagonal line with 0’s in all other spaces, except the fourth column. The numbers in the fourth column will be your solution for the variables x, y and z.
It is common to use fractions in scalar multiplication, because you often want to create that diagonal row of 1s. Get used to working with fractions. It will also be easier, for most steps in solving the matrix, to be able to write your fractions in improper form, and then convert them back to mixed numbers for the final solution. Therefore, the number 1 2/3 is easier to work with if you write it as 5/3. For example, the first row (R1) of our sample problem begins with the terms [3,1,-1,9]. The solution matrix should contain a 1 in the first position of the first row. In order to “change” our 3 into a 1, we can multiply the entire row by 1/3. Doing this will create the new R1 of [1,1/3,-1/3,3]. Be careful to keep any negative signs where they belong.
You can use some shorthand and indicate this operation as R2-R1=[0,-1,2,6]. Recognize that adding and subtracting are merely opposite forms of the same operation. You can either think of adding two numbers or subtracting the opposite. For example, if you begin with the simple equation 3-3=0, you could consider this instead as an addition problem of 3+(-3)=0. The result is the same. This seems basic, but it is sometimes easier to think of a problem in one form or the other. Just keep track of your negative signs.
Suppose you have a Row 1 of [1,1,2,6] and a Row 2 of [2,3,1,1]. You want to create a 0 term in the first column of R2. That is, you want to change the 2 into a 0. To do this, you need to subtract a 2. You can get a 2 by first multiplying Row 1 by the scalar multiplication 2, and then subtract the first row from the second row. In shorthand, you can think of this as R2-2*R1. First multiply R1 by 2 to get [2,2,4,12]. Then subtract this from R2 to get [(2-2),(3-2),(1-4),(1-12)]. Simplify this and your new R2 will be [0,1,-3,-11].
A common mistake occurs when conducting a combined multiplication and addition step in one move. Suppose, for example, you need to subtract double R1 from R2. When you multiply R1 by 2 to do this step, remember that you are not changing R1 in the matrix. You are only doing the multiplication to change R2. Copy R1 first in its original form, then make the change to R2.
1. Create a 1 in the first row, first column (R1C1). 2. Create a 0 in the second row, first column (R2C1). 3. Create a 1 in the second row, second column (R2C2). 4. Create a 0 in the third row, first column (R3C1). 5. Create a 0 in the third row, second column (R3C2). 6. Create 1 in the third row, third column (R3C3).
Create a 0 in the second row, third column (R2C3). Create a 0 in first row, third column (R1C3). Create a 0 in the first row, second column (R1C2).
Notice that multiplication and division are merely inverse functions of each other. We can say we are multiplying by 1/3 or dividing by 3, and the result is the same.
Copy down the unaffected row 3 as R3=[1,1,1,7]. Be very careful with subtracting negative numbers, to make sure you keep the signs correct. For now, leave the fractions in their improper forms. This will make later steps of the solution easier. You can simplify fractions in the final step of the problem.
Notice that as the left half of the row starts looking like the solution with the 0 and 1, the right half may start looking ugly, with improper fractions. Just carry them along for now. Remember to continue copying the unaffected rows, so R1=[1,1/3,-1/3,3] and R3=[1,1,1,7].
Continue to copy along R1=[1,1/3,-1/3,3] and R2=[0,1,-5/8,27/8]. Remember that you only change one row at a time.
Notice that the fractions, which appeared quite complicated in the previous step, have already begun to resolve themselves. Continue to carry along R1=[1,1/3,-1/3,3] and R2=[0,1,-5/8,27/8]. Notice that at this point, you have the diagonal of 1’s for your solution matrix. You just need to transform three more items of the matrix into 0’s to find your solution.
Copy along R1=[1,1/3,-1/3,3] and R3=[0,0,1,1].
Copy the unchanged R2=[0,1,0,4] and R3=[0,0,1,1].
1 0 0 2 0 1 0 4 0 0 1 1
Recall that the original equations for this problem were 3x+y-z=9, 2x-2y+z=-3, and x+y+z=7. When you replace the variables with their solved values, you get 32+4-1=9, 22-2*4+1=-3, and 2+4+1=7.
Because each equation simplifies to a true mathematical statement, your solutions are correct. If any of them did not resolve correctly, you would have to go back through your work and look for any errors. Some common mistakes occur in dropping negative signs along the way or confusing the multiplication and addition of fractions.
