Inverting the Coefficient Matrix: Solving Systems of Equations

It is a very good suggestion. I would have given the same if I had seen your question earlier.The answer is not x= -1/3 and y= 1/3. It is x= -1/3 and y= -1/3.Let me suggest you a method for finding the inverse of a 2x2 matrix. This method is much easier than finding the determinant and the cofactors.Let us take the same example. You have the matrix$$A= \left [\begin{array}{cc}2&4\\-1&-1\end{array}\right].$$You have to find the inverse matrix of A.Step 1: Take a matrix
  • #1
Danatron
25
0
Solve the following system of equations using the inverse of the coefficient matrix,
2x + 4y = -9
-x - y = 2

My attempt-

[2 4 [x = -9
-1 -1] y] = 2
|A| x b

|A| = -2-4=-6

[x = 1/-6 [-1 -4 [-9 1/-6 [ 1 = 0.03
y] = 1 2] 2] -5] = 0.166666

Would someone be able to confirm if the answer is correct please?

Thank you
 
Physics news on Phys.org
  • #2
sorry guys, i didn't type the matrices out like that! is there another way to type out a matrix on here?
 
  • #3
You can add matrices to the text with LaTeX commands. For example: ##\left(\begin{smallmatrix}2&4\\-1&-1\end{smallmatrix}\right)##. Right click on my matrix and choose "Show math as > TeX commands" to see how that is written in LaTeX code.

The idea in inverting the matrix is to form the double-wide matrix ##\left(\begin{smallmatrix}2&4&|\\-1&-1&|\end{smallmatrix}\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right)## and apply row operations until the left side becomes the identity matrix and the right side becomes the matrix inverse. Have you been taught how to apply matrix row operations in solving linear systems?
 
  • #4
Writing this using Latex, your matrix equation is
[tex]\begin{bmatrix}2 & 4 \\ -1 & -1 \end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}= \begin{bmatrix}-9 \\ 2\end{bmatrix}[/tex]

You give your solution as x= 0.03, y= 0.16666.

It's easy enough to check that your self isn't it?
2x+ 4y= 2(0.03)+ 4(0.1666)= 0.06+ 0.66664= 0.72664, not -9! (Not even close.)

You seem to be under the impression that the inverse matrix to [itex]\begin{bmatrix}-2 & 4 \\ -1 & -1\end{bmatrix}[/itex] is [itex](1/6)\begin{bmatrix}-1 & -4 \\ 1 & 2\end{bmatrix}[/itex]. It isn't as, again, you could have checked:
[tex]\begin{bmatrix}2 & 4 \\ -1 & -1 \end{bmatrix}\begin{bmatrix}-\frac{1}{6} & -\frac{2}{3} \\ \frac{1}{6} & \frac{1}{3}\end{bmatrix}[/tex][tex]= \begin{bmatrix}-\frac{2}{6}+ \frac{4}{6} & -\frac{4}{3}+ \frac{4}{3} \\ \frac{1}{6}- \frac{1}{6} & \frac{2}{3}- \frac{1}{3}\end{bmatrix}[/tex][tex]= \begin{bmatrix}-\frac{1}{3} & 0 \\ 0 & \frac{1}{3} \end{bmatrix}[/tex]
NOT the identity matrix (though pretty close)!

Looks to me like you calculated the determinant wrong. Be careful with the signs.

While your method of finding the determinant and cofactors will work to find the inverse matrix (it was the first method I learned), using the "augmented matrix" and row operations, as Hilbert2 suggests, is much easier and less error prone.
 
Last edited by a moderator:
  • Like
Likes 1 person
  • #5
Danatron said:
Solve the following system of equations using the inverse of the coefficient matrix,
2x + 4y = -9
-x - y = 2

My attempt-

[2 4 [x = -9
-1 -1] y] = 2
|A| x b

|A| = -2-4=-6

[x = 1/-6 [-1 -4 [-9 1/-6 [ 1 = 0.03
y] = 1 2] 2] -5] = 0.166666

Would someone be able to confirm if the answer is correct please?

Thank you

In small problems done by hand (and when all input data are rational numbers), avoid decimals and use rationals instead. So, write 1/6 instead of 0.16666, etc. It is easily do-able using a hand-held calculator: for example,
[tex]
\frac{a}{b} + \frac{c}{d}\frac{e}{f} = \frac{adf + bce}{bdf},[/tex]
etc.
 
  • #6
Danatron said:
Solve the following system of equations using the inverse of the coefficient matrix,
2x + 4y = -9
-x - y = 2

My attempt-

[2 4 [x = -9
-1 -1] y] = 2
|A| x b

|A| = -2-4=-6

[x = 1/-6 [-1 -4 [-9 1/-6 [ 1 = 0.03
y] = 1 2] 2] -5] = 0.166666

Would someone be able to confirm if the answer is correct please?
Thank you

There is a special rule for computing the inverse of a 2x2 matrix; it is worth remembering.
[tex] {\begin{bmatrix}a & b \\ c & d\end{bmatrix}}^{-1}
= \frac{1}{D} \begin{bmatrix}d & -b \\ -c & a\end{bmatrix}[/tex]
where
[tex] D = \left| \begin{array}{cc}a & b \\c & d\end{array} \right| = ad - bc[/tex]
is the determinant of the matrix.

In other words, you get the inverse by swapping the diagonal elements, changing the signs of the off-diagonal elements, and dividing by the determinant.
 
  • #8
HallsofIvy said:
Writing this using Latex, your matrix equation is
[tex]\begin{bmatrix}2 & 4 \\ -1 & -1 \end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}= \begin{bmatrix}-9 \\ 2\end{bmatrix}[/tex]

You give your solution as x= 0.03, y= 0.16666.

It's easy enough to check that your self isn't it?
2x+ 4y= 2(0.03)+ 4(0.1666)= 0.06+ 0.66664= 0.72664, not -9! (Not even close.)

You seem to be under the impression that the inverse matrix to [itex]\begin{bmatrix}-2 & 4 \\ -1 & -1\end{bmatrix}[/itex] is [itex](1/6)\begin{bmatrix}-1 & -4 \\ 1 & 2\end{bmatrix}[/itex]. It isn't as, again, you could have checked:
[tex]\begin{bmatrix}2 & 4 \\ -1 & -1 \end{bmatrix}\begin{bmatrix}-\frac{1}{6} & -\frac{2}{3} \\ \frac{1}{6} & \frac{1}{3}\end{bmatrix}[/tex][tex]= \begin{bmatrix}-\frac{2}{6}+ \frac{4}{6} & -\frac{4}{3}+ \frac{4}{3} \\ \frac{1}{6}- \frac{1}{6} & \frac{2}{3}- \frac{1}{3}\end{bmatrix}[/tex][tex]= \begin{bmatrix}-\frac{1}{3} & 0 \\ 0 & \frac{1}{3} \end{bmatrix}[/tex]
NOT the identity matrix (though pretty close)!

Looks to me like you calculated the determinant wrong. Be careful with the signs.

While your method of finding the determinant and cofactors will work to find the inverse matrix (it was the first method I learned), using the "augmented matrix" and row operations, as Hilbert2 suggests, is much easier and less error prone.

Thanks for the help.

Just one question, I am probably wrong but anyhow if i don't ask i won't learn.

If i plug the x= -1/3 & y= 1/3 into the equation to check then I am don't get -9?
2(-1/3) + 4(1/3) = 0.66666 recurring

what am i doing wrong?
 
  • #9
Several posters have told you what you have wrong. You calculated the determinant of coefficients wrong.$$
\left | \begin{array}{cc}
-2&4\\
-1&-1
\end{array}\right |= \text{?}$$
 
  • #10
so x= -1/3 and y= 1/3?
 
  • #11
"If i plug the x= -1/3 & y= 1/3 into the equation to check then I am don't get -9?
2(-1/3) + 4(1/3) = 0.66666 recurring"

You have already concluded that this answer is wrong.

Please read what was suggested by LCKurtz.
 

1. What is an inverse coefficient matrix?

An inverse coefficient matrix is the matrix that results from taking the inverse of a given coefficient matrix. In other words, it is the matrix that, when multiplied by the original coefficient matrix, results in an identity matrix.

2. Why is an inverse coefficient matrix important?

An inverse coefficient matrix is important because it allows for the solving of systems of linear equations. By multiplying the inverse matrix by the constant vector, the values of the variables in the system can be determined.

3. How is an inverse coefficient matrix calculated?

An inverse coefficient matrix can be calculated using various methods, such as Gauss-Jordan elimination, LU decomposition, or the adjugate matrix method. These methods involve a series of mathematical operations on the original coefficient matrix to obtain the inverse matrix.

4. What properties does an inverse coefficient matrix have?

An inverse coefficient matrix has several properties, including that it is symmetric, meaning that the inverse of an inverse matrix is the original matrix. It also has the property that when multiplied by the original matrix, the result is an identity matrix.

5. Are there any limitations to using an inverse coefficient matrix?

While an inverse coefficient matrix is a useful tool for solving systems of linear equations, there are some limitations to its use. One limitation is that not all matrices have an inverse. Additionally, the computation of an inverse matrix can be time-consuming and computationally expensive for large matrices.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
6
Views
545
  • Precalculus Mathematics Homework Help
Replies
3
Views
721
  • Precalculus Mathematics Homework Help
Replies
1
Views
459
  • Precalculus Mathematics Homework Help
Replies
4
Views
719
  • Precalculus Mathematics Homework Help
Replies
1
Views
896
  • Precalculus Mathematics Homework Help
Replies
2
Views
466
  • Precalculus Mathematics Homework Help
Replies
1
Views
779
  • Precalculus Mathematics Homework Help
Replies
2
Views
795
  • Precalculus Mathematics Homework Help
Replies
13
Views
164
Back
Top