সহসমীকরণ (সিস্টেম অফ ইকুয়েশনস)-কে ম্যাট্রিক্সের মাধ্যমে প্রকাশ

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Let's say we have the system of equations:

{x+y+z=62y+5z=−42x+5y−z=27\begin{cases} x + y + z &= 6 \\ 2y + 5z &= -4 \\ 2x + 5y - z &= 27 \\ \end{cases}

You can certainly solve this through elimination, but let's try a different way! Let's rewrite these equations using matrices.

We can split the coefficients on the left side of the equation into their own nice 3x3 matrix like so:

[11102525−1]\begin{bmatrix} 1&1&1 \\ 0&2&5 \\ 2&5&-1 \\ \end{bmatrix}

This may not make much sense yet, and that’s okay.

If you think about it, there must be a matrix operation that uses the coefficients and the values xx, yy, and zz to find the solution. The math is there. We just need the right tool.

The big question is this: which matrix operation gives us the correct values for xx, yy, and zz?

That’s what we’re about to figure out.

For now, let's write the entire equation in matrix form:

[11102525−1]⋅[6−427]\begin{bmatrix} 1&1&1 \\ 0&2&5 \\ 2&5&-1 \\ \end{bmatrix} \cdot \begin{bmatrix} 6\\-4\\27\\ \end{bmatrix}

If you know how matrix multiplication works, then you'll see why this works out the way it does!

We can write this equation as AX=BAX = B, where AA is the 3x3 matrix, XX is the variable matrix, and BB is the solution matrix.

We are trying to solve for XX. If this was a normal equation with normal variables, we would divide both sides by AA to solve for XX. Unfortunately, we can't really divide with matrices! Instead, we have to use something called the inverse of a matrix. An inverse matrix A−1A^{-1} satisfies the identity:

AA−1=A−1A=IAA^{-1} = A^{-1}A = I

where I is the identity matrix.

So going back to solving for our X matrix:

X=A−1BX = A^{-1}B

(Look up how to find the inverse of a matrix! Or just use a calculator...)

This expands to: X=−121[−276310−3−5−4−32]×[6−427]X = \frac{-1}{21} \begin{bmatrix} -27&6&3\\10&-3&-5\\-4&-3&2 \\ \end{bmatrix} \times \begin{bmatrix} 6\\-4\\27 \\ \end{bmatrix}

which by matrix multiplication yields:

X=−121[−105−6342]=[53−2]X = \frac{-1}{21} \begin{bmatrix} -105\\-63\\42 \end{bmatrix} = \begin{bmatrix} 5\\3\\-2 \\ \end{bmatrix}

And we're done! We can see that x=5x = 5, y=3y = 3, and z=−2z = -2 is the solution!

Video

(Video) How to Write an Augmented Matrix

by IntFinder

Here's a video by IntFinder on how you write a system of equations as an augmented matrix. An augmented matrix is just a matrix with a dashed line inside it.

Summary

Given the following system of equations: 4x−5y+z=0−2x−y+6=−19x+5y−5z=10\begin{align} 4x-5y+z &= 0 \\ -2x-y+6 &= -19 \\ x+5y-5z &=10 \\ \end{align}

Now that we have our augmented matrix, we need to make sure the system is in standard form.

In standard form, the variables go on the left side of the equation. The constant goes on the right side.

Right now, one equation is not set up that way. So we will rewrite the system a little bit to fix the second equation.

4x−5y+z=0−2x−y=−25x+5y−5z=10\begin{align} 4x-5y+z &= 0 \\ -2x-y &= -25 \\ x+5y-5z &=10 \\ \end{align}

Now, we can write our augmented matrix. The augmented matrix contains two parts: one part has the coefficient of the variables, while the second part has the constants.

The augmented matrix:
x y z RHS
4 -5 1 0
-2 -1 0 -25
1 5 -5 10

Image source: by Anusha Rahman

This is our final augmented matrix.

Illustrated · 2 min

Using Matrices to Represent Systems of Linear Equations

When we learned about elementary row operations in earlier lessons, we talked about using them later.

Well, this is that moment.

Quick Reminder: Elementary Row Operations

Elementary row operations are simple steps you can use to change a matrix without changing its solution.

There are three main types:

We use these operations to make matrices easier to work with and to help find solutions to systems of equations.

Now you’ll see how those skills help you solve real problems.

Another way that we can solve systems of linear equations is to use matrices, specifically the augmented matrix!

Augmented Matrices
For systems of equations, there's a coefficient matrix and a constant matrix

Image source: by Anusha Rahman

When we work with a system of equations, it helps to organize the information. One way to do this is by splitting the system into two parts: the coefficient matrix and the constant matrix.

The coefficient matrix is made up of all the numbers in front of the variables. These numbers tell us how much of each variable appears in each equation. The coefficient matrix does not include the variables themselves. They only include the numbers attached to them.

The constant matrix includes all the numbers on the right side of the equal sign. These values stay the same and do not have variables attached to them.

Together, these two matrices represent the original system of equations in a cleaner and more organized way. This setup makes it easier to apply matrix operations, like row operations, to solve the system. We can also use an inverse matrix or Cramer's Rule.

By separating the coefficients from the constants, we can focus on the structure of the system and solve it step by step.

In the Real World

We use augmented matrices in the real world when we need to solve many equations at the same time in an organized way. Engineers and scientists use them to model systems like traffic flow, electrical circuits, or computer graphics. Augmented matrices help turn messy information into a clear structure that is easier to solve and understand.

Type a concept, like “slope” or “photosynthesis”.

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