Question Video: Solving a Set of Simultaneous Equations Using Matrices Mathematics

Consider the system of equations 2𝑝 + 2π‘ž + 4π‘Ÿ = 4, βˆ’π‘ βˆ’ π‘ž βˆ’ π‘Ÿ = 14, 2𝑝 + 5π‘ž + 6π‘Ÿ = 10. Express the system as a single matrix equation.

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Video Transcript

Consider the system of equations. Express the system as a single matrix equation.

Remember, a matrix equation can be used to help us solve systems of linear equations. It will be of the form 𝐴𝐱 equals 𝐛, where 𝐴 is called the coefficient matrix, 𝐱 is a column vector containing the variables of our equation, and 𝐛 contains all of the constants in the equation. So let’s identify first the coefficient matrix. That’s the matrix that we’re going to call 𝐴. We have three equations with three variables. And so matrix 𝐴 is going to be a three-by-three matrix. The elements in the first row of matrix 𝐴 are the coefficients of 𝑝, π‘ž, and π‘Ÿ, respectively in our first equation.

We see that the coefficients are two, two, and four. And so that is the first row of our matrix. The elements in the second row of matrix 𝐴 are the coefficients of 𝑝, π‘ž, and π‘Ÿ in our second equation. Those are negative one, negative one, and negative one. And so those are the elements in the second row of our matrix. We’re going to repeat this process for the third row of our matrix, looking for the coefficients of 𝑝, π‘ž, and π‘Ÿ in our third equation. We now see that those are two, five, and six. And so we’ve identified our coefficient matrix. It’s two, two, four; negative one, negative one, negative one; and two, five, six.

We’re then going to look at the vector 𝐱. It’s a variable vector. And so we need to list the variables in our equation. Those are 𝑝, π‘ž, and π‘Ÿ. And it is in fact really important that we write them in that order. We need it to be the case that when we multiply the matrix 𝐴 by the column vector 𝐱, we end up with the original expressions two 𝑝 plus two π‘ž plus four π‘Ÿ, negative 𝑝 negative π‘ž negative π‘Ÿ, and so on. If we were to mix these variables around, we’re not going to end up with the exact same equation.

Finally, we’re interested in the constant vector 𝐛 that contains all of the constants from our equations. Once again, we need to give them in the correct order. And so we find vector 𝐛 is four, 14, 10. We write this as a single matrix equation by putting it back into the form 𝐴𝐱 equals 𝐛. And when we do, we see that the matrix equation we need is two, two, four; negative one, negative one, negative one; two, five, six times the vector 𝑝, π‘ž, π‘Ÿ equals the vector four, 14, 10.

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