Question Video: Finding an Unknown Matrix in a Matrix Equation

Find the matrix 𝐴 such that 𝐴[π‘₯₁ and π‘₯β‚‚ and π‘₯₃ and π‘₯β‚„] = [π‘₯₁ + 3π‘₯β‚‚ + 2π‘₯₃ and 2π‘₯₃ + π‘₯₁ and 6π‘₯₃ and π‘₯β‚„ + 3π‘₯β‚‚ + π‘₯₁].

03:03

Video Transcript

Find the matrix 𝐴 such that 𝐴 times the column matrix with elements π‘₯ one, π‘₯ two, π‘₯ three, and π‘₯ four is equal to the column matrix with elements π‘₯ one plus three π‘₯ two plus two π‘₯ three, two π‘₯ three plus π‘₯ one, six π‘₯ three, and π‘₯ four plus three π‘₯ two plus π‘₯ one.

We’re asked to find the matrix 𝐴 which when it multiplies to column matrix on the left is equal to the column matrix on the right. The first thing we can do is to work out the order of the matrix 𝐴. That’s how many columns and how many rows. And we can use the definition of matrix multiplication to help us here. This says that for matrices 𝐴 and 𝐡, matrix multiplication is well defined if 𝐴 has order π‘š by 𝑛 and 𝐡 has order 𝑛 by 𝑝 so that the product 𝐴𝐡, which we’ll call 𝐢, has order π‘š by 𝑝.

In our case, if we let the column matrix with elements π‘₯ one, π‘₯ two, π‘₯ three, and π‘₯ four equal to 𝐡 and the column matrix on our right-hand side be the matrix 𝐢, and while we don’t know the order of matrix 𝐴, we know that 𝐡 has order four by one and 𝐢 has order four by one. That is, they both have four rows and one column, so that 𝑛 is four and 𝑝 is one and π‘š is four. And since our matrix 𝐴 must have order π‘š by 𝑛, this has order four by four; that is, 𝐴 is a four-by-four matrix. It has four rows and four columns.

So making some room, our matrix equation is therefore of the form shown, where 𝐴 is a four-by-four matrix. To determine what the elements of 𝐴 are, it’s helpful if we align the variables in the same columns in our matrix on the right-hand side. There are four variables: π‘₯ one, π‘₯ two, π‘₯ three, and π‘₯ four. So we should leave a space in the first row for π‘₯ four at the end. Our π‘₯ one in the second row needs to come to the front. And π‘₯ three go under the π‘₯ three in the first row, and so on.

The real elements in the coefficient matrix 𝐴 are the coefficients of the variables in the corresponding row in our matrix 𝐢 on the right-hand side. So that in the first row, we have elements one, three, two, and zero since the coefficient of π‘₯ four in the first row of 𝐢 is zero. Our second row of 𝐴 then has elements one, zero, two, and zero. Our third row of 𝐴 has elements zero, zero, six, and zero. And finally, our fourth row will have elements one, three, zero, and one.

So that the matrix 𝐴 when it multiplies the column vector with elements π‘₯ one, π‘₯ two, π‘₯ three, and π‘₯ four is equal to the column vector with elements π‘₯ one plus three π‘₯ two plus two π‘₯ three, two π‘₯ three plus π‘₯ one, six π‘₯ three, and π‘₯ four plus three π‘₯ two plus π‘₯ one has elements one, three, two, zero in the first row; one, zero, two, zero in the second row; zero, zero, six, and zero in the third row; and one, three, zero, one in the fourth row. That’s a four-by-four matrix. And we can check that this is correct by performing matrix multiplication on the left-hand side and equating with the right-hand side.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.