Question Video: Solving a Matrix Equation Mathematics

Solve for the matrix 𝑋 in the matrix equation 𝑋 βˆ’ 𝐴^(𝑇) = 𝐡, where 𝐴 = [1, 2 and 7, 3] and 𝐡 = [1, 7 and 2, βˆ’3].

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Video Transcript

Solve for the matrix 𝑋 in the matrix equation 𝑋 minus the transpose of 𝐴 is equal to 𝐡, where 𝐴 is equal to one, two, seven, three and 𝐡 is equal to one, seven, two, negative three.

Since we have been asked to solve the equation for the matrix 𝑋, we should rearrange our equation so that it is in terms of 𝑋. We can simply add the transpose of 𝐴 to both sides of the equation. This will cancel out with the negative transpose of 𝐴 that we have on the left. What we will be left with is that 𝑋 is equal to 𝐡 plus the transpose of 𝐴.

Now, looking at the right-hand side of the equation, we already know what 𝐡 is. What we need to do is take the transpose of 𝐴 and then we have all of the components on the right. We’ve been given that 𝐴 is equal to one, two, seven, three. In order to find the transpose of this matrix, we simply take its rows and turn them into the columns of the transpose. Looking at the first row of 𝐴, we have one, two. So the first column of the transpose of 𝐴 will be one, two. Similarly, the second row of 𝐴 is seven, three. So the second column of the transpose of 𝐴 will be seven, three.

So here we have found the matrix of the transpose of 𝐴. We have that 𝑋 is equal to 𝐡 plus the transpose of 𝐴. In order to complete this sum, we need to add each of the corresponding elements of the two matrices. The first element of 𝐡 is one, and the first element of the transpose of 𝐴 is also one. So the first element of 𝑋 will be one plus one, which is just two. For the second element, we have seven plus seven, so that’s 14. For the third element, two plus two, which is just four. And for the final element, we have negative three plus three, which gives us zero.

Hence, we have reached our solution, which is that 𝑋 is equal to two, 14, four, zero.

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