Video: Operations on Matrices Involving the Transpose of a Matrix

Given that 𝐡 = (1, βˆ’3 and 7, βˆ’3), (𝐡 βˆ’ 𝐡^(𝑇))^(𝑇) = 𝐴, determine the value of π‘Žβ‚β‚‚ + π‘Žβ‚‚β‚.

03:37

Video Transcript

Given that 𝐡 is the matrix of elements one, negative three, seven, negative three and 𝐡 minus 𝐡 transpose all transpose is equal to 𝐴, determine the value of π‘Ž one two plus π‘Ž two one.

Let’s begin by clarifying some of the notation in the question. These superscript 𝑇s are used to denote the transpose of a matrix. The transpose of a matrix is the matrix found by swapping the rows and columns of that matrix around. For example, the first column of a matrix 𝐡 becomes the first row of the matrix 𝐡 transpose, and so on. We’ll look at how to find the transpose of these particular matrices during the question.

The notation lowercase π‘Ž, and then a subscript one and two means the element in the first row and second column of the matrix capital 𝐴. And in the same way, the notation lowercase π‘Ž and then a subscript two one means the element in the second row and first column of the matrix capital 𝐴.

We’re told that the matrix 𝐴 is equal to 𝐡 minus 𝐡 transpose all transpose. So, in order to determine the value of this sum, we need to find the matrix 𝐴. Let’s begin by finding the matrix 𝐡 transpose. So, this is the transpose of the matrix with elements one, negative three, seven, negative three.

The transpose is found by swapping the rows and columns around. So, the first row of the matrix 𝐡, that’s the row one, negative three becomes the first column of the matrix 𝐡 transpose. And then the second row of the matrix 𝐡 becomes the second column of the matrix 𝐡 transpose. So, 𝐡 transpose is the matrix of elements one, seven, negative three, negative three.

Next, we need to find the matrix 𝐡 minus 𝐡 transpose. So, we take the matrix we’ve just found for 𝐡 transpose and subtract it from our matrix 𝐡. To subtract two matrices, which must be of the same order, we subtract corresponding elements. So, in the first row and first column we have one minus one. In the first row and second column, we have negative three minus seven. In the second row first column, we have seven minus negative three. And in the second row second column, we have negative three minus negative three.

Evaluating each of these subtractions gives the matrix with elements zero, negative 10, 10, zero. Then the matrix 𝐴 is the transpose of this matrix, so the transpose of the matrix zero, negative 10, 10, zero.

The first row becomes the first column of the transpose matrix, giving zero, and then negative 10. And then the second row becomes the second column of the transpose matrix. So, the matrix 𝐡 minus 𝐡 transpose all transpose is equal to zero, 10, negative 10, zero.

Remember we are asked to determine the value of π‘Ž one two plus π‘Ž two one. The element in the first row and second column is equal to 10. And the element in the second row first column is equal to negative 10. So, the sum of these two elements, 10 plus negative 10, is equal to zero. We’ve found then that if 𝐡 is the matrix one, negative three, seven, negative three and 𝐴 is the matrix 𝐡 minus 𝐡 transpose all transpose, then the value of π‘Ž one two plus π‘Ž two one is equal to zero.

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