Question Video: Given an Equation a Matrix Satisfies Determine the Properties the Matrix Must Have | Nagwa Question Video: Given an Equation a Matrix Satisfies Determine the Properties the Matrix Must Have | Nagwa

# Question Video: Given an Equation a Matrix Satisfies Determine the Properties the Matrix Must Have Mathematics • First Year of Secondary School

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If π΄ is a square matrix, then the matrix π΄ + π΄^π‘ = οΌΏ. [A] a skew-symmetric matrix [B] a symmetric matrix [C] a diagonal matrix [D] a zero matrix

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### Video Transcript

If π΄ is a square matrix, then the matrix π΄ plus π΄ transpose is equal to blank. Option (A) a skew-symmetric matrix, option (B) a symmetric matrix, option (C) a diagonal matrix, or option (D) a zero matrix.

In this question, weβre given a square matrix π΄, and we need to determine what type of matrix π΄ plus π΄ transpose is going to be. And weβre given four possible options. Thereβs a few different ways of approaching this problem. For example, we could try inputting a square matrix π΄ into π΄ plus π΄ transpose. We could then see any interesting properties the result would have. For example, we could try the matrix π΄ to be the two-by-two matrix one, two, three, four. We want to know what the matrix one, two, three, four plus the transpose of this matrix would be equal to. To do this, we would need to start by recalling what we mean by the transpose.

When we take the transpose of a matrix, we want to write the rows as the columns and the columns as the rows. So in our two-by-two matrix one, two, three, four, we see the first row is one, two. We want this to be the first column in our transpose. So weβll write the first column as one, two. Next, we can see the second row is equal to three, four. So we want this to be equal to the second row in our transpose. So the transpose of this matrix is the two-by-two matrix one, three, two, four.

Now we want to add these two matrices together. And remember, to add two matrices of the same order together, we just add their entries together. Adding the entries of these two matrices together, we get the two-by-two matrix two, five, five, eight. And it can be difficult to see what properties this matrix might have. However, there is one thing we can notice. We can ask the question, what happens if we take the transpose of this matrix? We want to take the transpose of the two-by-two matrix two, five, five, eight. Remember, to do this, we want to write the rows as the columns and the columns as rows.

Letβs start with the first row. The first row of this matrix is two, five, so we want to write this as the first column which is two, five. Next, the second row is equal to five, eight. And we want this to be equal to the second column of our transpose, so the second column of our transpose will be five, eight. But now we can see something interesting. When we took the transpose of our matrix, we didnβt actually change our matrix. We ended up with exactly the same matrix. And this is actually a property of matrices. If a matrix π΅ is equal to its transpose, then we say that the matrix π΅ is a symmetric matrix.

And thereβs something interesting worth pointing out about this definition. Remember, when we take the transpose of a matrix, we switch the number of rows and columns around. In other words, an π-by-π matrix becomes an π-by-π matrix. And we know for two matrices to be equal, their orders must be equal. Therefore, if π΅ is equal to its own transpose, π΅ must also be a square matrix. So, by looking at our example, it appears that π΄ plus its own transpose should be a symmetric matrix. Letβs see if we can prove this.

Letβs start by letting π΄ be a square matrix, so its order is π-by-π, where the entry in row π and column π of matrix π΄ is given by lowercased π sub ππ. Then the first thing we can do is find the transpose of matrix π΄. To do this, we switch the rows and the columns around of matrix π΄. So in row π and column π of our transpose matrix, weβll have π sub ππ because we switched the rows and columns around in matrix π΄. And now that we have the entry in row π column π of matrix π΄ and the entry in row π column π of the transpose of π΄, we can add these together. Remember, to add two matrices of the same order together, we just need to add their entries together.

Of course, itβs worth pointing out here, because π΄ is a square matrix, we know that π΄ transpose will also be a square matrix, so we can just add their entries together, giving us the entry in row π and column π of π΄ plus π΄ transpose is π sub ππ plus π sub ππ. Now remember, weβre trying to prove that this is a symmetric matrix. We want to prove that this matrix is equal to its own transpose, and thereβs a few different ways of doing this. One way of doing this is to give this matrix a name. Letβs call this matrix π΅. Now we want to find the transpose of our matrix π΅. But what does this mean? The entry in row π and column π of π΅ transpose is going to be the entry in row π column π of matrix π΅. So, we want to know the entry in row π and column π of matrix π΅.

And of course, to do this, all we need to do is switch our πβs and πβs around. This will give us the entry in row π column π of matrix π΅. So, the entry in row π column π of π΅ transpose will be π sub ππ plus π sub ππ. Of course, we know addition is commutative, so we can switch the order of these two terms around. This gives us that π΅ transpose is π sub ππ plus π sub ππ. And now we can see that this is exactly the same as we have for matrix π΅. Therefore, because these are two matrices of the same order where all of their entries are equal, weβve shown that π΅ is equal to π΅ transpose.

And remember, our matrix π΅ was the blank entry in our question. And weβve just shown for any square matrix π΄, π΄ plus π΄ transpose has the property that itβs equal to its own transpose. In other words, this always gives us a symmetric matrix. Therefore, we were able to show if π΄ is any square matrix, then the matrix π΄ plus π΄ transpose is always equal to the symmetric matrix which was our option (B).

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