Question Video: Finding an Unknown Element of a Skew-Symmetric Matrix | Nagwa Question Video: Finding an Unknown Element of a Skew-Symmetric Matrix | Nagwa

Question Video: Finding an Unknown Element of a Skew-Symmetric Matrix Mathematics • First Year of Secondary School

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Given that the matrix 𝐡 = [0, βˆ’2π‘₯, βˆ’63 and 𝑧 + 14, 0, 3𝑧 and βˆ’3𝑦 + 6π‘₯, βˆ’12, 0] is skew-symmetric, find the value of π‘₯ + 𝑦 + 𝑧.

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Video Transcript

Given that the matrix 𝐡 equals zero, negative two π‘₯, negative 63, 𝑧 plus 14, zero, three 𝑧, negative three 𝑦 plus six π‘₯, negative 12, zero is skew-symmetric, find the value of π‘₯ plus 𝑦 plus 𝑧.

Given that this matrix is skew-symmetric, it must be true that 𝐡 transpose is equal to negative 𝐡. This holds for all skew-symmetric matrices. And it’s a really important property that we’re going to use to answer this question.

In order to use this, let’s first remind ourselves what 𝐡 transpose means. For an 𝑛-by-𝑛 matrix, 𝐡 defined by 𝐡 equals 𝐡 𝑖𝑗, the transpose of 𝐡, denoted 𝐡 T, is equal to 𝐡 𝑗𝑖. Notice how transposing a matrix swaps the row and column indices. This means that the rows become the columns and the columns become the rows.

So, in order to use this property, we’re going to have to transpose the matrix 𝐡. We do this by firstly taking the first row of matrix 𝐡, which is zero, negative two π‘₯, negative 63. And these entries become the first column of the matrix 𝐡 transpose, like so. Then, we take the entries in the second row of matrix 𝐡. That’s 𝑧 plus 14, zero, three 𝑧. And these entries become the second column of the matrix 𝐡 transpose. And finally, we take the entries in the third row of matrix 𝐡. That’s negative three 𝑦 plus six π‘₯, negative 12, and zero. And these entries become the entries of the third column in matrix 𝐡 transpose. So that gives us the matrix 𝐡 transpose.

So now, we’re going to use this property: 𝐡 transpose equals negative 𝐡. And remember, this holds true because we know that the matrix 𝐡 is skew-symmetric. Well, we just found the matrix 𝐡 transpose, and we’re going to set this equal to negative matrix 𝐡. Making this matrix negative changes the sign of each entry or term in the matrix. So the zero entries will stay the same, and these four entries are going to change sign. We’ve got to be careful with these two entries that have two parts with them because the negative applies to both of the values in the parentheses.

And now we can look at equating these two matrices. We can do this by comparing the entries in the same positions in each matrix. For instance, because these two matrices are equal, that must mean that 𝑧 plus 14 is equal to two π‘₯. It also means that negative three 𝑦 plus six π‘₯ is equal to 63 and that negative 12 is equal to negative three 𝑧. And that one is probably a good place to start. If negative 12 is equal to negative three 𝑧, then 𝑧 must be equal to four.

And then we can use that to solve the top equation: 𝑧 plus 14 equals two π‘₯. Substituting 𝑧 equals four gives us that four plus 14 equals two π‘₯, which is 18 equals two π‘₯. And that gives us that π‘₯ is equal to nine.

We can then use this to solve the second equation: negative three 𝑦 plus six π‘₯ equals 63. That gives us that negative three 𝑦 plus six multiplied by nine equals 63. And we know that six multiplied by nine is equal to 54. And then subtracting 54 from both sides of this equation gives us that negative three 𝑦 is equal to nine.

Remember that our question asked us to find the value of π‘₯ plus 𝑦 plus 𝑧. And we now know the values of π‘₯, 𝑦, and 𝑧. So π‘₯ plus 𝑦 plus 𝑧 equals nine minus three add four. And that gives us 10.

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