Question Video: Finding the Correct Expression for the Determinant of a Scalar Multiplied by a Matrix Mathematics

If 𝐴 is a matrix of size 𝑛 Γ— 𝑛, which of the following is equivalent to det (π‘˜π΄)? [A] π‘›π‘˜ det (𝐴) [B] π‘˜^(𝑛) det (𝐴) [C] det (𝐴) [D] π‘˜^(2𝑛) det (𝐴) [E] 1/π‘˜ det (𝐴)

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Video Transcript

If 𝐴 is a matrix of size 𝑛 by 𝑛, which of the following is equivalent to determinant of π‘˜ times 𝐴? Is it option (A) 𝑛 times π‘˜ multiplied by the determinant of 𝐴? Option (B) π‘˜ to the 𝑛th power multiplied by the determinant of 𝐴. Option (C) the determinant of 𝐴. Option (D) π‘˜ to the power of two 𝑛 times the determinant of 𝐴. Or is it option (E) one over π‘˜ multiplied by the determinant of 𝐴?

In this question, we’re given a square matrix 𝐴. And we’re asked to determine which of five given expressions is equivalent to the determinant of π‘˜ multiplied by 𝐴, where it’s worth noting that π‘˜ is a scalar value. We can answer this question directly by just recalling one of the properties of the determinant. We can recall for any square matrix 𝐴 of order 𝑛 by 𝑛 and scalar π‘˜, the determinant of π‘˜ times 𝐴 is equal to π‘˜ to the 𝑛th power multiplied by the determinant of 𝐴. And we can see that this is given in option (B).

Now, we could end here. However, it’s not very enlightening to answer a question just by recalling a property. So instead, let’s discuss why this property holds true. One way of proving this property is to use the definition of a determinant. We just need to show if we multiply a matrix by π‘˜, we can take out a factor of π‘˜ to the 𝑛th power to get an expression for the determinant of 𝐴. However, this proof is quite difficult. So instead, we’ll go over an easier proof where we use a different property.

We’re going to use the fact that if 𝐴 and 𝐡 are square matrices of the same order, then the determinant of 𝐴 times 𝐡 is equal to the determinant of 𝐴 multiplied by the determinant of 𝐡. To use this property to evaluate the determinant of π‘˜ times 𝐴, we’re going to need to rewrite our scalar multiplication in terms of matrix multiplication. And one way of doing this is to use the definition of matrix multiplication and the identity matrix.

Multiplying any matrix by π‘˜ times the identity matrix will multiply all of its entries by π‘˜, assuming that the orders match up. So we can rewrite this determinant as the determinant of π‘˜ multiplied by the identity matrix of order 𝑛 times 𝐴. Now, we can apply our property to write this as the product of two determinants. We get that this is equal to the determinant of π‘˜ multiplied by the identity matrix of order 𝑛 times the determinant of 𝐴.

And at this point, we can notice something interesting. π‘˜ multiplied by the identity matrix is a square matrix of order 𝑛 by 𝑛, where all of the entries on the main diagonal are π‘˜ and all of the entries not on the main diagonal are zero. In particular, we can notice this is a diagonal matrix because all of the entries not on the main diagonal are equal to zero. And now we can evaluate this determinant by recalling the determinant of any square diagonal matrix is the product of all entries on its main diagonal. In this case, we have 𝑛 entries on the main diagonal and all of these are equal to π‘˜. So, we get 𝑛 factors of π‘˜.

Therefore, the determinant of π‘˜ times 𝐴 is π‘˜ to the 𝑛th power multiplied by the determinant of 𝐴, which once again we can see is the answer given in option (B).

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