Science:Math Exam Resources/Courses/MATH221/December 2008/Question 03 (a)

MATH221 December 2008

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Question 03 (a)
Show that the determinant of the matrix $\displaystyle A={\begin{pmatrix}1&x&x^{2}\\1&y&y^{2}\\1&z&z^{2}\end{pmatrix}}$ is given by $\displaystyle (z-x)(y-x)(z-y)$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
How do elementary row operations, adding a multiple of one row to another or multiplying a row by a number c, affect the determinant of a matrix?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Recall two key properties of the determinant of a matrix: The determinant is unchanged if a multiple of one row is added to another. Multiplying a row by a number c changes the determinant by a factor of c. Start by performing two elementary row operations on A: Subtract the first row from the second row. Subtract the first row from the third row. The effect of these operations is to transform A to the matrix $B={\begin{pmatrix}1&x&x^{2}\\0&y-x&y^{2}-x^{2}\\0&z-x&z^{2}-x^{2}\end{pmatrix}}.$ By the above determinant property 1, $\det(A)=\det(B)$. The advantage gained in calculating $\det(B)$ is that its first column only contains a single non-zero entry, which is on the diagonal. Hence, when we calculate the determinant of $B$ we can reduce it to calculating the determinant of a 2 x 2 submatrix, $\det(B)=(1)\det {\begin{pmatrix}y-x&y^{2}-x^{2}\\z-x&z^{2}-x^{2}\end{pmatrix}}=\det {\begin{pmatrix}y-x&(y-x)(y+x)\\z-x&(z-x)(z+x)\end{pmatrix}}$ We can now use determinant property 2 to help us compute this 2 x 2 determinant, $\det {\begin{pmatrix}y-x&(y-x)(y+x)\\z-x&(z-x)(z+x)\end{pmatrix}}=(z-x)(y-x)\det {\begin{pmatrix}1&(y+x)\\1&(z+x)\end{pmatrix}}=(z-x)(y-x)((z+x)-(y+x))=(z-x)(y-x)(z-y).$ Therefore, $\det(A)=(z-x)(y-x)(z-y)$.

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