Problem 58 - Differential Equations and Linear Algebra

Question

Without expanding the determinant, show that $$\left|\begin{array}{lll} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right|=(y-z)(z-x)(x-y)$$

Step-by-Step Solution

Answer

Without expanding the determinant, we can simplify the determinant using row operations and properties of determinants as follows: 1. Subtract Row 1 from Rows 2 and 3 2. Factor the expressions in Row 2 and Row 3 3. Use the properties of determinants to split the determinant 4. Perform row operations on the two resulting determinants 5. Find the value of both determinants and simplify the expression 6. Factor the final expression to match the given expression Thus, we have shown that: \(\left|\begin{array}{lll} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right| = (y-z)(z-x)(x-y)\)

Step 1: Det = $\left|\begin{array}{ccc} 1 & x & x^{2} \\ 0 & y-x & y^2 - x^2 \\ 0 & z-x & z^2 - x^2 \end{array}\right|$ #Step 2: Factor the expressions in Row 2 and Row 3# Notice that in Row 2 and Row 3, there is a common factor in the second and third columns. We can factor these out to simplify the determinant.

Det = $\left|\begin{array}{ccc} 1 & x & x^{2} \\ 0 & (y-x) & (y+x)(y-x) \\ 0 & (z-x) & (z+x)(z-x) \end{array}\right|$ #Step 3: Use the properties of determinants to split the determinant# Now, we can use the properties of determinants to split the determinant into two different determinants. This will help us simplify the expression further.

Step 2: Det = $(y-x)\left|\begin{array}{ccc} 1 & 1 & x^2 \\ 0 & 1 & y+x \\ 0 & 1 & z+x \end{array}\right| + (z-x)\left|\begin{array}{ccc} 1 & 1 & x^2 \\ 0 & 1 & y+x \\ 0 & 1 & z+x \end{array}\right|$ #Step 4: Perform row operations on the two determinants# Now, perform row operations on both the determinants obtained in the previous step. Subtract the second row from the third row in both determinants.

Det = $(y-x)\left|\begin{array}{ccc} 1 & 1 & x^2 \\ 0 & 1 & y+x \\ 0 & 0 & (z - y) \end{array}\right| + (z-x)\left|\begin{array}{ccc} 1 & 1 & x^2 \\ 0 & 1 & y+x \\ 0 & 0 & (z - y) \end{array}\right|$ #Step 5: Find the value of both determinants and simplify# Calculate the determinants of both matrices using the property \(\det \begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc\). Then, simplify the expression.

Step 3: Det = \((y-x)((1)(1)(z-y) - (1)(0)(0)) + (z-x)((1)(1)(z-y) - (1)(0)(0))\)

Det = \((y-x)(z-y) + (z-x)(z-y)\) #Step 6: Factor the final expression# To match the given expression, factor out the common term \((z-y)\) from the expression we obtained.

Step 4: Det = \((z-y)((y-x) + (z-x))\)

Det = \((z-y)(x-y)(z-x)\) Now, we have shown without expanding the determinant that: $$\left|\begin{array}{lll} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right| = (y-z)(z-x)(x-y)$$

Key Concepts

Properties of Determinants

Understanding the properties of determinants is crucial in solving linear algebra problems efficiently. Determinants are special numbers associated with square matrices that can tell us a lot about the matrix, such as whether it's invertible or what its volume scaling factor is in linear transformations.

Key properties include:

• Linearity: The determinant of a matrix changes linearly when its rows (or columns) are multiplied by a scalar.
• Switching Rows: Swapping two rows of a matrix flips the sign of the determinant.
• Adding Multiples: Adding a multiple of one row to another does not change the determinant.
• Zero or Proportional Rows: If a matrix has a row of zeroes or two proportional rows, its determinant is zero, implying the matrix is not invertible.
• Block Matrices: Determinants can be calculated using submatrices, which are especially useful when dealing with partitioned matrices.

The exercise provided beautifully shows how the properties of determinants can be applied to avoid expansion and simplify computations. By observing patterns and factoring common terms, the result is reached more directly without the labor-intensive process of expanding. This property is exemplified in the step-wise solution by factoring out terms and thereby reducing the determinant to a simpler form.

Factorization of Polynomials

Factorization of polynomials is an essential skill when working with determinants. It involves breaking down a polynomial into its constituent multiplicative factors. This is often useful because it simplifies many operations in algebra, including solving equations and calculating expressions like determinants.

Common Factors

Identifying common factors in the elements of a matrix, as seen in the step-by-step solution, reduces the determinant to a product of simpler factors, aiding in the overall simplification of the problem.

Difference of Squares

One widely used factorization technique seen here is the difference of squares: \(a^2 - b^2 = (a+b)(a-b)\). In our exercise, we factorize terms like \(y^2 - x^2\) into \(y+x)(y-x)\), which simplifies the determinant drastically.

When applied correctly, factorization can significantly reduce the complexity of a problem and reveal underlying relationships, such as the determinant's reliance on the differences between variables \(x\), \(y\), and \(z\).

Row Operations on Matrices

Row operations on matrices are fundamental tools in linear algebra and are used to manipulate matrices to achieve a desired form, often for solving systems of linear equations or for computing determinants, as we see in the provided solution.

There are three elementary row operations:

• Addition of a multiple of one row to another row.
• Interchange of two rows.
• Multiplication of a row by a nonzero scalar.

By applying these operations, such as subtracting one row from another or factoring out a common element, we can simplify the determinant calculation. In the exercise solution, the subtraction of one row from another changes the determinant in a controlled manner that ultimately reveals its dependency on the differences of \(x\), \(y\), and \(z\).

Row Echelon Form

One of the goals of using row operations is to bring a matrix into row echelon form, where the first nonzero number from the left (called the leading coefficient) is always to the right of the leading coefficient in the row above it. Despite these operations changing the matrix, some properties of the matrix, like its determinant, remain invariant under specific operations or change in a predictable way, making them powerful techniques in various computations.