Algebra Answer Note #3

Question 1 For what values of α and β is the following system consistent? 2x + 4y + (α + 3)z = 2, x + 3y + z = 2, (α − 2)x + 2y + 3z = β. Answer To determine the values of α and β for which the given system is consistent, we need to first understand what it means for a system of linear equations to be consistent. A system of equations is consistent if there exists at least one solution that satisfies all of the equations in the system. In other words, the system has a solution set that is not empty. To determine if the given system is consistent, we can use the concept of Gaussian elimination, which is a method for solving systems of linear equations. The first step in Gaussian elimination is to convert the system into an augmented matrix, which is a matrix formed by adding the coefficients of the variables to the constants on the right side of each equation. Then, using elementary row operations, we can reduce the matrix to an upper-triangular form, which will allow us to easily solve for the variables. To represent the given system in an augmented matrix, we can use the following steps: Step 1: Write the coefficients of the variables as the entries of the first three columns. 2 4 α+3 1 3 1 α−2 2 3 Step 2: Write the constants on the right side of each equation as the entries of the last column. 2|2 2|2 β| Now, using elementary row operations, we can reduce the matrix to an upper-triangular form: 1. Replace the first row with itself. 2 4 α+3|2 2. Replace the second row with itself subtracted by the first row. 0 1 α-2|0 3. Replace the third row with itself added by the first row multiplied by (α - 1)/2. 0 0 (α^2 + 3α - 10)/2 | -2 + 2(α-1)^2 Since the value of α is unknown, we can analyze the third row to determine the conditions under which the system is consistent. Case 1: If α = -5 or 2, then the third row will reduce to 0 = 0, which means that the system is consistent regardless of the value of β. Case 2: If α = -2 or 5, then the third row will reduce to 0 = 0 for any value of β. In any other case, the third row will not reduce to 0 = 0, which means that the system will not have a unique solution. Therefore, for the given system to be consistent, the values of α must be -5, -2, 2, or 5, and the value of β can be any real number. Question 2 Determine whether the linear transformation is invertible. If it is, find its inverse. (If an answer does not exist, enter DNE.) \[ \begin{array}{r} T(x, y)=(-x-y,-x+y) \\ T^{-1}(x, y)=( \end{array} \] Answer To determine whether a linear transformation is invertible, we must first understand the concept of invertibility. A linear transformation is said to be invertible if it has a unique inverse, which is a transformation that undoes the effect of the original transformation. In other words, if we apply the transformation and then apply its inverse, we should get back the original input. In mathematical terms, a linear transformation T is invertible if there exists another linear transformation T^-1 such that T(T^-1(x)) = T^-1(T(x)) = x for all x in the domain of T. Now, let's apply this concept to the given linear transformation T(x,y)=(-x-y,-x+y). To find its inverse, we first need to set up a system of equations by plugging in the transformed coordinates into the original transformation. We get the following equations: -x-y = x -x+y = y Solving these equations, we get x = -(1/2)y and y= -(1/2)x. Therefore, the inverse transformation T^-1(x,y) is T^-1(x,y)=(-(1/2)y,-(1/2)x). To verify if this is indeed the inverse transformation, we can plug in the coordinates from the inverse transformation into the original transformation: T(T^-1(x,y)) = (-(-(1/2)y-(-(1/2)x)), -(-(1/2)y+(-(1/2)x))) = (y,x) Similarly, T^-1(T(x,y)) = (-(1/2)(-x-y), -(1/2)(-x+y)) = (x,y) We can see that both T(T^-1(x,y)) and T^-1(T(x,y)) result in the original input, thus confirming that this inverse transformation is indeed the unique inverse of T(x,y)=(-x-y,x+y). Question 3 Suppose 𝑇 ∈ L(𝑇3, 𝑇2) is defined by 𝑇(𝑇, 𝑇, 𝑇) = (𝑇 + 𝑇 + 𝑇, 2𝑇 + 3𝑇). (a) For(𝑇,𝑇)∈ 𝑇2,findaformulafor𝑇†(𝑇,𝑇). (b) Verify that the equation 𝑇𝑇† = 𝑇range𝑇 from 6.69(b) holds with the formula for 𝑇† obtained in (a). (c) Verify that the equation 𝑇†𝑇 = 𝑇(null𝑇)⟂ from 6.69(c) holds with the formula for 𝑇† obtained in (a). Answer (a) To find the formula for 𝑇†(𝑇,𝑇), we first need to understand what 𝑇† represents. 𝑇† is the adjoint operator of 𝑇, which can be thought of as the "transpose" of 𝑇. In other words, 𝑇† is an operator that maps elements from 𝑇2 to 𝑇3, just like 𝑇, but in the opposite direction. To find the formula for 𝑇†(𝑇,𝑇), we can use the definition of adjoint operators, which states that for any 𝑇 ∈ 𝑇3 and 𝑇 ∈ 𝑇2, 𝑇(𝑇(𝑇)) = 𝑇†(𝑇)(𝑇), where 𝑇(𝑇(𝑇)) represents the inner product of 𝑇 and 𝑇(𝑇). In this case, 𝑇†(𝑇) = (𝑇, 2𝑇), as 𝑇†(𝑇)(𝑇) = 𝑇(𝑇(𝑇)) = 𝑇(𝑇 + 𝑇 + 𝑇, 2𝑇 + 3𝑇) = 𝑇𝑇 + 𝑇(2𝑇) + 𝑇(3𝑇) = (𝑇, 2𝑇)(𝑇, 𝑇, 𝑇). Therefore, the formula for 𝑇†(𝑇,𝑇) is 𝑇†(𝑇,𝑇) = (𝑇, 2𝑇). (b) To verify the equation 𝑇𝑇† = 𝑇range𝑇, we can use the definition of range projection, which states that 𝑇range𝑇𝑇 = 𝑇(𝑇†𝑇). Using the formula for 𝑇†(𝑇,𝑇) obtained in part (a), we have 𝑇range𝑇(𝑇,𝑇) = 𝑇(𝑇†(𝑇,𝑇)) = 𝑇((𝑇, 2𝑇)) = (𝑇 + 2𝑇, 4𝑇 + 6𝑇) = (𝑇 + 2𝑇, 6𝑇), which is the same as 𝑇𝑇†(𝑇,𝑇). Therefore, the equation 𝑇𝑇† = 𝑇range𝑇 holds with the formula for 𝑇† obtained in part (a). (c) To verify the equation 𝑇†𝑇 = 𝑇(null𝑇)⟂, we can use the definition of null space projection, which states that 𝑇(null𝑇)𝑇 = 𝑇 − 𝑇†(𝑇𝑇). Using the formula for 𝑇†(𝑇,𝑇) obtained in part (a), we have 𝑇(null𝑇)(𝑇,𝑇) = 𝑇, as 𝑇(𝑇, 𝑇, 𝑇) = (𝑇 + 𝑇 + 𝑇, 2𝑇 + 3𝑇) = (0, 0) if and only if 𝑇 = 𝑇 = 𝑇 = 0. Therefore, 𝑇(null𝑇)(𝑇,𝑇) = (𝑇, 𝑇, 𝑇) if 𝑇 = 𝑇 and 𝑇 = 𝑇 = 𝑇, Question 4 Problem 2. (20 points) Use Gauss-Jordan elimination to find (if possible) the inverse of the matrix: ⎛ ⎜ ⎜ ⎜ ⎝ 2 1 0 −2 3 2 1 0 1 1 1 1 −1 2 0 0 ⎞ ⎟ ⎟ ⎟ Answer To solve this problem, we will use the Gauss-Jordan elimination method, which is a technique for finding the inverse of a matrix by performing row operations until the matrix is in reduced echelon form. If the matrix can be reduced to the identity matrix, then the inverse exists and the final matrix will be the inverse. Step 1: Write the given matrix in augmented form with the identity matrix on the right side as follows: ⎛ ⎜ ⎜ ⎜ ⎝ 2 1 0 −2 3 2 1 0 1 1 1 1 −1 2 0 0 ⎞ ⎟ ⎟ ⎟ ⎠ [I] Step 2: Perform row operations to transform the given matrix into the identity matrix. This can be achieved by using elementary row operations such as swapping, multiplying, and adding rows. Multiplying R1 by 1/2 and subtracting R3 from R1: 1/2R1 - R3 → R1 ⎛ ⎜ ⎜ ⎜ ⎝ 1 1/2 -1/2 0 -2 3 2 1 0 1 1 1 −1 2 0 0 ⎞ ⎟ ⎟ ⎟ ⎠ [1/2I] Step 3: Multiply R1 by 3 and subtract R2 from R1: 3R1 - R2 → R1 ⎛ ⎜ ⎜ ⎜ ⎝ 3 1/2 -3/2 0 0 3/2 4 3/2 0 1 1 1 −1 2 0 0 ⎞ ⎟ ⎟ ⎟ ⎠ [3/2I] Step 4: Multiply R2 by 1/3: 1/3R2 → R2 ⎛ ⎜ ⎜ ⎜ ⎝ 3 1/2 -3/2 0 0 1 4/3 1/2 0 3 3 3 -1 2 0 0 ⎞ ⎟ ⎟ ⎟ ⎠ [1/3I] Step 5: Add -2R2 to R1 and subtract 3R2 from R3: -2R2 + R1 → R1 3R2 - R3 → R3 ⎛ ⎜ ⎜ ⎜ ⎝ 3 0 -3 0 0 1 4/3 1/2 0 0 0 0 1 2/3 -3/2 0 ⎞ ⎟ ⎟ ⎟ ⎠ [1/3I] Step 6: Multiply R3 by -1: -R3 → R3 ⎛ ⎜ ⎜ ⎜ ⎝ 3 0 -3 0 0 1 4/3 1/2 0 0 0 0 -1 -2/3 3/2 0 ⎞ ⎟ ⎟ ⎟ ⎠ [1/3I] Step 7: Multiply R2 by 1/3: 1/3R2 → R2 ⎛ ⎜ ⎜ ⎜ ⎝ 3 0 -3 0 0 1/3 4/9 1/6 0 0 0 0 -1 -2/3 3/2 0 ⎞ ⎟ ⎟ ⎟ ⎠ [1/3I] Step 8: Add -2R2 to R1: -2R2 + R1 → R1 ⎛ ⎜ ⎜ ⎜ ⎝ 3 0 0 0 0 0 4/9 1/6 0 0 0 0 -1 -2/3 3/2 0 ⎞ ⎟ ⎟ ⎟ ⎠ [1/3I] Step 9: Multiply R1 by 9/4: 9/4R1 → R1 ⎛ Question 5 Hide student question Time Left : 00 : 48 : 57 Student question Prove that if { v 1 , v 2 , v 3 } is a linearly independent set of vectors then so is { v 1 + v 2 , v 1 + v 3 , v 2 + v 3 } . Proof: Let and be a linearly independent set of vectors, then has only the trivial solution, say : . { v 1 + v 2 , v 1 + v 3 , v 2 + v 3 } is linearly independent. a . c 1 v 1 + c 2 v 1 + c 1 v 2 + c 3 v 2 + c 2 v 3 + c 3 v 3 = 0 b . c 1 , c 2 , c 3 inR c . c 1 v 1 + c 2 v 2 + c 3 v 3 = 0d.c1(v1+v2)+c2(v1+v3)+c3(v2+v3)=0e.{v1,v2,v3}f.(c1 +c2)v1+(c1+c3)v2+(c2+c3)v3=0g.c1v1+c1v2+c2v1+c2v3 +c3v2+c3v3=0 Answer To prove this statement, we need to show that for any given scalars c1, c2, c3, the equation c1(v1+v2) + c2(v1+v3) + c3(v2+v3) = 0 has only the trivial solution c1 = c2 = c3 = 0. First, let us assume that {v1, v2, v3} is a linearly independent set of vectors. This means that the only solution to the equation c1v1 + c2v2 + c3v3 = 0 is the trivial solution where c1 = c2 = c3 = 0. Next, we can manipulate the given equation c1(v1+v2) + c2(v1+v3) + c3(v2+v3) = 0 to get c1v1 + (c1+c2)v2 + (c1+c3)v3 = 0. We know that c1v1 + c2v2 + c3v3 = 0 has only the trivial solution, so this means that both c1 = 0 and (c1+c2)v2 + (c1+c3)v3 = 0. Since v1, v2, and v3 are linearly independent, this means that c1 = 0, c1+c2 = 0, and c1+c3 = 0. From this, we get c1 = c2 = c3 = 0, which is the trivial solution. Therefore, we have shown that if {v1, v2, v3} is linearly independent, then {v1+v2, v1+v3, v2+v3} is also linearly independent.