Problem 10 - Differential Equations and Linear Algebra

Question

Let \(A=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 5 & 7\end{array}\right]\) and \(B=\left[\begin{array}{rl}3 & b \\ -4 & a \\ a & b\end{array}\right].\) (a) Compute \(A B\) and determine the values of \(a\) and \(b\) such that \(A B=I_{2}.\) (b) Using the values of \(a\) and \(b\) obtained in (a), compute BA.

Step-by-Step Solution

Answer

To summarize, we computed the matrix product AB and found the values of \(a=2\) and \(b=-1\) such that \(AB = I_{2}\). We then calculated the matrix product BA using these values, and found that \(BA=I_{2}\) as well.

Step 1: Compute the matrix product AB

To compute the matrix product AB, we will follow the standard rules of matrix multiplication. Recall the rule for matrix multiplication: (AB)ij = Σ (Aik * Bkj) where k runs from 1st to the last element of the row/column. In this case, A is a 2x3 matrix and B is a 3x2 matrix, so the resulting matrix product will be a 2x2 matrix. AB = \[\left[\begin{array}{lll}1 & 2 & 3 \\\ 2 & 5 & 7\\\end{array}\right] \left[\begin{array}{rl}3 & b \\\ -4 & a \\\ a & b\\\end{array}\right] \] The matrix product will have the following form: AB = \[\left[\begin{array}{ll}(1*3 + 2*(-4) + 3*a) & (1*b + 2*a + 3*b) \\\ (2*3 + 5*(-4) + 7*a) & (2*b + 5*a + 7*b) \\\end{array}\right] \] Calculate the entries: (1*3 + 2*(-4) + 3*a) = 3 - 8 + 3a = -5 + 3a (1*b + 2*a + 3*b) = b + 2a + 3b = 2a + 4b (2*3 + 5*(-4) + 7*a) = 6 - 20 + 7a = -14 + 7a (2*b + 5*a + 7*b) = 2b + 5a + 7b = 5a + 9b Now, AB = \[\left[\begin{array}{ll} -5 + 3a & 2a + 4b \\\ -14 + 7a & 5a + 9b \\\end{array}\right] \]

Step 2: Find the values of a and b such that AB = I2

Now, we want to find the values of a and b such that the matrix product AB is equal to the identity matrix I2. The identity matrix I2 is given by: I2 = \[\left[\begin{array}{ll} 1 & 0 \\\ 0 & 1 \\\end{array}\right] \] To find the values of a and b that satisfy AB = I2, we can set the entries of AB equal to the entries of I2: -5 + 3a = 1 => 3a = 6 => a = 2 2a + 4b = 0 => 2*2 + 4b = 0 => 4 + 4b = 0 => b = -1 So, we find that a = 2 and b = -1.

Step 3: Compute the matrix product BA with the values of a and b

Now that we have the values of a and b, we can compute the matrix product BA. Using the given values of a and b, we have: A = \[\left[\begin{array}{lll}1 & 2 & 3 \\\ 2 & 5 & 7\end{array}\right] \] B = \[\left[\begin{array}{rl}3 & -1 \\\ -4 & 2 \\\ 2 & -1\end{array}\right] \] Now, calculate the matrix product BA. Since A is a 2x3 matrix and B is a 3x2 matrix, the resulting matrix product will be a 2x2 matrix: BA = \[\left[\begin{array}{ll} (1*3 + 2*(-4) + 3*2) & (1*(-1) + 2*2 + 3*(-1)) \\\ (2*3 + 5*(-4) + 7*2) & (2*(-1) + 5*2 + 7*(-1)) \\\end{array}\right] \] Calculate the entries: (1*3 + 2*(-4) + 3*2) = 3 - 8 + 6 = 1 (1*(-1) + 2*2 + 3*(-1)) = -1 + 4 - 3 = 0 (2*3 + 5*(-4) + 7*2) = 6 - 20 + 14 = 0 (2*(-1) + 5*2 + 7*(-1)) = -2 + 10 - 7 = 1 Thus, BA = \[\left[\begin{array}{ll} 1 & 0 \\\ 0 & 1 \\\end{array}\right] \] Find out that BA is also equal to the identity matrix I2.

Key Concepts

Understanding Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra that combines two matrices to produce a third matrix. Here are some essential details:

• The product of an m × n matrix A and an n × p matrix B is an m × p matrix C.
• Each element cᵢⱼ of the matrix C is computed by taking the dot product of the i-th row of A and the j-th column of B.
• If A is a matrix with dimensions that do not align properly with B, then the multiplication is not defined.

When working through matrix multiplication problems, it's crucial to perform each multiplication and addition step carefully to avoid errors. In the context of the given exercise, we find that computing AB requires multiplying each row of A with each column of B, resulting in a 2 × 2 matrix since A is 2 × 3 and B is 3 × 2. It's similar to following a recipe where combining specific ingredients in the right order and proportion creates a specific dish—in this case, the 'dish' is the product matrix C.

The Determinant of a Matrix

The determinant is a scalar value that can be computed from the elements of a square matrix. It has numerous important properties and applications within linear algebra, including solving systems of linear equations, finding the inverse of a matrix, and evaluating if a matrix is invertible.

• The determinant can only be calculated for square matrices (n × n).
• The symbol used to denote the determinant of a matrix A is det(A) or |A|.
• A key property of the determinant is that if the determinant of a matrix is zero, the matrix does not have an inverse and is considered singular or non-invertible.

For matrix multiplication problems involving the determinant, you may be asked to verify certain properties, such as the determinant of the product of two matrices being equal to the product of their determinants, represented as det(AB) = det(A)det(B).

Invertible Matrices and Their Criteria

In linear algebra, an invertible matrix, also known as a non-singular matrix or a nondegenerate matrix, is a matrix that has an inverse. The inverse of a matrix A is denoted by A^-1 and is unique for each invertible matrix.

• A matrix is invertible if it is a square matrix and its determinant is not equal to zero (det(A) ≠ 0).
• Finding A^-1 enables us to solve matrix equations such as AX = B by multiplying both sides by A^-1, resulting in X = A^-1B.
• In the context of the exercise, finding that the product AB equals the identity matrix suggests that B is the inverse of A with the specific values of a and b that were calculated.

An invertible matrix can be thought of as an 'undo' function for matrix transformation—it rolls back what the original matrix did. It's a powerful concept and is widely used for solving equations, in optimizations, and more broadly, any time you need to reverse the effect of a linear transformation.