Math Problem Statement

If A = [[1, 2, 1], [2, 3, -1], [0, -1, 3]] and B = [[4, 2, 1], [-3, 0, 5], [-1, 1, 6]], find the following: AB, BA, A^-1, B^-1, (AB)^-1, (BA)^-1.

Solution

To solve the given matrix problem, we will follow these steps:

You are given two matrices:

B = \begin{pmatrix} 4 & 2 & 1 \\ -3 & 0 & 5 \\ -1 & 1 & 6 \end{pmatrix}.$$ We are required to: 1. Compute the product $$AB$$ and $$BA$$. 2. Find the inverses $$A^{-1}$$ and $$B^{-1}$$. 3. Use the results to compute $$(AB)^{-1}$$ and $$(BA)^{-1}$$. Let's start with the matrix multiplication $$AB$$ and $$BA$$. ### Step 1: Multiply $$A$$ and $$B$$ to find $$AB$$ The product of two matrices $$A$$ and $$B$$ is obtained by the formula: $$AB = \begin{pmatrix} a_{11}b_{11} + a_{12}b_{21} + a_{13}b_{31} & a_{11}b_{12} + a_{12}b_{22} + a_{13}b_{32} & a_{11}b_{13} + a_{12}b_{23} + a_{13}b_{33} \\ a_{21}b_{11} + a_{22}b_{21} + a_{23}b_{31} & a_{21}b_{12} + a_{22}b_{22} + a_{23}b_{32} & a_{21}b_{13} + a_{22}b_{23} + a_{23}b_{33} \\ a_{31}b_{11} + a_{32}b_{21} + a_{33}b_{31} & a_{31}b_{12} + a_{32}b_{22} + a_{33}b_{32} & a_{31}b_{13} + a_{32}b_{23} + a_{33}b_{33} \end{pmatrix}$$ ### Step 2: Multiply $$B$$ and $$A$$ to find $$BA$$ Similarly, you will find the matrix $$BA$$ using the same multiplication procedure but reversing the order. ### Step 3: Calculate inverses Once we compute the products $$AB$$ and $$BA$$, we'll proceed to find the inverses $$A^{-1}$$ and $$B^{-1}$$ using the formula for the inverse of a 3x3 matrix. The inverse of a matrix $$A$$ is given by: $$A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A),$$ where $$\text{det}(A)$$ is the determinant of $$A$$, and $$\text{adj}(A)$$ is the adjugate of $$A$$. I'll calculate each of these values for you now.Here are the results of the calculations: 1. **Product $$AB$$:** $$AB = \begin{pmatrix} -3 & 3 & 17 \\ 0 & 3 & 11 \\ 0 & 3 & 13 \end{pmatrix}$$ 2. **Product $$BA$$:** $$BA = \begin{pmatrix} 8 & 13 & 5 \\ -3 & -11 & 12 \\ 1 & -5 & 16 \end{pmatrix}$$ 3. **Inverse of $$A$$ ($$A^{-1}$$):** $$A^{-1} = \begin{pmatrix} -1.333 & 1.167 & 0.833 \\ 1 & -0.5 & -0.5 \\ 0.333 & -0.167 & 0.167 \end{pmatrix}$$ 4. **Inverse of $$B$$ ($$B^{-1}$$):** $$B^{-1} = \begin{pmatrix} -1.667 & -3.667 & 3.333 \\ 4.333 & 8.333 & -7.667 \\ -1 & -2 & 2 \end{pmatrix}$$ 5. **Inverse of $$AB$$ ($$(AB)^{-1}$$):** $$(AB)^{-1} = \begin{pmatrix} -0.333 & -0.667 & 1 \\ 0 & 2.167 & -1.833 \\ 0 & -0.5 & 0.5 \end{pmatrix}$$ 6. **Inverse of $$BA$$ ($$(BA)^{-1}$$):** $$(BA)^{-1} = \begin{pmatrix} 6.444 & 12.944 & -11.722 \\ -3.333 & -6.833 & 6.167 \\ -1.444 & -2.944 & 2.722 \end{pmatrix}$$ Would you like further details on any step? Here are some related questions: 1. How do you calculate the determinant of a 3x3 matrix? 2. What is the significance of matrix inverses in linear algebra? 3. Can two matrices always be multiplied together? Why or why not? 4. How do you interpret matrix multiplication geometrically? 5. How can the properties of inverse matrices be used in solving systems of linear equations? **Tip:** The inverse of a matrix is unique only if the matrix is square and its determinant is non-zero.

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Math Problem Analysis

Mathematical Concepts

Matrix Multiplication
Matrix Inversion
Linear Algebra

Formulas

Matrix Multiplication Formula: C = A * B
Inverse of a 3x3 Matrix: A^-1 = (1/det(A)) * adj(A)

Theorems

Properties of Matrix Multiplication
Inverse of a Matrix
Determinant and Adjugate Matrices

Suitable Grade Level

Undergraduate (Linear Algebra Course)