Question

\(\mathrm{A}_{1}, \mathrm{~A}_{2}, \mathrm{~A}_{3} \ldots \mathrm{A}_{\mathrm{n}}\) and \(\mathrm{B}_{1}, \mathrm{~B}_{2}, \mathrm{~B}_{3} \ldots . \mathrm{B}_{\mathrm{n}}\) are non-singular square matrices order \(\mathrm{n}\) such that \(\mathrm{A}_{1} \mathrm{~B}_{1}=\mathrm{I}_{\mathrm{n}}, \mathrm{A}_{2} \mathrm{~B}_{2}=\mathrm{In}, \mathrm{A}_{3} \mathrm{~B}_{3}=\mathrm{In}-\mathrm{A}_{\mathrm{n}} \mathrm{B}_{\mathrm{n}}=\mathrm{I}_{\mathrm{n}}\), then \(\left(\mathrm{A}_{1} \mathrm{~A}_{2} \mathrm{~A}_{3}-\mathrm{A}_{\mathrm{n}}\right)^{-1}=\) (1) \(\mathrm{B}_{1} \mathrm{~B}_{2} \mathrm{~B}_{3}-\ldots \mathrm{B}_{\mathrm{n}}\) (2) \(\mathrm{B}_{1}^{-1} \mathrm{~B}_{2}^{-1} \mathrm{~B}_{3}^{-1}-\mathrm{B}_{\mathrm{n}}^{-1}\) (3) \(\mathrm{B}_{\mathrm{n}} \mathrm{B}_{\mathrm{n}-1} \mathrm{~B}_{\mathrm{n}-2}-\ldots_{1}\) (4) \(\mathrm{B}_{\mathrm{n}}{ }^{-1} \mathrm{~B}_{\mathrm{n}-1}{ }^{-1} \mathrm{~B}_{\mathrm{n}-2}{ }^{-1}-\mathrm{B}_{1}^{-1}\)

Short answer

Answer: The inverse of the product of matrices A is \(B_nB_{n-1}\ldots B_1\).

Step-by-step solution

Write the problem as a single equation

First, let's define the equation we want to solve: \((A_1A_2A_3\ldots A_n)^{-1}\)

Apply the property of inverse matrices

For any pair of non-singular matrices A and B such that \(AB = I_n\), we can use the property \((AB)^{-1} = B^{-1}A^{-1}\). This can be applied in our equation as follows: \((A_1A_2A_3\ldots A_{n-1}A_n)^{-1} = (A_n(A_1A_2A_3\ldots A_{n-1}))^{-1} = A_{n-1}^{-1}\ldots A_1^{-1}A_n^{-1}\)

Substitute given relationships between A and B matrices

We can now substitute the given relationships between A and B matrices into our equation: \(A_{n-1}^{-1}\ldots A_1^{-1}A_n^{-1} = B_{n-1}\ldots B_1B_n\)

Rearrange the order of matrices B

Note that the order of the product matters in matrix multiplication. We can rearrange the order of the product of matrices B to match the given options: \(B_nB_{n-1}\ldots B_1 = (A_1A_2A_3\ldots A_n)^{-1}\)

Identify the correct option

Now we can see that the correct option is (3): $\mathrm{B}_{\mathrm{n}} \mathrm{B}_{\mathrm{n}-1} \mathrm{~B}_{\mathrm{n}-2}-\ldots_{1}$

Key concepts

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra that involves the multiplication of two matrices to produce a new matrix. This operation is not as straightforward as simple arithmetic multiplication. The key idea is that to multiply two matrices, the number of columns in the first matrix must equal the number of rows in the second matrix.
Matrix multiplication is defined mathematically as follows: if we have two matrices, \[A = \begin{bmatrix}a_{11} & a_{12} \a_{21} & a_{22}\end{bmatrix}, B = \begin{bmatrix}b_{11} & b_{12} \b_{21} & b_{22}\end{bmatrix},\]then their product, AB is given by: \[AB = \begin{bmatrix}a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22}\end{bmatrix}\]This operation is not commutative, meaning \(AB eq BA\), so the order in which matrices are multiplied is important. You can't just switch them around.
In our exercise, recognizing the order of multiplication helps to find the correct inverse once relationships are known, leading to the correct identification of option (3).

Non-singular Matrices

Non-singular matrices are also called invertible or regular matrices. These matrices have a unique property: there exists another matrix, called the inverse, which when multiplied with the original matrix, yields the identity matrix. A square matrix is non-singular if it is not zero and its determinant is not zero.
The identity matrix, denoted as \(I_n\) for an \(n \times n\) matrix, acts like the number 1 in scalar multiplication. For any non-singular matrix \(A\), when \(AA^{-1} = A^{-1}A = I_n\), where \(A^{-1}\) is the inverse matrix of \(A\). This fundamental property is essential in solving the exercise. Each \(A_i\) and \(B_i\) in the exercise satisfies this, i.e., \(A_iB_i = I_n\), allowing us to manipulate them into a desired form and verify that statement is true for inverses as required in the solution.

Matrix Properties

Matrices, much like numbers, have various properties that guide how they can be manipulated and solved within equations. Understanding these properties is crucial to solving more complex problems like the exercise provided.
- **Associativity:** The product of matrices is associative, meaning for any matrices \(A\), \(B\), and \(C\), we have \((AB)C = A(BC)\). This property allows grouping of matrices differently when performing multiplication.- **Distributive Law:** Matrix multiplication is distributive over addition, meaning \(A(B + C) = AB + AC\). This allows breaking down complex equations into simpler parts.- **Inverse Property:** For non-singular matrices, \((AB)^{-1} = B^{-1}A^{-1}\) shows that the inverse of a product is the reverse product of the inverses. This was crucial in our exercise, where the order was significant in determining the correct sequence of inverses.- **Transpose and Symmetry:** Some matrices exhibit symmetry properties where the transpose of a matrix reflects it over its main diagonal. While not strictly needed here, the ability to recognize symmetric or transposable matrices can simplify solutions.
Grasping these properties ensures that matrix problems are tackled accurately and allows leveraging these in contexts like matrix inverses and solving linear equations effectively.