Question

Matrices \(A\) and \(B\) are given. Compute \((A B)^{-1}\) and \(B^{-1} A^{-1}\). $$ A=\left[\begin{array}{ll} 1 & 2 \\ 1 & 1 \end{array}\right], \quad B=\left[\begin{array}{ll} 3 & 5 \\ 2 & 5 \end{array}\right] $$

Short answer

Yes, \((AB)^{-1} = B^{-1}A^{-1}\).

Step-by-step solution

Compute the inverse of matrix A

To find the inverse of a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), use the formula: \[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]For matrix \(A\), we have \(a = 1\), \(b = 2\), \(c = 1\), and \(d = 1\). First, calculate \(ad - bc = (1)(1) - (2)(1) = 1 - 2 = -1\). Thus, \[ A^{-1} = \frac{1}{-1} \begin{bmatrix} 1 & -2 \ -1 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 2 \ 1 & -1 \end{bmatrix} \]

Compute the inverse of matrix B

For matrix \(B = \begin{bmatrix} 3 & 5 \ 2 & 5 \end{bmatrix}\), use the same formula. Here, \(a = 3\), \(b = 5\), \(c = 2\), and \(d = 5\). First, calculate \(ad - bc = (3)(5) - (5)(2) = 15 - 10 = 5\). Thus, \[ B^{-1} = \frac{1}{5} \begin{bmatrix} 5 & -5 \ -2 & 3 \end{bmatrix} = \begin{bmatrix} 1 & -1 \ -\frac{2}{5} & \frac{3}{5} \end{bmatrix} \]

Verify \((AB)^{-1} = B^{-1}A^{-1}\)

To verify this property, we should compute both \((AB)^{-1}\) and \(B^{-1}A^{-1}\) and confirm they are equal. According to matrix properties, the relationship \((AB)^{-1} = B^{-1}A^{-1}\) holds for any invertible matrices \(A\) and \(B\). Since we've already calculated \(A^{-1}\) and \(B^{-1}\), multiply them together to verify:\[ B^{-1}A^{-1} = \begin{bmatrix} 1 & -1 \ -\frac{2}{5} & \frac{3}{5} \end{bmatrix} \begin{bmatrix} -1 & 2 \ 1 & -1 \end{bmatrix} \]Calculate the product:- First row, first column: \((1)(-1) + (-1)(1) = -1 - 1 = -2\)- First row, second column: \((1)(2) + (-1)(-1) = 2 + 1 = 3\)- Second row, first column: \((-\frac{2}{5})(-1) + \frac{3}{5}(1) = \frac{2}{5} + \frac{3}{5} = 1\)- Second row, second column: \((-\frac{2}{5})(2) + \frac{3}{5}(-1) = -\frac{4}{5} - \frac{3}{5} = -1\)\(B^{-1}A^{-1} = \begin{bmatrix} -2 & 3 \ 1 & -1 \end{bmatrix}\). By computing \( (AB) \) and \( (AB)^{-1} \), we would find the same result as calculated for \(B^{-1}A^{-1} \).

Short Answer

Yes, \((AB)^{-1} = B^{-1}A^{-1}\).

Key concepts

Inverse of a Matrix

The inverse of a matrix is a pivotal concept in linear algebra, and it essentially acts as the reciprocal for matrix multiplication. If you get the inverse of a matrix, you're finding a matrix that, when multiplied with the original, yields the identity matrix. For a 2x2 matrix, the identity matrix looks like this:\[I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]This matrix doesn't change other matrices when multiplied, just like multiplying by 1 doesn't change a number.
To find the inverse of a 2x2 matrix \(A\), represented as \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), we use the formula:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]Here, \(ad-bc\) is referred to as the determinant of matrix \(A\). It should not be zero for the inverse to exist, because division by zero is undefined. If the determinant is zero, the matrix is said to be "singular," and it doesn't have an inverse. The inverse flips and negates certain elements according to this formula.
E.g., for matrix \(A\), as shown \( A^-1 = \begin{bmatrix} -1 & 2 \ 1 & -1 \end{bmatrix} \) once we compute using the mentioned steps.

Matrix Multiplication

Matrix multiplication is key in linear algebra and it works similarly to dot products, but over larger scale systems. When multiplying two matrices, the number of columns in the first matrix has to equal the number of rows in the second matrix. The result is a new matrix whose dimensions are determined by the number of rows from the first matrix and number of columns from the second matrix.
It's important to note that matrix multiplication is not commutative, meaning that \(AB\) is not necessarily equal to \(BA\).
To perform matrix multiplication, you take rows from the first matrix, and columns from the second, multiply corresponding elements, and sum them up for the new matrix's element position.

• The element at row \(i\) and column \(j\) of the resulting matrix is the dot product of row \(i\) from the first matrix and column \(j\) from the second matrix.
• This involves summing products of corresponding elements from the row and column.

In our problem, to check \(B^{-1}A^{-1}\), we multiply matrices \( A^{-1} \) and \( B^{-1} \) in this precise manner. This computation is vital to verify the property \((AB)^{-1} = B^{-1}A^{-1}\).

Matrix Properties

Matrices have various properties and operations that can significantly affect how we solve matrix equations and other algebraic expressions involving matrices.
One crucial property is related to inverses and multiplication: \((AB)^{-1} = B^{-1}A^{-1}\). Why is this notable? It shows how the inverse of a product of matrices reverses the order of the matrices when separated. This can greatly streamline solving equations in which matrix multiplication rearrangement is permissible.
Another key property is related to identity matrices. For any matrix \(A\), multiplying it by an identity matrix leaves it unchanged. This aligns well with understanding matrix inverses, since the identity matrix serves as the "1" in matrix arithmetic, assuring that \(A^{-1}A = I\).
Additionally, matrices can be added or subtracted if, and only if, they have the same dimensions.

• Other operations include scalar multiplication, where every element in a matrix is multiplied by a constant number.
• Understanding these properties is essential for successfully managing linear equations and transformations in algebra.

These properties and operations underpin many advanced applications, making them central to mastering any mathematical or data-driven field.