Question

Matrices \(A\) and \(B\) are given. Solve the matrix equation \(A X=B\). $$ A=\left[\begin{array}{ll} 2 & 2 \\ 3 & 1 \end{array}\right], \quad B=I_{2} $$

Short answer

Matrix \(X\) is the inverse of \(A\): \(\begin{bmatrix} -\frac{1}{4} & \frac{1}{2} \\ \frac{3}{4} & -\frac{1}{2} \end{bmatrix}\).

Step-by-step solution

Understand the Problem

We need to find matrix \(X\) such that \(AX = B\), where \(A\) is a given matrix, and \(B\) is the identity matrix \(I_2\), which is a 2x2 identity matrix \(\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\).

Determine Inverse of Matrix A

To find \(X\), we need to isolate it by solving for \(X = A^{-1}B\). The first step here is to compute the inverse of matrix \(A\). A 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\) has an inverse given by \(A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\), provided that the determinant \(ad-bc\) is not zero.

Calculate Determinant of A

The determinant of matrix \(A = \begin{bmatrix} 2 & 2 \ 3 & 1 \end{bmatrix}\) is calculated as follows: \(\text{det}(A) = (2)(1) - (3)(2) = 2 - 6 = -4\). Since the determinant is not zero, \(A\) is invertible.

Find Inverse of A

Using the inverse formula, we find:\[ A^{-1} = \frac{1}{-4} \begin{bmatrix} 1 & -2 \ -3 & 2 \end{bmatrix} = \begin{bmatrix} -\frac{1}{4} & \frac{1}{2} \ \frac{3}{4} & -\frac{1}{2} \end{bmatrix} \]

Multiply A^{-1} with B (Identity Matrix)

Now, compute \(X = A^{-1}B\). Since \(B = I_2\), \(X = A^{-1}I_2 = A^{-1}\), thus \(X\) is the same as \(A^{-1}\).\[ X = \begin{bmatrix} -\frac{1}{4} & \frac{1}{2} \ \frac{3}{4} & -\frac{1}{2} \end{bmatrix} \]

Verify the Solution

Verify by multiplying \(A\) with \(X\) to confirm we get identity matrix \(B\).\[ AX = \begin{bmatrix} 2 & 2 \ 3 & 1 \end{bmatrix} \begin{bmatrix} -\frac{1}{4} & \frac{1}{2} \ \frac{3}{4} & -\frac{1}{2} \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \] This confirms the solution is correct.

Key concepts

Matrix Inverse

Solving matrix equations often involves finding the inverse of a matrix. To solve an equation like \(AX = B\), we need to isolate \(X\), and that requires the inverse of matrix \(A\). However, not all matrices have an inverse. For a matrix to have an inverse, it must be square (same number of rows and columns) and have a non-zero determinant. More on determinants later!

For a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the formula for the inverse is:

• \(A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\)

Here, the inverse is derived by swapping the diagonal elements (\(a\) with \(d\)), changing the signs of the off-diagonal elements (\(b\) and \(c\)), and dividing by the determinant \(ad-bc\). It's crucial to remember that the determinant must not be zero, as dividing by zero is undefined. In the given problem, the determinant was -4, which allowed us to compute the inverse.

Determinant of a Matrix

The determinant of a matrix is a special number that can be calculated from its elements. For a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant is found using

• \(det(A) = ad - bc\)

The determinant provides essential information about the matrix. Specifically, it tells us if a matrix is invertible. If the determinant is zero, the matrix doesn't have an inverse. In our problem, we calculated the determinant of \(A = \begin{bmatrix} 2 & 2 \ 3 & 1 \end{bmatrix}\) as \(-4\). Since \(-4\) is not zero, \(A\) has an inverse.

Understanding the role of the determinant can also help you delve deeper into the matrix's properties, like its scaling factor. In geometry, it can indicate the area or volume distortion caused by the transformation represented by the matrix.

Identity Matrix

An identity matrix is a special type of square matrix. It acts like 1 does in multiplication for real numbers. The identity matrix is denoted as \(I\) and has 1s on the diagonal and 0s elsewhere. For a 2x2 identity matrix, it looks like this:

• \(I_2 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\)

The identity matrix has unique properties:

• Multiplying any matrix \(A\) by the identity matrix of compatible size (\(I\)) results in matrix \(A\). That means \(AI = A\) and \(IA = A\).
• It is a neutral element for matrix multiplication.

In our exercise, we set \(B\) as the identity matrix \(I_2\). When calculating \(X = A^{-1}B\), having \(B = I_2\) simplifies the multiplication because multiplying by the identity matrix does not change \(A^{-1}\). Hence, \(X = A^{-1}\). This functionality makes the identity matrix very useful in solving matrix equations.