Question

Matrices \(A\) and \(B\) are given. Solve the matrix equation \(A X=B\). $$ \begin{array}{l} A=\left[\begin{array}{ll} -1 & -2 \\ -2 & -3 \end{array}\right] \\ B=\left[\begin{array}{lll} 13 & 4 & 7 \\ 22 & 5 & 12 \end{array}\right] \end{array} $$

Short answer

Matrix \( X \) is \( \begin{bmatrix} -5 & 2 & -3 \\ -4 & -3 & -2 \end{bmatrix} \).

Step-by-step solution

Analyze the equation

We need to solve the matrix equation \( AX = B \), where \( A \) and \( B \) are given matrices. Our goal is to find matrix \( X \).

Check matrix dimensions

Matrix \( A \) is a 2x2 matrix, and matrix \( B \) is a 2x3 matrix. To multiply these matrices, \( X \) must be a 2x3 matrix since it should match the dimensions of \( B \).

Find the inverse of matrix A

First, we need to find the inverse of matrix \( A \) (let's denote it as \( A^{-1} \)) to solve for \( X \). The inverse of a matrix \( A \) is found using the formula: \[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \]Where \( \text{det}(A) \) is the determinant of \( A \), and \( \text{adj}(A) \) is the adjoint of \( A \).

Calculate the determinant of A

The determinant of \( A \) is calculated as follows:\[ \text{det}(A) = (-1)(-3) - (-2)(-2) = 3 - 4 = -1 \]

Calculate the adjugate of A

The adjugate of matrix \( A \) is obtained by swapping the elements on the main diagonal and changing the signs of the other two elements:\[ \text{adj}(A) = \begin{bmatrix} -3 & 2 \ 2 & -1 \end{bmatrix} \]

Form the inverse of A

Using the determinant and the adjugate matrix, compute the inverse of \( A \):\[ A^{-1} = \frac{1}{-1} \begin{bmatrix} -3 & 2 \ 2 & -1 \end{bmatrix} = \begin{bmatrix} 3 & -2 \ -2 & 1 \end{bmatrix} \]

Solve for X

Now that we have \( A^{-1} \), solve for \( X \) by multiplying \( A^{-1} \) with \( B \):\[ X = A^{-1}B = \begin{bmatrix} 3 & -2 \ -2 & 1 \end{bmatrix} \begin{bmatrix} 13 & 4 & 7 \ 22 & 5 & 12 \end{bmatrix} \]Perform the matrix multiplication:- First row: \[ (3)(13) + (-2)(22) = 39 - 44 = -5 \] \[ (3)(4) + (-2)(5) = 12 - 10 = 2 \] \[ (3)(7) + (-2)(12) = 21 - 24 = -3 \]- Second row: \[ (-2)(13) + (1)(22) = -26 + 22 = -4 \] \[ (-2)(4) + (1)(5) = -8 + 5 = -3 \] \[ (-2)(7) + (1)(12) = -14 + 12 = -2 \]Thus, matrix \( X \) is:\[ X = \begin{bmatrix} -5 & 2 & -3 \ -4 & -3 & -2 \end{bmatrix} \]

Confirm the solution

Verify by multiplying \( A \) and \( X \) to check if the result is \( B \).\[ AX = \begin{bmatrix} -1 & -2 \ -2 & -3 \end{bmatrix} \begin{bmatrix} -5 & 2 & -3 \ -4 & -3 & -2 \end{bmatrix} \]Perform matrix multiplication:- First row: \[ (-1)(-5) + (-2)(-4) = 5 + 8 = 13 \] \[ (-1)(2) + (-2)(-3) = -2 + 6 = 4 \] \[ (-1)(-3) + (-2)(-2) = 3 + 4 = 7 \]- Second row: \[ (-2)(-5) + (-3)(-4) = 10 + 12 = 22 \] \[ (-2)(2) + (-3)(-3) = -4 + 9 = 5 \] \[ (-2)(-3) + (-3)(-2) = 6 + 6 = 12 \]These match \( B \), confirming our solution is correct.

Key concepts

Matrix Multiplication

Matrix multiplication is a way to combine two matrices to produce a third matrix. This operation is fundamental in linear algebra. For two matrices to be multiplied, the number of columns in the first matrix must be equal to the number of rows in the second matrix. For example, in our matrix equation \( AX = B \), matrix \( A \) is a 2x2 matrix, and \( X \) is a 2x3 matrix. The product \( A \cdot X \) will be a 2x3 matrix, since the inner dimensions match (both being 2).

• Each element in the product matrix is computed by taking the dot product of the corresponding row from the first matrix and the column from the second matrix.
• This involves multiplying elements pairwise and then summing them up.

Remember, matrix multiplication is not commutative, meaning \( A \cdot B eq B \cdot A \). It's important to keep the order consistent when performing this operation.

Matrix Inverse

The matrix inverse is essentially the matrix version of division. Not all matrices have inverses, but when they do, multiplying a matrix by its inverse results in the identity matrix. This is somewhat similar to the number 1 for real numbers, as multiplying any number by 1 results in the same number.

• For a square matrix \( A \) (i.e., having the same number of rows and columns), the inverse is denoted \( A^{-1} \).
• Matrix \( A \) is invertible if and only if its determinant is non-zero.
• The inverse of a matrix \( A \) allows us to solve the equation \( AX = B \) by rewriting it as \( X = A^{-1}B \).

In our example exercise, since matrix \( A \) has an inverse (because its determinant is \(-1\), not zero), we can use this inverse to find \( X \).

Determinant of a Matrix

The determinant is a special number that can be calculated from a square matrix. It provides important information about the matrix, like whether it has an inverse.

• A 2x2 matrix \( A \), shown as \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), has a determinant calculated as \( ad - bc \).
• If the determinant is zero, the matrix is singular, meaning it does not have an inverse.
• In our case, matrix \( A \) has a determinant of \(-1\), indicating that it is invertible.

Determining whether a matrix has a non-zero determinant is crucial before attempting to find its inverse.

Adjugate of a Matrix

The adjugate (or adjoint) of a matrix is used in finding the inverse of a matrix. Specifically, it's a matrix formed by taking the transpose of the cofactor matrix.

• For a 2x2 matrix, the adjugate is found by swapping the main diagonal elements and changing the sign of the other two elements.
• If \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), then \( \text{adj}(A) = \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \).
• This matrix comes into play in the formula for the inverse: \( A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \).

Using the adjugate is essential when computing the inverse, and it simplifies these calculations considerably.