Question

Matrices \(A\) and \(B\) are given. Solve the matrix equation \(A X=B\). $$ \begin{array}{l} A=\left[\begin{array}{cc} -3 & -6 \\ 4 & 0 \end{array}\right] \\ B=\left[\begin{array}{cc} 48 & -30 \\ 0 & -8 \end{array}\right] \end{array} $$

Short answer

The matrix \(X\) is \(\begin{bmatrix} 0 & -2 \\ -8 & 5 \end{bmatrix}\).

Step-by-step solution

Verify Matrix Dimensions

First, ensure the matrices are compatible for multiplication and the equation can be solved. Matrix \(A\) is a \(2 \times 2\) matrix, and matrix \(B\) is also \(2 \times 2\). Thus, both have compatible dimensions, and \(X\) will also be a \(2 \times 2\) matrix.

Find the Inverse of Matrix A

Calculate the inverse of matrix \(A\). The inverse of a matrix \(A\) is given by: \[A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]Matrix \(A\) is \(\begin{bmatrix} -3 & -6 \ 4 & 0 \end{bmatrix}\), so the determinant is \(ad-bc = (-3)(0) - (4)(-6) = 24\). The inverse is:\[A^{-1} = \frac{1}{24}\begin{bmatrix} 0 & 6 \ -4 & -3 \end{bmatrix} = \begin{bmatrix} 0 & \frac{1}{4} \ -\frac{1}{6} & -\frac{1}{8} \end{bmatrix}\]

Multiply the Inverse of A by B

Calculate \(X = A^{-1}B\). Use the inverse calculated in the previous step:\(X = \begin{bmatrix} 0 & \frac{1}{4} \ -\frac{1}{6} & -\frac{1}{8} \end{bmatrix} \begin{bmatrix} 48 & -30 \ 0 & -8 \end{bmatrix}\)Multiplying these matrices:\[X = \begin{bmatrix} 0 \cdot 48 + \frac{1}{4} \cdot 0 & 0 \cdot (-30) + \frac{1}{4}(-8) \ -\frac{1}{6} \cdot 48 + -\frac{1}{8} \cdot 0 & -\frac{1}{6}(-30) + -\frac{1}{8}(-8) \end{bmatrix}\]Calculating the resulting matrix:\[X = \begin{bmatrix} 0 & -2 \ -8 & 5 \end{bmatrix}\]

Verify Your Solution

Multiply the solution for \(X\) back by \(A\) to ensure it results in \(B\):\(AX = \begin{bmatrix} -3 & -6 \ 4 & 0 \end{bmatrix} \begin{bmatrix} 0 & -2 \ -8 & 5 \end{bmatrix}\)\[AX = \begin{bmatrix} (-3)(0) + (-6)(-8) & (-3)(-2) + (-6)(5) \ (4)(0) + (0)(-8) & (4)(-2) + (0)(5) \end{bmatrix} = \begin{bmatrix} 48 & -30 \ 0 & -8 \end{bmatrix}\]The result is indeed matrix \(B\), confirming that the solution \(X\) is correct.

Key concepts

Matrix Multiplication

Matrix multiplication is a key mathematical operation, especially useful when dealing with linear transformations and systems of equations. When multiplying two matrices, say matrix \(A\) with dimensions \(m \times n\) and matrix \(B\) with dimensions \(n \times p\), the resulting product is a new matrix \(C\) with dimensions \(m \times p\). The number of columns in the first matrix must equal the number of rows in the second matrix for multiplication to be defined.

For each element \(c_{ij}\) in the resulting matrix \(C\), we sum the products of corresponding elements from the \(i\)-th row of matrix \(A\) and the \(j\)-th column of matrix \(B\). This can be formally described by:

• \(c_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + ... + a_{in}b_{nj}\)

This operation is not commutative, meaning that \(A \times B eq B \times A\) in general. This uniqueness is essential when working with matrix equations, as it determines the order and feasibility of matrix operations.

Matrix Inverse

The inverse of a matrix is a concept that many students find tricky, but it's quite straightforward once you break it down. Given a square matrix \(A\), the inverse of \(A\), denoted as \(A^{-1}\), is a matrix that, when multiplied with \(A\), yields the identity matrix \(I\). The identity matrix acts like the number one in regular arithmetic.

The formula for finding the inverse of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is:

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

The term \(ad-bc\) is called the determinant, and it's crucial because the inverse exists only when the determinant is non-zero. If the determinant is zero, the matrix is said to be singular, and it doesn't have an inverse. When dealing with equations like \(AX = B\), knowing \(A^{-1}\) is vital, as it allows us to find \(X\) by computing \(X = A^{-1}B\).

Determinate of a Matrix

The determinant is a special number calculated from a square matrix, which tells us several things about the matrix, including whether it is invertible. For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant \(D\), is given by the formula:

\[D = ad - bc\]
The determinant acts as a scalar value that helps in many practical applications, such as determining the area or volume in geometry, or assessing the solvability of systems of linear equations.

One of the key roles of the determinant is in finding the matrix inverse. If the determinant is zero, the matrix is singular, meaning it does not have an inverse, and linear systems involving this matrix cannot be uniquely solved. In our exercise, a positive determinant (like 24) allowed us to compute the inverse, thus solving the matrix equation. Understanding determinants also aids in grasping fundamental concepts in linear algebra and beyond. It's a building block for more advanced topics like eigenvalues and eigenvectors.