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Matrices \(A\) and \(B\) are given. Solve the matrix equation \(A X=B\). $$ \begin{array}{l} A=\left[\begin{array}{ccc} -2 & 0 & 4 \\ -5 & -4 & 5 \\ -3 & 5 & -3 \end{array}\right] \\ B=\left[\begin{array}{ccc} -18 & 2 & -14 \\ -38 & 18 & -13 \\ 10 & 2 & -18 \end{array}\right] \end{array} $$

Short Answer

Expert verified
\(X = A^{-1} B\), calculated as \(X = \frac{1}{26} \begin{bmatrix} 41 & -1 & 20 \\ 9 & 6 & 20 \\ 5 & -8 & 8 \end{bmatrix} B\).

Step by step solution

01

Check Dimensions

First, check the dimensions of matrices \(A\) and \(B\) to ensure the equation \(A X = B\) is valid to solve for \(X\). Here, \(A\) is a 3x3 matrix and \(B\) is also a 3x3 matrix, which means \(X\), the unknown matrix, will also be a 3x3 matrix.
02

Find the Inverse of A

To solve for \(X\), we need to find \(A^{-1}\), the inverse of matrix \(A\). Use the formula for the inverse of a 3x3 matrix, which involves finding the cofactor matrix, transposing it, and dividing by the determinant of \(A\). Compute the determinant of \(A\), which is calculated to be 26.
03

Calculate Minors and Cofactors

Find the minors and cofactors of each element of \(A\). For example, the first cofactor is calculated with a minor of \(\begin{vmatrix} -4 & 5 \ 5 & -3 \end{vmatrix} = (-4)(-3) - (5)(5) = -41\). Repeat this process for each element of \(A\) to form the cofactor matrix.
04

Transpose the Cofactor Matrix

Transpose the cofactor matrix by flipping it over its diagonal. This forms what is known as the adjugate matrix. For our matrix \(A\), this results in: \(\begin{bmatrix} 41 & -1 & 20 \ 9 & 6 & 20 \ 5 & -8 & 8 \end{bmatrix}\).
05

Form the Inverse of A

Divide the adjugate matrix by the determinant of \(A\), which is 26. Therefore, \(A^{-1} = \frac{1}{26} \begin{bmatrix} 41 & -1 & 20 \ 9 & 6 & 20 \ 5 & -8 & 8 \end{bmatrix}\).
06

Solve for X by Matrix Multiplication

To find \(X\), multiply \(A^{-1}\) by \(B\). Perform the matrix multiplication, aligning rows of \(A^{-1}\) with columns of \(B\). The resulting matrix \(X\) is \(1/26 \begin{bmatrix} -18 & 2 & -14 \ -38 & 18 & -13 \ 10 & 2 & -18 \end{bmatrix}\), where every element of \(B\) has been processed through \(A^{-1}\).
07

Verify the Solution

Multiply \(A\) and \(X\) to verify that the product equals \(B\). If \(A X = B\), then \(X\) is correctly solved.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Inversion
Matrix inversion is a useful operation primarily applied in solving linear equations of the form \(AX = B\). This operation is only feasible if matrix \(A\) is square and its determinant is non-zero. To find the inverse of a 3x3 matrix, which is needed here, we follow several steps:
  • Calculate the determinant of the matrix.
  • Find the cofactor matrix, consisting of minors for each element.
  • Transpose the cofactor matrix to form the adjugate.
  • Divide the adjugate by the determinant to obtain the inverse.
For matrix \(A\), these steps yield the inverse \(A^{-1}\), which is pivotal in rewriting the matrix equation \(A X = B\) as \(X = A^{-1} B\). This makes solving for \(X\) straightforward.
Matrix Multiplication
Matrix multiplication involves combining rows from the first matrix with columns from the second matrix. It's a fundamental operation used when solving matrix equations like \(AX = B\) by taking \(X = A^{-1}B\).
To perform matrix multiplication:
  • Align each row of the first matrix with each column of the second matrix.
  • Multiply corresponding elements and sum the results to form a single element in the resulting matrix.
This operation results in matrix \(X\) in our exercise, showing how matrices transform values from one space to another.
Matrix Determinant
A matrix determinant is a critical scalar value that aids in understanding properties of the matrix, such as invertibility. For a 3x3 matrix, the determinant is calculated by expanding along any row or column using minors and cofactors.
To calculate the determinant of matrix \(A\):
  • Identify a row or column and find all associated minors.
  • Formulate cofactors by applying the correct sign to each minor.
  • Sum the products of elements and their cofactors.
If the determinant is non-zero, like the 26 in our exercise, the matrix is invertible and further operations are possible.
Cofactor Matrix
The cofactor matrix is derived from the original matrix and consists of cofactors, which are signed minors. These values play a role in forming the adjugate matrix, crucial for finding the inverse of a matrix.
Steps to form a cofactor matrix include:
  • Calculate minors for each element of the matrix.
  • Apply a checkerboard pattern of signs to turn minors into cofactors.
The cofactor matrix, when transposed, becomes the adjugate matrix, which, after division by the determinant, results in the inverse matrix. In our problem, creating this matrix allowed for the computation of \(A^{-1}\), essential for solving the matrix equation \(AX = B\).

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Most popular questions from this chapter

Matrices \(A\) and \(B\) are defined. (a) Give the dimensions of \(A\) and \(B\). If the dimensions properly match, give the dimensions of \(A B\) and \(B A\). (b) Find the products \(A B\) and \(B A\), if possible. $$ \begin{array}{l} A=\left[\begin{array}{cc} -5 & 2 \\ -5 & -2 \\ -5 & -4 \end{array}\right] \\ B=\left[\begin{array}{ccc} 0 & -5 & 6 \\ -5 & -3 & -1 \end{array}\right] \end{array} $$

Four pairs of vectors \(\vec{x}\) and \(\vec{y}\) are given below. For each pair, compute \(\|\vec{x}\|\), \(\|\vec{y}\|,\) and \(\|\vec{x}+\vec{y}\| .\) Use this information to answer: Is it always, sometimes, or never true that \(\|\vec{x}\|+\|\vec{y}\|=\|\vec{x}+\vec{y}\| ?\) If it always or never true, explain why. If it is sometimes true, explain when it is true. (a) \(\vec{x}=\left[\begin{array}{l}1 \\ 1\end{array}\right], \vec{y}=\left[\begin{array}{l}2 \\ 3\end{array}\right]\) (b) \(\vec{x}=\left[\begin{array}{c}1 \\ -2\end{array}\right], \vec{y}=\left[\begin{array}{c}3 \\ -6\end{array}\right]\) (c) \(\vec{x}=\left[\begin{array}{c}-1 \\ 3\end{array}\right], \vec{y}=\left[\begin{array}{l}2 \\ 5\end{array}\right]\) (d) \(\vec{x}=\left[\begin{array}{l}2 \\ 1\end{array}\right], \vec{y}=\left[\begin{array}{l}-4 \\ -2\end{array}\right]\)

Matrices \(A\) and \(B\) are defined. (a) Give the dimensions of \(A\) and \(B\). If the dimensions properly match, give the dimensions of \(A B\) and \(B A\). (b) Find the products \(A B\) and \(B A\), if possible. $$ \begin{array}{l} A=\left[\begin{array}{ll} 1 & 4 \\ 7 & 6 \end{array}\right] \\ B=\left[\begin{array}{cccc} 1 & -1 & -5 & 5 \\ -2 & 1 & 3 & -5 \end{array}\right] \end{array} $$

A matrix \(A\) and vector \(\vec{b}\) are given. (a) Solve the equation \(A \vec{x}=\vec{O}\) (b) Solve the equation \(A \vec{x}=\vec{b}\). In each of the above, be sure to write your answer in vector format. Also, when possible, give 2 particular solutions to each equation. $$ \begin{array}{l} A=\left[\begin{array}{cccc} -4 & 2 & -5 & 4 \\ 0 & 1 & -1 & 5 \end{array}\right], \\ \vec{b}=\left[\begin{array}{c} -3 \\ -2 \end{array}\right] \end{array} $$

A matrix \(A\) is given. Sketch \(\vec{x}, \vec{y}, A \vec{x}\) and \(A \vec{y}\) on the same Cartesian axes, where $$\vec{x}=\left[\begin{array}{l} 1 \\ 1 \end{array}\right] \text { and } \vec{y}=\left[\begin{array}{c} -1 \\ 2 \end{array}\right]$$ $$ A=\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right] $$

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