Question

Matrices \(A\) and \(\vec{b}\) are given. (a) Give \(\operatorname{det}(A)\) and \(\operatorname{det}\left(A_{i}\right)\) for all \(i\). (b) Use Cramer's Rule to solve \(A \vec{x}=\vec{b}\). If Cramer's Rule cannot be used to find the solution, then state whether or not a solution exists. \(A=\left[\begin{array}{ccc}4 & 9 & 3 \\ -5 & -2 & -13 \\ -1 & 10 & -13\end{array}\right]\) \(\vec{b}=\left[\begin{array}{c}-28 \\ 35 \\\ 7\end{array}\right]\)

Short answer

Solutions: \(x_1 = 6.19\), \(x_2 = -7.33\), \(x_3 = -7.56\).

Step-by-step solution

Calculate the Determinant of A

The matrix \(A\) is a 3x3 matrix given by \(A = \left[ \begin{array}{ccc} 4 & 9 & 3 \ -5 & -2 & -13 \ -1 & 10 & -13 \end{array} \right]\). To find the determinant of a 3x3 matrix, we use the formula:\[\det(A) = a(ei-fh) - b(di-fg) + c(dh-eg)\]where the elements are defined as:- \(a=4,\ b=9,\ c=3\)- \(d=-5,\ e=-2,\ f=-13\)- \(g=-1,\ h=10,\ i=-13\)Substitute these values into the formula:\[\det(A) = 4((-2)(-13) - (-13)(10)) - 9((-5)(-13) - (-13)(-1)) + 3((-5)(10) - (-2)(-1))\]Solving this yields:\[\det(A) = 4(26 + 130) - 9(65 + 13) + 3(-50 - 2) = 4 \times 156 - 9 \times 78 + 3 \times (-52)\]\[\det(A) = 624 - 702 - 156 = -234\]

Verify the Use of Cramer's Rule

Cramer's Rule can be applied to determine \(\vec{x}\) when \(A\) is invertible, i.e., when \(\det(A) eq 0\). Since \(\det(A) = -234 eq 0\), Cramer's Rule is applicable.

Form Matrices A_i by Replacing Columns

To use Cramer's Rule, form matrices \(A_i\) where \(A_i\) is the matrix obtained from \(A\), replacing its \(i\)-th column with \(\vec{b}\). We will compute \(\det(A_1)\), \(\det(A_2)\), and \(\det(A_3)\).- **Matrix \(A_1\):** Replace the first column of \(A\) with \(\vec{b}\) to get: \[ A_1 = \left[ \begin{array}{ccc} -28 & 9 & 3 \ 35 & -2 & -13 \ 7 & 10 & -13 \end{array} \right] \]- **Matrix \(A_2\):** Replace the second column of \(A\) with \(\vec{b}\) to get: \[ A_2 = \left[ \begin{array}{ccc} 4 & -28 & 3 \ -5 & 35 & -13 \ -1 & 7 & -13 \end{array} \right] \]- **Matrix \(A_3\):** Replace the third column of \(A\) with \(\vec{b}\) to get: \[ A_3 = \left[ \begin{array}{ccc} 4 & 9 & -28 \ -5 & -2 & 35 \ -1 & 10 & 7 \end{array} \right] \]

Calculate the Determinants of A_i

Now, calculate the determinants of \(A_1\), \(A_2\), and \(A_3\):- **\(\det(A_1)\)**: \[ \det(A_1) = -28((-2)(-13) - 10(-13)) - 9(35(-13) - 7(-13)) + 3(35 \times 10 - 7 \times -2) \] Calculating this step by step results in: \[ \det(A_1) = -1448 \]- **\(\det(A_2)\)**: \[ \det(A_2) = 4(35 \times -13 - 7 \times -13) + 28(-5 \times -13 - (-13)(-1)) + 3(-5 \times 7 - 35 \times -1) \] Calculating this yields: \[ \det(A_2) = 1716 \]- **\(\det(A_3)\)**: \[ \det(A_3) = 4(-2 \times 7 - 10 \times 35) + 9(-5 \times 7 - (-1 \times 35)) + 28(-5 \times 10 - (-2 \times -1)) \] Solve this to find: \[ \det(A_3) = 1768 \]

Apply Cramer's Rule to Solve for x

Cramer's Rule gives \(x_i = \frac{\det(A_i)}{\det(A)}\). Use it to calculate \(x_1\), \(x_2\), and \(x_3\).- **\(x_1\)**: \[ x_1 = \frac{\det(A_1)}{\det(A)} = \frac{-1448}{-234} = \frac{1448}{234} = 6.19 \]- **\(x_2\)**: \[ x_2 = \frac{\det(A_2)}{\det(A)} = \frac{1716}{-234} = -7.33 \]- **\(x_3\)**: \[ x_3 = \frac{\det(A_3)}{\det(A)} = \frac{1768}{-234} = -7.56 \]

Capture the Solution

The solutions according to Cramer's Rule are:- \(x_1 = 6.19\)- \(x_2 = -7.33\)- \(x_3 = -7.56\)

Key concepts

Determinants

Determinants are a vital component in matrix algebra, particularly when analyzing matrices. Essentially, the determinant is a scalar value that can be computed from a square matrix. This single number provides critical insights into the matrix's properties, notably whether or not the set of vectors represented by a matrix are linearly independent.
To compute the determinant of a 3x3 matrix, we use the formula \[\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\] where each letter corresponds to the position of elements in the matrix. For example, in matrix \( A = \begin{bmatrix} 4 & 9 & 3 \ -5 & -2 & -13 \ -1 & 10 & -13 \end{bmatrix}\), the determinant provides essential information such as whether the matrix is invertible. A non-zero determinant, like \( \text{det}(A) = -234 \), tells us the matrix is invertible, which in this context, allows us to apply methods like Cramer's Rule.

Matrix Algebra

Matrix algebra is a collection of mathematical tools used to handle matrices, which are essential for managing complex systems in various applications such as engineering and physics. In our problem, we deal with a matrix equation of the form \( A \vec{x} = \vec{b} \), where the goal is to find the vector \( \vec{x} \). Matrices provide a structured, efficient approach to representing and solving sets of linear equations.
Key operations in matrix algebra include addition, subtraction, and multiplication of matrices, as well as calculating determinants and inverses. It is important to follow specific rules, such as ensuring the matrices are of compatible sizes for addition or multiplication. Analyzing the structure in terms of rows and columns allows us to utilize these tools effectively. Using these operations, more complex techniques such as Cramer's Rule or Gaussian elimination can be performed to solve linear systems.

Matrix Inversion

Matrix inversion is the process of finding another matrix (the inverse) which, when multiplied by the original matrix, results in the identity matrix. An identity matrix is essentially a square matrix with ones on the diagonal and zeros everywhere else. If \( A \) is a matrix, its inverse \( A^{-1} \) should satisfy \( A \cdot A^{-1} = I \), where \( I \) is the identity matrix.
If a matrix is invertible, it means that the matrix equations can be uniquely solved. The invertibility of the matrix is directly linked to its determinant; specifically, a matrix is invertible if and only if its determinant is non-zero. Thus, for our given matrix \( A \), with \( \det(A) = -234 eq 0 \), an inverse can exist, opening pathways for solutions to the linear equations using techniques such as Cramer's Rule or matrix inversion itself.

Solving Linear Equations

Solving linear equations involves finding the values of unknown variables that satisfy the given equations. In the realm of matrices, this process is organized and effective, allowing for the handling of multiple equations simultaneously, using approaches like Cramer's Rule.
Cramer's Rule is a method that uses determinants to find solutions to linear equations. For a system represented by \( A \vec{x} = \vec{b} \), and where \( A \) is a square invertible matrix, Cramer's Rule provides a formula for each variable \( x_i \): \( x_i = \frac{\det(A_i)}{\det(A)} \). Each \( A_i \) is obtained by substituting a column of \( A \) with \( \vec{b} \).
This method is especially useful for theoretical solutions, as it directly links the structure of the system to its solution via determinants. In our problem, the solutions \( x_1 = 6.19 \), \( x_2 = -7.33 \), and \( x_3 = -7.56 \) were found using this method, reflecting how determinants play a crucial role in solving linear equations.