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Chapter 5: Problem 53

Verify that $$ \boldsymbol{A}=\left[\begin{array}{rrr} 0 & 1 & -3 \\ 2 & 1 & 0 \\ 1 & -2 & 1 \end{array}\right] $$ has an inverse $$ \mathbf{A}^{-1}=\frac{1}{13}\left[\begin{array}{rrr} 1 & 5 & 3 \\ -2 & 3 & -6 \\ -5 & 1 & -2 \end{array}\right] $$ and hence solve the equation $$ A C A=\left[\begin{array}{lll} 2 & 1 & 1 \\ 0 & 2 & 3 \\ 2 & 1 & 0 \end{array}\right] $$

Short Answer

Expert verified
Matrix \( \mathbf{A} \) has an inverse; solve the equation to get \( C = \frac{1}{13} \left[\begin{array}{rrr} 2 & 13 & -4 \\ -16 & 9 & -18 \\ -3 & 9 & -3 \end{array}\right] \).

Step by step solution

01

Verify A has an Inverse

Multiply matrix \( \mathbf{A} \) by \( \mathbf{A}^{-1} \) to check if the product is the identity matrix, \( I \). The product \( \mathbf{A} \cdot \mathbf{A}^{-1} \) should equal the 3x3 identity matrix.
02

Calculate A * A^{-1}

Compute \( \mathbf{A} \cdot \mathbf{A}^{-1} = \left[\begin{array}{rrr}0 & 1 & -3 \2 & 1 & 0 \1 & -2 & 1\end{array}\right] \cdot \frac{1}{13}\left[\begin{array}{rrr}1 & 5 & 3 \-2 & 3 & -6 \-5 & 1 & -2\end{array}\right] \). Perform the matrix multiplication.
03

Verify Identity Matrix Result

The resulting product is \( \left[\begin{array}{ccc}1 & 0 & 0 \0 & 1 & 0 \0 & 0 & 1\end{array}\right] \), which is the 3x3 identity matrix, confirming \( \mathbf{A}^{-1} \) is correct.
04

Set up the Equation \( A C A = B \)

Rewrite the equation to solve: \( AC = B \cdot A^{-1} \). Given \[ B = \left[\begin{array}{lll} 2 & 1 & 1 \ 0 & 2 & 3 \ 2 & 1 & 0 \end{array}\right] \] and \( \mathbf{A} \), compute \( B \cdot \mathbf{A}^{-1} \).
05

Compute \( B \cdot A^{-1} \)

Calculate\[ \left[\begin{array}{lll} 2 & 1 & 1 \ 0 & 2 & 3 \ 2 & 1 & 0 \end{array}\right] \cdot \frac{1}{13} \left[\begin{array}{rrr} 1 & 5 & 3 \ -2 & 3 & -6 \ -5 & 1 & -2 \end{array}\right] \]. Simplify to find \( C \).
06

Simplify and Solve for C

Perform the matrix multiplication to obtain \[ C = \frac{1}{13} \left[\begin{array}{rrr} 2 & 13 & -4 \ -16 & 9 & -18 \ -3 & 9 & -3 \end{array}\right] \]. Check if the solution satisfies \( AC = B \cdot A^{-1} \).

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Key Concepts

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

Matrix Multiplication
Matrix multiplication is the process of calculating the product of two matrices. The product of an m x n matrix and an n x p matrix results in an m x p matrix. Unlike simple arithmetic, matrix multiplication is not commutative; meaning that \( A \times B eq B \times A \). This is crucial in verifying the existence of an inverse matrix.

Here's how to perform matrix multiplication:
  • Identify the rows of the first matrix and the columns of the second matrix.
  • For each element in the resulting matrix, calculate it by taking the dot product of the corresponding row and column.
  • Add the products for each row-column pair to get the resulting element.
For example, if you multiply matrix \( \mathbf{A} \) by \( \mathbf{A}^{-1} \), you systematically apply this method to get each element of the identity matrix.
Identity Matrix
An identity matrix is a special kind of square matrix. It acts as the neutral element in matrix multiplication, much like the number 1 in ordinary multiplication. When any matrix is multiplied by the identity matrix of compatible dimensions, it remains unchanged.

The structure of an identity matrix \( I_n \) is such that:
  • The diagonal elements are 1.
  • All off-diagonal elements are 0.
For example, a 3x3 identity matrix looks like this:\[ I_3 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \]

When verifying if \( \mathbf{A}^{-1} \) is an inverse of \( \mathbf{A} \), compute \( \mathbf{A} \cdot \mathbf{A}^{-1} \). It should result in the identity matrix, confirming that \( \mathbf{A}^{-1} \) is indeed the correct inverse.
Inverse Matrix
The inverse of a matrix is a matrix that, when multiplied by the original matrix, yields the identity matrix. Only square matrices may have inverses. Notably, a matrix must be non-singular (determinant is not zero) to possess an inverse.

To find an inverse matrix \( \mathbf{A}^{-1} \) of a 3x3 matrix \( \mathbf{A} \), ensure that:
  • You calculate the determinant of the matrix \( \mathbf{A} \).
  • If the determinant is non-zero, apply the formula or method suited to your matrix dimension to compute the inverse.
Multiplying \( \mathbf{A} \) by \( \mathbf{A}^{-1} \) should return the identity matrix \( I \). In our solved example, multiplying \( \mathbf{A} \) and \( \mathbf{A}^{-1} \) does yield \( I_3 \), thus confirming \( \mathbf{A}^{-1} \) is an accurate inverse of \( \mathbf{A} \).
Linear Equations
Solving linear equations with matrices can be done efficiently using matrix operations. The given exercise pertains to the equation \( AC = B \cdot A^{-1} \). Here, \( A \) and \( B \) are matrices, and the goal is to solve for matrix \( C \).

To solve such equations:
  • Construct matrices \( A \), \( B \), and find \( A^{-1} \).
  • Perform matrix multiplication to simplify and isolate \( C \). In this exercise, multiplying \( B \) by \( A^{-1} \) simplifies the process.

The solution involves calculating \( C \) using the matrix equation \( C = B \cdot A^{-1} \). This technique effectively solves matrix equations and is widely applicable in various fields requiring linear equations resolution.

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Most popular questions from this chapter

Find all the \(2 \times 2\) matrices that commute (that is \(\boldsymbol{A B}=\boldsymbol{B A}\) ) with \(\left[\begin{array}{rr}1 & -1 \\ 0 & 2\end{array}\right]\)

Given the matrix $$ \boldsymbol{A}=\lambda\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]+\mu\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right]+v\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right] $$ (a) find the value of \(\lambda, \mu, v\) so that \(\boldsymbol{A}=\left[\begin{array}{ll}0 & -1 \\ 0 & 3\end{array}\right]\); (b) show that no solution is possible if $$ \mathbf{A}=\left[\begin{array}{lr} 1 & -1 \\ 1 & 0 \end{array}\right] $$

A popular method of numerical integration - see the work in Chapter 8 - involves Gaussian integration; it is used in finite-element calculations which are well used in most of engineering. As a simple example, the numerical integral over the interval \(-1 \leqslant x \leqslant 1\) is written $$ \int_{-1}^{1} f(x) \mathrm{d} x=C_{1} f\left(x_{1}\right)+C_{2} f\left(x_{2}\right) $$ and the formula is made exact for the four functions \(f=1, f=x, f=x^{2}\) and \(f=x^{3}\), so it must be accurate for all cubics. This leads to the four equations $$ \begin{gathered} C_{1}+C_{2}=2 \\ C_{1} x_{1}+C_{2} x_{2}=0 \\ C_{1} x_{1}^{2}+C_{2} x_{2}^{2}=\frac{2}{3} \\ C_{1} x_{1}^{3}+C_{2} x_{2}^{3}=0 \end{gathered} $$ Use Gaussian elimination to reduce the equations and hence deduce that the equations are only consistent if \(x_{1}\) and \(x_{2}\) are chosen at the 'Gauss' points \(\pm \frac{1}{\sqrt{3}}\)

Find all the eigenvalues and eigenvectors of the matrix $$ \left[\begin{array}{llll} 1 & 0 & 0 & 2 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 2 & 0 & 0 & 1 \end{array}\right] $$

The matrices $$ \boldsymbol{A}=\left[\begin{array}{lll} 1 & 2 & 1 \\ 3 & 0 & 2 \end{array}\right], \quad \boldsymbol{B}=\left[\begin{array}{lll} 4 & 1 & 3 \\ 0 & 2 & 1 \end{array}\right] $$ and $$ \boldsymbol{C}=\left[\begin{array}{ll} 1 & 2 \\ 3 & 1 \\ 2 & 3 \end{array}\right] $$ are given. (a) Which of the following make sense: \(A B, A C\), \(\boldsymbol{B C}, \boldsymbol{A B}^{\mathrm{T}}, \boldsymbol{A C}^{\mathrm{T}}\) and \(\boldsymbol{B C}^{\mathrm{T}}\) ? (b) Evaluate those products that do exist. (c) Evaluate \(\left(\boldsymbol{A}^{\mathrm{T}} \boldsymbol{B}\right) \boldsymbol{C}\) and \(\boldsymbol{A}^{\mathrm{T}}(\boldsymbol{B C})\) and show that they are equal.
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