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Chapter 7: Problem 19

Either compute the inverse of the given matrix, or else show that it is singular. \(\left(\begin{array}{rrrr}{1} & {-1} & {2} & {0} \\ {-1} & {2} & {-4} & {2} \\\ {1} & {0} & {1} & {3} \\ {-2} & {2} & {0} & {-1}\end{array}\right)\)

Short Answer

Expert verified
Question: Calculate the inverse of the given matrix or show it is singular (non-invertible): $\left(\begin{array}{rrrr}{1} & {-1} & {2} & {0} \\\ {-1} & {2} & {-4} & {2} \\\ {1} & {0} & {1} & {3} \\\ {-2} & {2} & {0} & {-1}\end{array}\right)$ Answer: The inverse of the given matrix is: $$ A^{-1}=\left(\begin{array}{cccc}{\frac{-8}{33}} & {\frac{-3}{33}} & {\frac{24}{33}} & {\frac{-23}{33}} \\\ {\frac{-23}{33}} & {\frac{2}{33}} & {0} & {\frac{-2}{33}} \\\ {\frac{24}{33}} & {\frac{-11}{33}} & {\frac{-4}{33}} & {\frac{-2}{33}} \\\ {\frac{4}{33}} & {\frac{-1}{33}} & {0} & {\frac{-8}{33}}\end{array}\right) $$

Step by step solution

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01

Confirm the matrix is square

The given matrix has 4 rows and 4 columns, so it is a square matrix: $\left(\begin{array}{rrrr}{1} & {-1} & {2} & {0} \\\ {-1} & {2} & {-4} & {2} \\\ {1} & {0} & {1} & {3} \\\ {-2} & {2} & {0} & {-1}\end{array}\right)$
02

Calculate the determinant

We need to calculate the determinant of the matrix to check if it's non-zero. If it's non-zero, then the matrix is invertible. We will use the Laplace Expansion method to find the determinant. $\det(A)= \left(\begin{array}{rrrr}{1} & {-1} & {2} & {0} \\\ {-1} & {2} & {-4} & {2} \\\ {1} & {0} & {1} & {3} \\\ {-2} & {2} & {0} & {-1}\end{array}\right)\( = \)\left|\begin{array}{cccc}{1} & {-1} & {2} & {0} \\\ {-1} & {2} & {-4} & {2} \\\ {1} & {0} & {1} & {3} \\\ {-2} & {2} & {0} & {-1}\end{array}\right|$ Now we expand the determinant along the first row: \(= 1\left|\begin{array}{ccc}{2} & {-4} & {2} \\\ {0} & {1} & {3} \\\ {2} & {0} & {-1} \end{array}\right| - (-1)\left|\begin{array}{ccc}{-1} & {-4} & {2} \\\ {1} & {1} & {3} \\\ {-2} & {0} & {-1} \end{array}\right| + 2\left|\begin{array}{ccc}{-1} & {2} & {2} \\\ {1} & {0} & {3} \\\ {-2} & {2} & {-1} \end{array}\right| + 0\left|\begin{array}{ccc}{-1} & {2} & {-4} \\\ {1} & {0} & {1} \\\ {-2} & {2} & {0} \end{array}\right|\) Calculate the smaller 3x3 determinants: \(= 1\left[2(3)(-1) - (1)(1)(2)\right] - (-1)\left[(-1)(1)(-1) - (3)(-4)(2)\right] + 2\left[(-1)(0)(-1) - (3)(2)(2)\right]\) \(= 1\left[-6-2\right] - (-1)\left[-1 + 24\right] + 2\left[0-12\right]\) \(= 1(-8) - (-1)(23) + 2(-24)\) \(= -8 + 23 - 48\) \(= 15 - 48\) \(= -33\) So, the determinant of the given matrix is -33.
03

Calculate the inverse using the Adjugate method

Since the determinant of the given matrix A is non-zero (-33), the matrix is invertible. Now we will calculate the inverse of A using the Adjugate method. First, we need to find the matrix of minors. Then, we convert the matrix of minors into the matrix of cofactors by applying the checkerboard pattern of positive and negative signs. We will then transpose the resulting matrix, which is the Adjugate of A, and divide each element by the determinant. $$ \textbf{Matrix of Minors} \left(\begin{array}{cccc}{8} & {-23} & {-24} & {4} \\\ {-3} & {-2} & {11} & {-1} \\\ {24} & {0} & {4} & {0} \\\ {-23} & {2} & {-2} & {8}\end{array}\right) $$ Next, we change the signs according to the checkerboard pattern. $$ \textbf{Matrix of Cofactors} \left(\begin{array}{cccc}{8} & {23} & {-24} & {-4} \\\ {3} & {-2} & {11} & {1} \\\ {-24} & {0} & {4} & {0} \\\ {23} & {2} & {2} & {8}\end{array}\right) $$ Now, we transpose this matrix. $$ \textbf{Adj A}=\left(\begin{array}{cccc}{8} & {3} & {-24} & {23} \\\ {23} & {-2} & {0} & {2} \\\ {-24} & {11} & {4} & {2} \\\ {-4} & {1} & {0} & {8}\end{array}\right) $$ Finally, we divide each element of the adjugate by the determinant. $$ A^{-1}=\frac{1}{-33}\left(\begin{array}{cccc}{8} & {3} & {-24} & {23} \\\ {23} & {-2} & {0} & {2} \\\ {-24} & {11} & {4} & {2} \\\ {-4} & {1} & {0} & {8}\end{array}\right) $$ So, the inverse of the given matrix A is: $$ A^{-1}=\left(\begin{array}{cccc}{\frac{-8}{33}} & {\frac{-3}{33}} & {\frac{24}{33}} & {\frac{-23}{33}} \\\ {\frac{-23}{33}} & {\frac{2}{33}} & {0} & {\frac{-2}{33}} \\\ {\frac{24}{33}} & {\frac{-11}{33}} & {\frac{-4}{33}} & {\frac{-2}{33}} \\\ {\frac{4}{33}} & {\frac{-1}{33}} & {0} & {\frac{-8}{33}}\end{array}\right) $$

Key Concepts

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

Square Matrix
A square matrix is a matrix that has the same number of rows and columns, creating a perfect square format. Imagine a grid of numbers organized neatly into rows and columns, and if the count of rows equals the count of columns, you've got a square matrix. The simplest examples are the 2x2 or 3x3 matrices, but they can be larger, like 4x4 matrices, depending on the complexity of the problem.

In the given exercise, we encounter a 4x4 matrix, which is indeed a square matrix since it has 4 rows and 4 columns. Determining whether a matrix is square is the first step in knowing if we can calculate certain properties like the determinant or the inverse of the matrix.
Determinant Calculation
The determinant of a square matrix is a special scalar value that provides insight into various properties of the matrix. For example, it can tell you if the matrix is invertible, which means there's a unique matrix that, when multiplied by the original, yields the identity matrix.

Calculating the determinant isn't always straightforward, especially as the size of the matrix increases. For larger matrices, such as the 4x4 matrix in our exercise, it often involves breaking down the matrix into smaller parts and methodically working through them to arrive at a single number.
Laplace Expansion
When calculating the determinant of a larger square matrix, the Laplace expansion method is a common technique. This method involves expanding the determinant along a row or column (typically, the one with the most zeros to simplify calculations). You then calculate the determinants of the smaller submatrices that are formed by removing the row and column of each element chosen for the expansion.

In our exercise, Laplace expansion is used to find the determinant of the 4x4 matrix. The expansion is done along the first row, resulting in the calculation of several 3x3 determinant subproblems. The signs of these smaller determinants alternate based on their position in the matrix, following a checkerboard pattern of positive and negative signs.
Adjugate Matrix
The adjugate matrix, also known as the adjoint, is a critical component in finding the inverse of a square matrix when the determinant is nonzero. The adjugate matrix is the transpose of the cofactor matrix, which means that you switch its rows and columns.

Creating the adjugate matrix is a process with multiple steps: you begin by finding the matrix of minors, apply the checkerboard pattern to get the cofactor matrix, and then transpose it. In the context of our exercise, once the adjugate matrix of the given matrix is determined, it's used alongside the determinant to calculate the matrix inverse.
Cofactor Matrix
Lastly, the cofactor matrix is derived from the matrix of minors by applying a checkerboard pattern of signs. Each element of the cofactor matrix corresponds to a minor, which is the determinant of a smaller matrix formed by removing the row and column of the element, multiplied by \( (-1)^{i+j} \), where \(i\) and \(j\) are the row and column indices respectively.

In the solution to our exercise, after calculating the matrix of minors, we flip the sign of certain elements according to their position to get the matrix of cofactors. This step is pivotal for forming the adjugate matrix, which is utilized to compute the inverse of the original matrix.

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Most popular questions from this chapter

A mass \(m\) on a spring with constant \(k\) satisfies the differential equation (see Section 3.8 ) \(m u^{\prime \prime}+k u=0\) where \(u(t)\) is the displacement at time \(t\) of the mass from its equilibrium position. (a) Let \(x_{1}=u\) and \(x_{2}=u^{\prime}\); show that the resulting system is \(\mathbf{x}^{\prime}=\left(\begin{array}{rr}{0} & {1} \\ {-k / m} & {0}\end{array}\right) \mathbf{x}\) (b) Find the eigenvalues of the matrix for the system in part (a). (c) Sketch several trajectories of the system. Choose one of your trajectories and sketch the corresponding graphs of \(x_{1}\) versus \(t\) and of \(x_{2}\) versus \(t\), Sketch both graphs on one set of axes. (d) What is the relation between the eigenvalues of the coefficient matrix and the natural frequency of the spring-mass system?

Consider a \(2 \times 2\) system \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}\). If we assume that \(r_{1} \neq r_{2}\), the general solution is \(\mathbf{x}=c_{1} \xi^{(1)} e^{t_{1}^{\prime}}+c_{2} \xi^{(2)} e^{\prime 2},\) provided that \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly independent In this problem we establish the linear independence of \(\xi^{(1)}\) and \(\xi^{(2)}\) by assuming that they are linearly dependent, and then showing that this leads to a contradiction. $$ \begin{array}{l}{\text { (a) Note that } \xi \text { (i) satisfies the matrix equation }\left(\mathbf{A}-r_{1} \mathbf{I}\right) \xi^{(1)}=\mathbf{0} ; \text { similarly, note that }} \\ {\left(\mathbf{A}-r_{2} \mathbf{I}\right) \xi^{(2)}=\mathbf{0}} \\ {\text { (b) Show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right) \xi^{(1)}=\left(r_{1}-r_{2}\right) \mathbf{\xi}^{(1)}} \\\ {\text { (c) Suppose that } \xi^{(1)} \text { and } \xi^{(2)} \text { are linearly dependent. Then } c_{1} \xi^{(1)}+c_{2} \xi^{(2)}=\mathbf{0} \text { and at least }}\end{array} $$ $$ \begin{array}{l}{\text { one of } c_{1} \text { and } c_{2} \text { is not zero; suppose that } c_{1} \neq 0 . \text { Show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2} \boldsymbol{\xi}^{(2)}\right)=\mathbf{0}} \\ {\text { and also show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2} \boldsymbol{\xi}^{(2)}\right)=c_{1}\left(r_{1}-r_{2}\right) \boldsymbol{\xi}^{(1)} \text { . Hence } c_{1}=0, \text { which is }} \\\ {\text { a contradiction. Therefore } \xi^{(1)} \text { and } \boldsymbol{\xi}^{(2)} \text { are linearly independent. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Modify the argument of part (c) in case } c_{1} \text { is zero but } c_{2} \text { is not. }} \\ {\text { (e) Carry out a similar argument for the case in which the order } n \text { is equal to } 3 \text { ; note that }} \\ {\text { the procedure can be extended to cover an arbitrary value of } n .}\end{array} $$

Find the solution of the given initial value problem. Draw the corresponding trajectory in \(x_{1} x_{2} x_{3}\) - space and also draw the graph of \(x_{1}\) versus \(t .\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rrr}{-\frac{5}{2}} & {1} & {1} \\ {1} & {-\frac{5}{2}} & {1} \\ {1} & {1} & {-\frac{5}{2}}\end{array}\right) \mathbf{x}, \quad \mathbf{x}(0)=\left(\begin{array}{r}{2} \\ {3} \\\ {-1}\end{array}\right) $$

Verify that the given vector satisfies the given differential equation. \(\mathbf{x}^{\prime}=\left(\begin{array}{rrr}{1} & {1} & {1} \\ {2} & {1} & {-1} \\ {0} & {-1} & {1}\end{array}\right) \mathbf{x}, \quad \mathbf{x}=\left(\begin{array}{r}{6} \\ {-8} \\ {-4}\end{array}\right) e^{-t}+2\left(\begin{array}{r}{0} \\ {1} \\ {-1}\end{array}\right) e^{2 t}\)

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{-1} & {-4} \\ {1} & {-1}\end{array}\right) \mathbf{x} $$
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