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Chapter 5: Problem 2

For the given matrices \(A\) find \(A^{-1}\) if it exists and verify that \(A A^{-1}=\) \(A^{-1} A=I\). If \(A^{-1}\) does not exist explain why. (a) \(A=\left(\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right)\) (b) \(A=\left(\begin{array}{ll}0 & 1 \\ 0 & 2\end{array}\right)\) (c) \(A=\left(\begin{array}{ll}1 & c \\ 0 & 1\end{array}\right)\) (d) \(A=\left(\begin{array}{cc}a & b \\ b & a\end{array}\right),\) where \(|a| \neq|b|\).

Short Answer

Expert verified
(a) \(A^{-1} = \frac{1}{3}\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\); (b) not invertible; (c) \(A^{-1} = \begin{pmatrix} 1 & -c \\ 0 & 1 \end{pmatrix}\); (d) \(A^{-1} = \frac{1}{a^2-b^2}\begin{pmatrix} a & -b \\ -b & a \end{pmatrix}\)."

Step by step solution

01

Determine if Matrix A is Invertible (Part a)

For matrix \(A=\begin{pmatrix} 2 & -1 \ -1 & 2 \end{pmatrix}\), calculate the determinant using the formula \(\det(A) = ad - bc\) where \(a, b, c,\) and \(d\) are the elements of the matrix. Here, \(a = 2, b = -1, c = -1, d = 2\). Thus, \(\det(A) = (2)(2) - (-1)(-1) = 4 - 1 = 3\). Since the determinant is non-zero, \(A\) is invertible.
02

Find the Inverse (Part a)

The inverse of a 2x2 matrix \(A=\begin{pmatrix} a & b \ c & d \end{pmatrix}\) is given by \(A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\). For \(A=\begin{pmatrix} 2 & -1 \ -1 & 2 \end{pmatrix}\), \(A^{-1} = \frac{1}{3} \begin{pmatrix} 2 & 1 \ 1 & 2 \end{pmatrix}\).
03

Verify the Inverse (Part a)

Multiply \(A\) and \(A^{-1}\) to confirm that \(AA^{-1} = I\). Calculate \(\begin{pmatrix} 2 & -1 \ -1 & 2 \end{pmatrix} \begin{pmatrix} \frac{2}{3} & \frac{1}{3} \ \frac{1}{3} & \frac{2}{3} \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\), thus \(AA^{-1} = I\). Verify \(A^{-1}A = I\) similarly.
04

Determine if Matrix A is Invertible (Part b)

For matrix \(A=\begin{pmatrix} 0 & 1 \ 0 & 2 \end{pmatrix}\), the determinant is \(\det(A) = (0)(2) - (1)(0) = 0\). Since the determinant is zero, \(A\) is not invertible.
05

Describe Non-Invertibility (Part b)

Since the determinant of matrix \(A=\begin{pmatrix} 0 & 1 \ 0 & 2 \end{pmatrix}\) is zero, it is a singular matrix and does not have an inverse.
06

Determine if Matrix A is Invertible (Part c)

For matrix \(A=\begin{pmatrix} 1 & c \ 0 & 1 \end{pmatrix}\), calculate the determinant: \(\det(A) = (1)(1) - (c)(0) = 1\). Since the determinant is non-zero, \(A\) is invertible.
07

Find the Inverse (Part c)

Use the formula for a 2x2 matrix. The inverse \(A^{-1}\) is \(\begin{pmatrix} 1 & -c \ 0 & 1 \end{pmatrix}\), derived by taking \(\frac{1}{1}\) times \(\begin{pmatrix} 1 & -c \ 0 & 1 \end{pmatrix}\), which results in the identity matrix when multiplied with \(A\).
08

Verify the Inverse (Part c)

Check that multiplying \(A\) and \(A^{-1}\) gives the identity matrix: \(\begin{pmatrix} 1 & c \ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & -c \ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\). Verify \(A^{-1}A = I\) similarly.
09

Determine if Matrix A is Invertible (Part d)

For matrix \(A=\begin{pmatrix} a & b \ b & a \end{pmatrix}\), calculate the determinant: \(\det(A) = (a)(a) - (b)(b) = a^2 - b^2\). For invertibility, \(a^2 eq b^2\) given \(|a| eq |b|\). Therefore, \(A\) is invertible.
10

Find the Inverse (Part d)

Use the formula for 2x2 matrices again, assuming \(a^2 eq b^2\). The inverse \(A^{-1}\) is \(\frac{1}{a^2 - b^2}\begin{pmatrix} a & -b \ -b & a \end{pmatrix}\).
11

Verify the Inverse (Part d)

Show that multiplying \(A\) with \(A^{-1}\) gives the identity matrix: \(\begin{pmatrix} a & b \ b & a \end{pmatrix}\begin{pmatrix} \frac{a}{a^2-b^2} & \frac{-b}{a^2-b^2} \ \frac{-b}{a^2-b^2} & \frac{a}{a^2-b^2} \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\). Verify similarly that \(A^{-1}A = I\).

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Key Concepts

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

Determinant Calculation
The determinant is a critical value used to determine whether a matrix is invertible. For a 2x2 matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant \( \det(A) \) is calculated using the formula \( ad - bc \).

The determinants of matrices help identify whether a matrix is invertible or singular. A non-zero determinant means the matrix is invertible, while a zero determinant indicates a singular matrix.

To find determinants:
  • Identify the elements \( a, b, c, \) and \( d \) from the matrix.
  • Apply the formula \( ad - bc \).
For example, given a matrix \( A = \begin{pmatrix} 2 & -1 \ -1 & 2 \end{pmatrix} \), the determinant is calculated as \( (2)(2) - (-1)(-1) = 4 - 1 = 3 \). This non-zero value denotes that the matrix is invertible.
Identity Matrix
The identity matrix serves as the multiplicative identity in matrix algebra, much like the number 1 in regular arithmetic. For any matrix \( A \), when multiplied by an identity matrix, the result is \( A \) itself. In a 2x2 format, the identity matrix is \( \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \).

An identity matrix maintains the dimensions of the matrix it is being multiplied with, ensuring the product remains unchanged.

When checking for matrix invertibility by computing \( AA^{-1} = I \) or \( A^{-1}A = I \), you aim to find whether the original matrix when multiplied by its inverse results in an identity matrix.
  • Easy operation where matrix properties remain unaffected.
  • The identity matrix is diagonal with ones along the main diagonal and zeroes elsewhere.
This property is crucial during the verification of inverses and solving linear equations.
Singular Matrix
A singular matrix is one that lacks an inverse due to its determinant being zero. This matrix type signifies an inability to solve certain algebraic equations using standard matrix methods.

For example, with matrix \( A = \begin{pmatrix} 0 & 1 \ 0 & 2 \end{pmatrix} \), the determinant \( \det(A) = 0 \), making \( A \) a singular matrix.

When a matrix is singular:
  • The system of equations leads to infinite solutions or no solution.
  • The matrix ranks less than typical invertible matrices.
Detecting singular matrices is key as they flag potential problems in linear systems without retraceable solutions.
Invertible Matrices
Invertible matrices, or non-singular matrices, are those which have a determinant that is not zero. This allows them to have an inverse, which is crucial in matrix calculations such as solving system equations, transformations, and more.

The inverse of a 2x2 matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is calculated using the formula \( A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \).
  • Invertible matrices solve equations efficiently.
  • They hold a one-to-one correspondence in mappings.
These matrices ensure that equalities \( AA^{-1} = I \) and \( A^{-1}A = I \) validate the identity matrix, confirming the invertibility.

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Most popular questions from this chapter

(a) Let \(A, B,\) and \(D\) be \(n \times n\) matrices. Assume that \(B\) is invertible. If \(A=B D B^{-1},\) prove by induction that \(A^{m}=B D^{m} B^{-1}\) is true for \(m \geq 1\) (b) Given that \(A=\left(\begin{array}{ll}-8 & 15 \\ -6 & 11\end{array}\right)=B\left(\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right) B^{-1}\) where \(B=\) \(\left(\begin{array}{ll}5 & 3 \\ 3 & 2\end{array}\right)\) what is \(A^{10} ?\)

Recall that \(p(x)=x^{2}-5 x+6\) is called a polynomial, or more specifically, a polynomial over \(\mathbb{R},\) where the coefficients are elements of \(\mathbb{R}\) and \(x \in \mathbb{R}\). Also, think of the method of solving, and solutions of, \(x^{2}-5 x+6=0\). We would like to define the analogous situation for \(2 \times 2\) matrices. First define where \(A\) is a \(2 \times 2\) matrix \(p(A)=A^{2}-5 A+6 I\). Discuss the method of solving and the solutions of \(A^{2}-5 A+6 I=0\).

Prove: If the determinant of a matrix \(A\) is zero, then \(A\) does not have an inverse. Hint: Use the indirect method of proof and exercise 5 .

For the given matrices \(A\) find \(A^{-1}\) if it exists and verify that \(A A^{-1}=\) \(A^{-1} A=I .\) If \(A^{-1}\) does not exist explain why. (a) \(A=\left(\begin{array}{ll}1 & 3 \\ 2 & 1\end{array}\right)\) (b) \(A=\left(\begin{array}{ll}6 & -3 \\ 8 & -4\end{array}\right)\) (c) \(A=\left(\begin{array}{cc}1 & -3 \\ 0 & 1\end{array}\right)\) (d) \(A=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\) (e) Use the definition of the inverse of a matrix to find \(A^{-1}: A=\) $$ \left(\begin{array}{ccc} 3 & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & -5 \end{array}\right) $$

(a) If $$ A=\left(\begin{array}{cc} 2 & 1 \\ 1 & -1 \end{array}\right), X=\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right), \text { and } B=\left(\begin{array}{l} 3 \\ 1 \end{array}\right) $$ \(2 x_{1}+x_{2}=3\) show that \(A X=B\) is a way of expressing the system \(x_{1}-x_{2}=1\) using matrices. (b) Express the following systems of equations using matrices: (i) \(2 x_{1}-x_{2}=4\) $$ x_{1}+x_{2} \quad=3 $$ $$ x_{1}+x_{2}=0 $$ (iii) \(x_{2} \quad=5\) $$ x_{1}+x_{2}+2 x_{3}=1 \quad x_{1} \quad+3 x_{3}=6 $$ (ii) \(x_{1}+2 x_{2}-x_{3}=-1\) \(x_{1}+3 x_{3}+x_{3}=5\)
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