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Chapter 10: Problem 56

Let \(\left.A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right) . \text { Compute } A^{-1} . \text {(Assume that } a d-b c \neq 0 .\right)\).

Short Answer

Expert verified
The inverse is \( A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \).

Step by step solution

01

Verify Matrix Condition

First, check if the matrix \(A\) is invertible by ensuring the determinant \(ad - bc eq 0\). This condition confirms that the matrix has an inverse. Since the problem states this condition is met, proceed to find the inverse.
02

Determine the Determinant

Calculate the determinant of matrix \(A\). For the matrix \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant \(\text{det}(A)\) is \(ad - bc\).
03

Using the Inverse Formula

Apply the formula for the inverse of a 2x2 matrix. The inverse \( A^{-1} \) is given by:\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \]
04

Substitute Values into Formula

Substitute the variables from matrix \(A\) into the inverse matrix formula:\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \]This expression represents the inverse matrix of \(A\).

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Key Concepts

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

Determinant of a Matrix
Calculating the determinant of a matrix is a crucial first step in understanding whether a matrix can be inverted. The determinant is a special number that can tell us a lot about the properties of a matrix.
If we have a matrix
  • For a 2x2 matrix, the determinant is found using the formula \( ad - bc \).
  • It's important to remember that the order of multiplication and subtraction here is essential to get the correct result.
  • This number needs to be other than zero for the matrix to be invertible. If the determinant is zero, it means the rows or columns of the matrix are linearly dependent, resulting in no unique inverse existing.
In this context, the determinant acts like a gatekeeper; if it is not zero, you are cleared to proceed with finding an inverse. This special characteristic highlights the determinant's importance in linear algebra.
2x2 Matrix
A 2x2 matrix is one of the simplest forms of matrices and serves as the building block for many matrix-related concepts.
Matrix can be represented as:
  • Each component of the matrix represents an entry, denoted by letters such as \( a, b, c, \) and \( d \).
  • The simplicity of a 2x2 matrix allows for straightforward computations, like easily calculating the determinant with the formula \( ad - bc \).
  • A neat property of a 2x2 matrix is that its inverse, if it exists, can be directly computed using a specific formula tailored for these matrices, making them very manageable.
Understanding how to manipulate a 2x2 matrix opens the door to working with more complex matrix operations.
Invertible Matrix
An invertible matrix, also known as a non-singular or non-degenerate matrix, is a matrix that has an inverse.
This means there exists another matrix that, when multiplied with the original matrix, yields the identity matrix.
  • This property is critical for solving systems of linear equations, as it ensures solutions exist and are unique.
  • The condition for invertibility hinges on the determinant: it must not be zero. When \( ad - bc eq 0 \), a 2x2 matrix is invertible.
  • For a matrix \( A \), its inverse is notated as \( A^{-1} \) and is calculated using the inverse formula: \[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \]
This concept of invertibility is foundational in linear algebra, enabling the transformation and manipulation of linear systems with precision.

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Most popular questions from this chapter

A curve \(y=x^{3}+A x^{2}+B x+C\) passes through the three points \((1,-2),(2,3),\) and (3,20) (a) Write down a system of three linear equations satisfied by \(A, B,\) and \(C\) (b) Solve the system in part (a). (c) With the values for \(A, B,\) and \(C\) that you found in part (b), use a graphing utility to draw the curve \(y=x^{3}+A x^{2}+B x+C,\) checking to see that it appears to pass through the three given points.

Use a graphing utility to compute the matrix products. $$\left(\begin{array}{rrr} 12 & -10 & 13 \\ 5 & 7 & 25 \\ -8 & 9 & 28 \end{array}\right)\left(\begin{array}{rrr} -11 & 31 & 6 \\ 0 & 1 & -14 \\ 41 & 12 & -17 \end{array}\right)$$

Use a graphing utility to compute the matrix products. $$\left(\begin{array}{rrrr} -6 & 9 & -5 & 1 \\ 9 & -1 & -5 & 2 \\ -5 & -5 & 9 & -3 \\ 1 & 2 & -3 & 1 \end{array}\right)\left(\begin{array}{cccc} 0.5 & 1 & 1.5 & 2 \\ 1 & 2 & 3.5 & 5.5 \\ 1.5 & 3.5 & 7 & 12.5 \\ 2 & 5.5 & 12.5 & 25 \end{array}\right)$$

Use Cramers rule to solve those systems for which \(D \neq 0 .\) In cases where \(D=0,\) use Gaussian elimination or matrix methods. $$\left\\{\begin{aligned} 12 x &-11 z=13 \\ 6 x+6 y-4 z &=26 \\ 6 x+2 y-5 z &=13 \end{aligned}\right.$$

Show that $$\left|\begin{array}{lll} a_{1}+A_{1} & b_{1} & c_{1} \\ a_{2}+A_{2} & b_{2} & c_{2} \\ a_{3}+A_{3} & b_{3} & c_{3} \end{array}\right|=\left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|+\left|\begin{array}{lll} A_{1} & b_{1} & c_{1} \\ A_{2} & b_{2} & c_{2} \\ A_{3} & b_{3} & c_{3} \end{array}\right|$$
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