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Chapter 9: Problem 59

Use the determinant to find the inverse of \(A\) $$ A=\left[\begin{array}{rr} 2 & 1 \\ 3 & -1 \end{array}\right] $$

Short Answer

Expert verified
The inverse of matrix \(A\) is \(\begin{bmatrix} \frac{1}{5} & \frac{1}{5} \\ \frac{3}{5} & -\frac{2}{5} \end{bmatrix}\).

Step by step solution

01

Calculate the Determinant

To find the inverse of a matrix using the determinant, we first need to compute the determinant of matrix \(A\). The formula for a \(2 \times 2\) matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\) is \(\text{det}(A) = ad - bc\). For the given matrix \(A = \begin{bmatrix} 2 & 1 \ 3 & -1 \end{bmatrix}\), \(a = 2\), \(b = 1\), \(c = 3\), and \(d = -1\). Substitute these values into the formula: \(\text{det}(A) = (2)(-1) - (1)(3) = -2 - 3 = -5\).
02

Check Determinant for Invertibility

For a matrix to be invertible, its determinant must be non-zero. Here, \(\text{det}(A) = -5\), which is not zero. Thus, matrix \(A\) is invertible.
03

Apply the Inverse Formula for a 2x2 Matrix

The inverse of a \(2 \times 2\) matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\) is given by \(A^{-1} = \frac{1}{\text{det}(A)}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\). We found \(\text{det}(A) = -5\), so \(A^{-1} = \frac{1}{-5}\begin{bmatrix} -1 & -1 \ -3 & 2 \end{bmatrix}\).
04

Simplify the Inverse Matrix

Finally, apply the scalar multiplication to compute \(A^{-1}\): \[A^{-1} = \begin{bmatrix} \frac{-1}{-5} & \frac{-1}{-5} \ \frac{-3}{-5} & \frac{2}{-5} \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & \frac{1}{5} \ \frac{3}{5} & -\frac{2}{5} \end{bmatrix}\].

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Key Concepts

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

Determinant
The determinant of a matrix is a special number that can tell us a lot about the matrix itself. For a 2x2 matrix, the determinant is calculated by a formula:
  • If our matrix is \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), then its determinant is given by \( \text{det}(A) = ad - bc \).
In our example matrix \( A = \begin{bmatrix} 2 & 1 \ 3 & -1 \end{bmatrix} \), we calculate it as \[ (2)(-1) - (1)(3) = -2 - 3 = -5. \]The determinant, \(-5\), is a crucial value because it helps us know if the matrix can be inverted.
2x2 Matrix
A 2x2 matrix is a simple yet powerful structure in linear algebra. It has two rows and two columns, making it:
  • Easy to work with in examples and problems.
  • A basic building block for understanding larger matrix concepts.
The general form is:\[A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\]In practice, like in our exercise, the matrix order tells us the size, and using the matrix we can solve various problems like finding the inverse, determining linear transformations, and more.
Invertibility
A matrix is considered invertible (or non-singular) if you can find another matrix that, when multiplied with the original, gives the identity matrix. For a 2x2 matrix, this boils down to checking its determinant:
  • The matrix is invertible if its determinant is not zero.
  • In our case, the determinant of matrix \( A \) is \(-5\), which is non-zero; this tells us that matrix \( A \) has an inverse.
The identity matrix for a 2x2 structure looks like this:\[I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]Successfully finding an inverse means finding matrix \( A^{-1} \) such that \( A \cdot A^{-1} \) equals \( I \).
Scalar Multiplication
Scalar multiplication involves multiplying a matrix by a single number, called a scalar. When finding inverses or transformations, this step ensures that each element in the matrix is adjusted correctly. Here's how we handle it in our context:
  • After computing the inverse matrix using formulae, each component is multiplied by \( \frac{1}{\text{det}(A)} \).
So for our matrix \( A \), we initially found an inverse matrix form:\[\begin{bmatrix} -1 & -1 \ -3 & 2 \end{bmatrix}\]Then, applying scalar multiplication by \( \frac{1}{-5} \) changed it to:\[\begin{bmatrix} \frac{1}{5} & \frac{1}{5} \ \frac{3}{5} & -\frac{2}{5} \end{bmatrix}\]This process finalizes the inverse, dividing each element by the determinant, making it suitable for use in further calculations.

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Most popular questions from this chapter

Find the transpose of $$ A=\left[\begin{array}{r} 2 \\ -3 \\ 5 \end{array}\right] $$

Reduce the system of linear equations to upper triangular form and solve. $$ \begin{array}{l} 7 x-y=4 \\ 3 x+2 y=1 \end{array} $$

$$ A=\left[\begin{array}{rr} -1 & 0 \\ 1 & 2 \end{array}\right], \quad B=\left[\begin{array}{rr} 2 & 3 \\ -1 & 1 \end{array}\right], \quad C=\left[\begin{array}{rr} 1 & 2 \\ 0 & -1 \end{array}\right] $$ $$ \text { Show that }(A B) C=A(B C) \text { . } $$

$$ A=\left[\begin{array}{rr} -1 & 0 \\ 1 & 2 \end{array}\right], \quad B=\left[\begin{array}{rr} 2 & 3 \\ -1 & 1 \end{array}\right], \quad C=\left[\begin{array}{rr} 1 & 2 \\ 0 & -1 \end{array}\right] $$ $$ \text { Show that } A C \neq C A \text { . } $$

Let \(A=\left[\begin{array}{lll}1 & 4 & -2\end{array}\right]\) and \(B=\left[\begin{array}{r}-1 \\ 2 \\ 3\end{array}\right]\) (a) Compute \(A B\). (b) Compute \(B A\).
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