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For each matrix, find \(A^{-1}\) if it exists. Do not use a calculator. $$A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]$$

Short Answer

Expert verified
The inverse matrix is \(A^{-1} = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 1 & -1 & 1 \end{array} \right]\).

Step by step solution

01

Understand Invertibility Criteria

For a matrix to have an inverse, it must be square and its determinant must be non-zero. We first check if the given matrix is square and proceed to find its determinant.
02

Calculate the Determinant

The determinant of a 3x3 matrix \(A = \left[ \begin{array}{ccc} 1 & 0 & 0 \ 1 & 1 & 0 \ 0 & 1 & 1 \end{array} \right]\) is calculated as follows: \[ \text{det}(A) = 1(1 \cdot 1 - 0 \cdot 1) - 0(1 \cdot 1 - 0 \cdot 0) + 0(1 \cdot 1 - 1 \cdot 0) = 1 \cdot 1 = 1 \]. Since the determinant is 1, which is not zero, the matrix is invertible.
03

Create an Augmented Matrix

To find the inverse, set up the augmented matrix \([A | I]\), where \(I\) is the identity matrix: \[ \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 0 \1 & 1 & 0 & 0 & 1 & 0 \0 & 1 & 1 & 0 & 0 & 1 \end{array} \right] \].
04

Apply Row Operations to Reach Row Echelon Form

Using row operations, transform the left side to the identity matrix. Row operations are reversible changes:1. \(R_2 - R_1 \to R_2\): \[ \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 0 \ 0 & 1 & 0 & -1 & 1 & 0 \ 0 & 1 & 1 & 0 & 0 & 1 \end{array} \right] \]2. \(R_3 - R_2 \to R_3\): \[ \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 0 \ 0 & 1 & 0 & -1 & 1 & 0 \ 0 & 0 & 1 & 1 & -1 & 1 \end{array} \right] \].
05

Obtain the Identity Matrix on the Left Side

Continue row operations to turn the left into the identity matrix:1. \(R_1 - 0(R_2) - 0(R_3) \to R_1\) confirms \(R_1\) unchanged.2. \(R_2 + R_1 \to R_2\): \[ \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 0 \ 0 & 1 & 0 & 0 & 1 & 0 \ 0 & 0 & 1 & 1 & -1 & 1 \end{array} \right] \].
06

Extract the Inverse Matrix

After converting the left side to the identity matrix via row operations, what's left on the right side is the inverse of the original matrix:\[ A^{-1} = \left[ \begin{array}{ccc} 1 & 0 & 0 \ -1 & 1 & 0 \ 1 & -1 & 1 \end{array} \right] \].

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Key Concepts

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

Determinant Calculation
Calculating the determinant is essential to find out whether a matrix is invertible. The determinant is a special number calculated from a square matrix. If the determinant is non-zero, the matrix can be inverted. For a 3x3 matrix, the determinant is given by the formula:
  • Multiply the diagonal elements from top left to bottom right and subtract the product of the other diagonals.
  • Determinant for matrix \[ A = \begin{bmatrix} 1 & 0 & 0 \ 1 & 1 & 0 \ 0 & 1 & 1 \end{bmatrix} \] is calculated as follows:
  • \[ \text{det}(A) = 1(1 \cdot 1 - 0 \cdot 1) - 0(1 \cdot 1 - 0 \cdot 0) + 0(1 \cdot 1 - 1 \cdot 0) = 1 \cdot 1 = 1.\]

This result shows us that the determinant is 1, which confirms that the matrix is invertible.
Row Operations
Row operations are steps we use to simplify matrices. They help transform the matrix into a form that's easier to work with. Important types of row operations include:
  • Swapping two rows.
  • Multiplying a row by a non-zero scalar.
  • Adding or subtracting a multiple of one row to another.

In matrix inversion, the goal is to use these operations to change the left side of the augmented matrix into an identity matrix. By doing so, the operations on the right side reveal the inverse matrix. These operations are purposely reversible, ensuring the overall matrix does not change its inherent properties.
In the example provided, specific operations such as \(R_2 - R_1 \rightarrow R_2\) and \(R_3 - R_2 \rightarrow R_3\) are applied to reach a desired row form. This moves the original matrix's elements toward forming the identity on its side.
Augmented Matrix
An augmented matrix is used when we want to apply row operations to find solutions or invert matrices. It combines the original matrix with another, aiding in manipulation and calculation. To form an augmented matrix:
  • Place the original matrix, \(A\), on the left.
  • Position the identity matrix, \(I\), on the right to get \([A | I]\).

The idea is to use row operations so that the left half transforms into the identity matrix while the right half changes into the inverse of \(A\). In our context, this is represented as:\[\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 0 \1 & 1 & 0 & 0 & 1 & 0 \0 & 1 & 1 & 0 & 0 & 1 \end{array} \right]\]
This setup aids in simultaneously transforming an equation system and directly finding the inverse.
Identity Matrix
The identity matrix is an important concept in linear algebra. It's a square matrix where all the elements on the main diagonal are ones, and all other elements are zeros. In matrix terms, it's the equivalent of the number 1 in basic arithmetic.
  • This matrix does not affect other matrices when involved in multiplication.
  • It's denoted typically as \(I\) and sized appropriately to fit the context, such as \(I_3\) for a 3x3 matrix.

When finding an inverse, achieving the identity matrix on one side of an augmented matrix means the right side has transformed into the inverse matrix. Through row operations, transforming the left side of the augmented matrix to the identity allows for the extraction of the inverse matrix from the right side. For instance, reaching:\[\left[ \begin{array}{ccc} 1 & 0 & 0 \0 & 1 & 0 \0 & 0 & 1 \end{array} \right]\]means success in retrieving \(A^{-1}\) from the augmented part.

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Most popular questions from this chapter

Graph the solution set of each system of inequalities by hand. $$\begin{aligned} &y \geq 3^{x}\\\ &y \geq 2 \end{aligned}$$

Solve each system graphically. Check your solutions. Do not use a calculator. $$\begin{array}{r}3 x-y=4 \\\x+y=0\end{array}$$

Graph the solution set of each system of inequalities by hand. $$\begin{aligned} 2 x+3 y & \leq 12 \\ 2 x+3 y & > -6 \\ 3 x+y & < 4 \\ x & \geq 0 \\ y & \geq 0 \end{aligned}$$

To analyze population dynamics of the northern spotted owl, mathematical ecologists divided the female owl population into three categories: juvenile (up to 1 year old), subadult (1 to 2 years old), and adult (over 2 years old). They concluded that the change in the makeup of the northern spotted owl population in successive years could be described by the following matrix equation. $$\left[\begin{array}{c} j_{n+1} \\ s_{n+1} \\ a_{n+1} \end{array}\right]=\left[\begin{array}{rrr} 0 & 0 & 0.33 \\ 0.18 & 0 & 0 \\ 0 & 0.71 & 0.94 \end{array}\right]\left[\begin{array}{c} j_{n} \\ s_{n} \\ a_{n} \end{array}\right]$$ The numbers in the column matrices give the numbers of females in the three age groups after \(n\) years and \(n+1\) years. Multiplying the matrices yields the following. $$\begin{aligned} &j_{n+1}=0.33 a_{n}\\\ &s_{n+1}=0.18 j_{n}\\\ &a_{n+1}=0.71 s_{n}+0.94 a_{n} \end{aligned}$$ (Source: Lamberson, R. H., R. McKelvey, B. R. Noon, and C. Voss, "A Dynamic Analysis of Northern Spotted Owl Viability in a Fragmented Forest Landscape," Conservation Biology, Vol. \(6, \text { No. } 4 .)\) (a) Suppose there are currently 3000 female northern spotted owls: 690 juveniles, 210 subadults, and 2100 adults. Use the preceding matrix equation to determine the total number of female owls for each of the next 5 years. (b) Using advanced techniques from linear algebra, we can show that, in the long run, $$ \left[\begin{array}{c} j_{n+1} \\ s_{n+1} \\ a_{n+1} \end{array}\right]=0.98359\left[\begin{array}{c} j_{n} \\ s_{n} \\ a_{n} \end{array}\right] $$ What can we conclude about the long-term fate of the northern spotted owl? (c) In this model, the main impediment to the survival of the northern spotted owl is the number 0.18 in the second row of the \(3 \times 3\) matrix. This number is low for two reasons: The first year of life is precarious for most animals living in the wild, and juvenile owls must eventually leave the nest and establish their own territory. If much of the forest near their original home has been cleared, then they are vulnerable to predators while searching for a new home. Suppose that, due to better forest management, the number 0.18 can be increased to \(0.3 .\) Rework part (a) under this new assumption.

Graph the solution set of each system of inequalities by hand. $$\begin{array}{r} x \geq 0 \\ x+y \leq 4 \\ 2 x+y \leq 5 \end{array}$$

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