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Chapter 9: Problem 11

Use the Gauss-Jordan method to find \(\mathbf{A}^{-1}\), if it exists. Check your answers by using a graphing calculator to find \(\mathbf{A}^{-1} \mathbf{A}\) and \(\mathbf{A} \mathbf{A}^{-1}\). $$\mathbf{A}=\left[\begin{array}{rrr}1 & -4 & 8 \\ 1 & -3 & 2 \\ 2 & -7 & 10\end{array}\right]$$

Short Answer

Expert verified
The inverse of \(\mathbf{A}\) does not exist because it has a row of zeroes.

Step by step solution

01

- Augment the Matrix

Begin by writing the augmented matrix of \(\mathbf{A}\) with the identity matrix \(\mathbf{I}\) on its right: \[\left[\begin{array}{rrr|rrr}1 & -4 & 8 & 1 & 0 & 0 \ 1 & -3 & 2 & 0 & 1 & 0 \ 2 & -7 & 10 & 0 & 0 & 1\end{array}\right]\]
02

- Row Operations to Form Row Echelon Form

Use row operations to transform the matrix to row echelon form. First, subtract the first row from the second row: \[ R_2 = R_2 - R_1: \left[\begin{array}{rrr|rrr}1 & -4 & 8 & 1 & 0 & 0 \ 0 & 1 & -6 & -1 & 1 & 0 \ 2 & -7 & 10 & 0 & 0 & 1\end{array}\right] \]
03

- Row Operations to Form Row Echelon Form (continued)

Continue with row operations. Subtract 2 times the first row from the third row: \[ R_3 = R_3 - 2R_1: \left[\begin{array}{rrr|rrr}1 & -4 & 8 & 1 & 0 & 0 \ 0 & 1 & -6 & -1 & 1 & 0 \ 0 & 1 & -6 & -2 & 0 & 1\end{array}\right] \]
04

- Making Row 3 Unique

To make the third row unique, subtract the second row from the third row: \[ R_3 = R_3 - R_2: \left[\begin{array}{rrr|rrr}1 & -4 & 8 & 1 & 0 & 0 \ 0 & 1 & -6 & -1 & 1 & 0 \ 0 & 0 & 0 & -1 & -1 & 1\end{array}\right] \]
05

- Obtaining Reduced Row Echelon Form

Continue row operations until reaching reduced row echelon form. Notice \(\mathbf{A}\) has a row of zeroes, which means \(\mathbf{A}^{-1}\) doesn't exist.
06

Conclusion

Since \(\mathbf{A}\) doesn't have full rank (it has a row of zeroes), \(\mathbf{A}^{-1}\) does not exist.

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Key Concepts

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

Matrix Inversion
Matrix inversion is the process of finding the matrix \(\mathbf{A}^{-1}\), such that when multiplied with the original matrix \(\mathbf{A}\), it produces the identity matrix \(\mathbf{I}\). In mathematical terms, this can be expressed as \(\mathbf{A} \mathbf{A}^{-1} = \mathbf{I}\). Only square matrices (same number of rows and columns) can have an inverse, and not all square matrices are invertible. A matrix that can be inverted is called non-singular or invertible. \To find the inverse of a matrix using the Gauss-Jordan method, follow these steps: \
  • Augment the original matrix \(\mathbf{A}\) with the identity matrix \(\mathbf{I}\), resulting in an augmented matrix.
  • Use row operations to transform the left side of the augmented matrix to the identity matrix.
  • If successful, the right side of the augmented matrix will become the inverse of \(\mathbf{A}\).
  • If you cannot form the identity matrix due to a row of zeroes or other issues, the matrix is not invertible.
Row Echelon Form
Row echelon form (REF) is a form of a matrix used to solve systems of linear equations. In this form:
  • All the rows consisting entirely of zeros are at the bottom.
  • The leading entry (first non-zero number from the left) in each non-zero row after the first occurs to the right of the leading entry in the preceding row.
  • The leading entry in any non-zero row is 1.
  • The leading entry is the only non-zero number in its column.
Achieving REF involves using row operations: swapping rows, multiplying a row by a non-zero scalar, and adding or subtracting multiples of rows from each other. It's an essential step in Gaussian elimination.
Reduced Row Echelon Form
Reduced row echelon form (RREF) takes row echelon form a step further. In addition to the criteria for REF, RREF requires:
  • Each leading 1 is the only non-zero entry in its column.
  • All entries above and below each leading 1 are zeros.
The RREF of a matrix is unique, meaning every matrix has exactly one reduced row echelon form. This form makes it easier to identify the solutions to a system of linear equations. Conversion to RREF from REF involves additional row operations to ensure that all columns with a leading entry contain zeros elsewhere.
Rank of a Matrix
The rank of a matrix is the dimension of the vector space spanned by its rows or columns. It represents the number of linearly independent rows or columns in the matrix. For a matrix \(\mathbf{A}\), the rank tells us:
  • If it's full rank, meaning its rows or columns span the maximum possible space, the matrix is invertible.
  • If it has less than full rank, it indicates dependencies among rows or columns, implying the matrix is singular (not invertible).
In the example exercise, the matrix \(\mathbf{A}\) ended up with a row of zeroes in its process to RREF, showing it has less than full rank and confirming that \(\mathbf{A}^{-1}\) does not exist.

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Most popular questions from this chapter

Simplify. Write answers in the form \(a+b i,\) where a and \(b\) are real numbers. $$(3-4 i)-(-2-i)$$

Decompose into partial fractions. $$\frac{9 x^{3}-24 x^{2}+48 x}{(x-2)^{4}(x+1)}$$

Solve the system of equations using the inverse of the coefficient matrix of the equivalent matrix equation. $$\begin{aligned}&2 x-3 y=7\\\&4 x+y=-7\end{aligned}$$

a system of equations is given, together with the inverse of the coefficient matrix. Use the inverse of the coefficient matrix to solve the system of equations. $$\begin{aligned} 11 x+3 y &=-4 \\ 7 x+2 y &=5 \end{aligned}$$ $$\mathbf{A}^{-1}=\left[\begin{array}{rr}2 & -3 \\ -7 & 11\end{array}\right]$$

a system of equations is given, together with the inverse of the coefficient matrix. Use the inverse of the coefficient matrix to solve the system of equations. $$\begin{aligned} y-z &=-4 \\ 4 x+y &=-3 \\ 3 x-y+3 z &=1 \end{aligned}$$ $$\mathbf{A}^{-1}=\frac{1}{5}\left[\begin{array}{rrr}-3 & 2 & -1 \\ 12 & -3 & 4 \\ 7 & -3 & 4\end{array}\right]$$
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