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In Exercises \(19-28,\) find \(A^{-1}\) by forming \([A | I]\) and then using row operations to obtain \([I | B],\) where \(A^{-1}=[B]\). Check that \(A A^{-1}=I\) and \(A^{-1} A=I\) $$A=\left[\begin{array}{lll}3 & 0 & 0 \\\0 & 6 & 0 \\\0 & 0 & 9\end{array}\right]$$

Short Answer

Expert verified
The inverse of the given matrix A is \( A^{-1} = \left[ \begin{array}{lll} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{6} & 0 \\ 0 & 0 & \frac{1}{9} \end{array} \right] \)

Step by step solution

01

Form the Augmented Matrix [A|I]

The augmented matrix is formed by enriching the original matrix \( A \) with the identity matrix on the right: \[ \left[ \begin{array}{lll|lll} 3 & 0 & 0 & 1 & 0 & 0 \\ 0 & 6 & 0 & 0 & 1 & 0 \\ 0 & 0 & 9 & 0 & 0 & 1 \end{array} \right] \]
02

Apply Row Operations to form [I|B]

Since given matrix is a diagonal matrix, dividing each row by it's corresponding diagonal element will result in: \[ \left[ \begin{array}{lll|lll} 1 & 0 & 0 & \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 & 0 & \frac{1}{6} & 0 \\ 0 & 0 & 1 & 0 & 0 & \frac{1}{9} \end{array} \right] \] This gives us [I|B] where \( B={A^{-1}}={\left[ \begin{array}{lll} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{6} & 0 \\ 0 & 0 & \frac{1}{9} \end{array} \right]} \)
03

Verify that AA^{-1}=I and A^{-1}A=I

This can be verified by multiplying the obtained \( B \) with the original matrix \( A \). We find that both the products \( AB =\left[ \begin{array}{lll} 3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 9 \end{array} \right] \left[ \begin{array}{lll} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{6} & 0 \\ 0 & 0 & \frac{1}{9} \end{array} \right] = \left[ \begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] = I\] and \[ BA=\left[ \begin{array}{lll} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{6} & 0 \\ 0 & 0 & \frac{1}{9} \end{array} \right] \left[ \begin{array}{lll} 3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 9 \end{array} \right] = \left[ \begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] = I\] are equal to the identity matrix.

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Key Concepts

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

Augmented Matrix
An augmented matrix is a crucial concept in linear algebra, especially when finding the inverse of a matrix. It is essentially a composite matrix that combines two matrices: the matrix you are trying to inverse, denoted as \(A\), and the identity matrix of the same size as \(A\). By appending the identity matrix to \(A\), you create a larger matrix written as \([A | I]\).

This process sets the stage for computing an inverse matrix by transforming \(A\) into an identity matrix through row operations. The transformed identity portion on the right becomes the inverse of \(A\). This augmented form is particularly helpful for visualizing changes that occur during matrix operations.
Row Operations
Row operations are the backbone of matrix manipulation and serve as a tool to modify matrices into a desired form. They include three primary types: swapping two rows, multiplying a row by a non-zero scalar, and adding or subtracting the multiple of one row to another.

In the process of finding the inverse of a matrix, row operations are used to systematically simplify the augmented matrix \([A|I]\) into the form \([I|B]\). Performing these operations correctly will change the left side of the augmented matrix to the identity matrix, which allows the right side, \(B\), to reveal the inverse of \(A\).
  • Swapping ensures the matrix can be simplified hierarchically.
  • Multiplying alters row values for easy transitions.
  • Adding or subtracting assists in creating zeros strategically.
Identity Matrix
The identity matrix is a special type of square matrix that has ones on the diagonal and zeros elsewhere. For a matrix \(A\) of size \(n \times n\), the identity matrix \(I\) is also of the same size. It acts as a multiplicative identity element for matrices, much like the number 1 for real numbers.

When calculating the inverse of a matrix, the identity matrix plays a central role as the transformation target. Initially appended to form \([A | I]\), the goal is to transform the matrix \(A\) to resemble \(I\) through row operations. Once this is accomplished, the opposite side becomes the inverse matrix. Multiply any square matrix by the identity, and you are left with the original matrix unchanged.
Inverse Matrix Computation
Inverse matrix computation is a structured process to find a matrix that, when multiplied with the original matrix, results in the identity matrix. For any invertible matrix \(A\), there exists an inverse matrix \(A^{-1}\) such that both \(A \cdot A^{-1} = I\) and \(A^{-1} \cdot A = I\).

To compute \(A^{-1}\), you first form the augmented matrix \([A | I]\) and apply row operations until \(A\) is transformed into the identity matrix \(I\). The computations done during row operations transpose the appended identity matrix into the inverse matrix \(B\) such that \(A^{-1} = B\).

It is important to note that not every matrix has an inverse, and a matrix must be non-singular (with a non-zero determinant) to have an inverse. Checking your results by performing \(A \cdot A^{-1}\) and \(A^{-1} \cdot A\) ensures correctness, as both should return \(I\).

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Most popular questions from this chapter

The figure shows the letter \(L\) in a rectangular coordinate system. (GRAPH CANNOT COPY) The figure can be represented by the matrix $$B=\left[\begin{array}{llllll}0 & 3 & 3 & 1 & 1 & 0 \\\0 & 0 & 1 & 1 & 5 & 5\end{array}\right]$$ Each column in the matrix describes a point on the letter. The order of the columns shows the direction in which a pencil must move to draw the letter. The \(L\) is completed by connecting the last point in the matrix, \((0,5),\) to the starting point, \((0,0) .\) Use these ideas to solve Exercises \(53-60 .\) The figure shows the letter \(L\) in a rectangular coordinate system. (GRAPH CANNOT COPY) The figure can be represented by the matrix $$B=\left[\begin{array}{llllll}0 & 3 & 3 & 1 & 1 & 0 \\\0 & 0 & 1 & 1 & 5 & 5\end{array}\right]$$ Each column in the matrix describes a point on the letter. The order of the columns shows the direction in which a pencil must move to draw the letter. The \(L\) is completed by connecting the last point in the matrix, \((0,5),\) to the starting point, \((0,0) .\) Use these ideas to solve Exercises \(53-60 .\) The figure shows the letter \(L\) in a rectangular coordinate system. (GRAPH CANNOT COPY) The figure can be represented by the matrix $$B=\left[\begin{array}{llllll}0 & 3 & 3 & 1 & 1 & 0 \\\0 & 0 & 1 & 1 & 5 & 5\end{array}\right]$$ Each column in the matrix describes a point on the letter. The order of the columns shows the direction in which a pencil must move to draw the letter. The \(L\) is completed by connecting the last point in the matrix, \((0,5),\) to the starting point, \((0,0) .\) Use these ideas to solve Exercises \(53-60 .\) $$\text { a. If } A=\left[\begin{array}{rr}1 & 0 \\\0 & -1\end{array}\right], \text { find } A B$$ b. Graph the object represented by matrix \(A B\). What effect does the matrix multiplication have on the letter \(L\) represented by matrix \(B\) ?

What is meant by the order of a matrix? Give an example with your explanation.

What is the multiplicative identity matrix?

Let $$\begin{aligned}&A=\left[\begin{array}{ll}1 & 0 \\\0 & 1\end{array}\right], \quad B=\left[\begin{array}{rr}1 & 0 \\\0 & -1\end{array}\right], \quad C=\left[\begin{array}{rr}-1 & 0 \\ 0 & 1\end{array}\right]\\\&D=\left[\begin{array}{rr}-1 & 0 \\\0 & -1\end{array}\right]\end{aligned}$$ Use any three of the matrices to verify a distributive property.

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. Two \(2 \times 2\) invertible matrices can have a matrix sum that is not invertible.

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