Problem 28
Question
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 \(A^{-1} A=I\) $$A=\left[\begin{array}{rrrr}2 & 0 & 0 & 1 \\\0 & 1 & 0 & 0 \\\0 & 0 & -1 & 0 \\\0 & 0 & 0 & 2 \end{array}\right]$$
Step-by-Step Solution
Verified Answer
The inverse matrix \(A^{-1}\) is \(\left[\begin{array}{cccc}0.5 & 0 & 0 & 0\ 0 & 1 & 0 & 0 \0 & 0 & -1 & 0 \0 & 0 & 0 & 0.5 \end{array}\right]\)
1Step 1: Form the Augmented Matrix [A | I]
Form the augmented matrix by appending the identity matrix \(I\) to right of the given matrix \(A\), to create the following matrix: \[\left[\begin{array}{cccc|cccc}2 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 \0 & 0 & 0 & 2 & 0 & 0 & 0 & 1\end{array}\right]\]
2Step 2: Apply Row Operations
Apply a row operation to convert the main diagonal into ones to obtain the resulting matrix: \[\left[\begin{array}{cccc|cccc}1 & 0 & 0 & 0.5 & 0.5 & 0 & 0 & 0 \0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \0 & 0 & 1 & 0 & 0 & 0 & -1 & 0 \0 & 0 & 0 & 1 & 0 & 0 & 0 & 0.5\end{array}\right]\] This operation involves dividing the first row by 2 and the last row by 2, and multiplying third row by -1.
3Step 3: Get the Inverse Matrix
After applying the row operations, the matrix takes the form \([I | B]\), where \(B\) is the inverse of matrix \(A\). So, \(A^{-1} = B = \left[\begin{array}{cccc}0.5 & 0 & 0 & 0\ 0 & 1 & 0 & 0 \0 & 0 & -1 & 0 \0 & 0 & 0 & 0.5 \end{array}\right]\)
4Step 4: Verify the results
The final step involves verifying that the inverse is correct by checking whether \(AA^{-1} = A^{-1}A = I\). Calculating both products will give us the identity matrix, confirming that the calculations are correct.
Key Concepts
Augmented MatrixRow OperationsIdentity Matrix
Augmented Matrix
An augmented matrix is a combination of two matrices put side by side, usually to solve a system of linear equations or in this case, to find the inverse of a matrix. It is formed by taking an original matrix, known as matrix A, and appending another matrix, often the identity matrix, to its right. When working to find the inverse, we write the augmented matrix as \[A | I\], where I represents the identity matrix.
- The augmented matrix condenses the information of two matrices into one convenient notation, simplifying calculations.
- It is a particularly useful tool when applying row operations, as it keeps track of changes made during the process applied to both sides equally.
Row Operations
Row operations are essential maneuvers used to manipulate matrices in various matrix algebra problems, such as determining the inverse of a matrix. There are three fundamental types of row operations: swapping rows, multiplying a row by a non-zero scalar, and adding or subtracting a multiple of one row to another.
Types of Row Operations
- Row Swapping: Switching the positions of two rows.
- Scalar Multiplication: Multiplying all elements of a row by a non-zero constant.
- Row Addition/Subtraction: Adding or subtracting the elements of one row from those of another row, often after multiplying by a scalar.
Identity Matrix
The identity matrix, denoted by I, is a square matrix with ones on the main diagonal and zeros everywhere else. In linear algebra, it is the equivalent of the number 1 in multiplication since any matrix multiplied by the identity matrix will result in the original matrix. Key characteristics of the identity matrix include:
- It acts as a neutral element in matrix multiplication.
- When finding the inverse of a matrix, the goal is to transform the original matrix into the identity matrix through row operations.
Other exercises in this chapter
Problem 28
Find (if possible) the following matrices: \(a, A B\) \(\boldsymbol{b}, B A\) $$A=\left[\begin{array}{rr}3 & -2 \\\1 & 5\end{array}\right], \quad B=\left[\begin
View solution Problem 28
Evaluate each determinant. $$\left|\begin{array}{rrr}1 & 2 & 3 \\\2 & 2 & -3 \\\3 & 2 & 1\end{array}\right|$$
View solution Problem 28
solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{array}{l} 3 x+y-z=
View solution Problem 29
Find (if possible) the following matrices: \(a, A B\) \(\boldsymbol{b}, B A\) $$A=\left[\begin{array}{lllll}1 & 2 & 3 & 4\end{array}\right], \quad B=\left[\begi
View solution