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Math  /  Algebra

QuestionFind A−1A^{-1} by forming [A∥][A \|] and then using row operations to obtain [∥B][\| B], where A−1=[B]A^{-1}=[B]. Check that AA−1=IA A^{-1}=I and A−1A=IA^{-1} A=I. A=[44−103−1−1−41]A=\left[\begin{array}{rrr} 4 & 4 & -1 \\ 0 & 3 & -1 \\ -1 & -4 & 1 \end{array}\right]
Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. A−1=A^{-1}= □\square (Type an integer or a simplified fraction for each matrix element.) B. The matrix does not have an inverse.

Studdy Solution

STEP 1

1. The matrix A A is a square matrix, which is a necessary condition for the existence of an inverse.
2. The inverse of a matrix A A , denoted A−1 A^{-1} , exists if and only if A A is non-singular, i.e., its determinant is non-zero.
3. We will use the augmented matrix method to find A−1 A^{-1} .

STEP 2

1. Form the augmented matrix [A∣I][A \mid I], where I I is the identity matrix.
2. Use row operations to transform [A∣I][A \mid I] into [I∣B][I \mid B], where B=A−1 B = A^{-1} .
3. Verify that AA−1=I A A^{-1} = I and A−1A=I A^{-1} A = I .

STEP 3

Form the augmented matrix [A∣I][A \mid I]:
[A∣I]=[44−110003−1010−1−41001][A \mid I] = \left[\begin{array}{ccc|ccc} 4 & 4 & -1 & 1 & 0 & 0 \\ 0 & 3 & -1 & 0 & 1 & 0 \\ -1 & -4 & 1 & 0 & 0 & 1 \end{array}\right]

STEP 4

Use row operations to transform [A∣I][A \mid I] into [I∣B][I \mid B].
1. Make the first pivot 1 by swapping Row 1 and Row 3:
[−1−4100103−101044−1100]\left[\begin{array}{ccc|ccc} -1 & -4 & 1 & 0 & 0 & 1 \\ 0 & 3 & -1 & 0 & 1 & 0 \\ 4 & 4 & -1 & 1 & 0 & 0 \end{array}\right]
2. Multiply Row 1 by −1-1 to make the pivot positive:
[14−100−103−101044−1100]\left[\begin{array}{ccc|ccc} 1 & 4 & -1 & 0 & 0 & -1 \\ 0 & 3 & -1 & 0 & 1 & 0 \\ 4 & 4 & -1 & 1 & 0 & 0 \end{array}\right]
3. Eliminate the first column below the pivot by replacing Row 3 with Row 3 minus 4 times Row 1:
[14−100−103−10100−123104]\left[\begin{array}{ccc|ccc} 1 & 4 & -1 & 0 & 0 & -1 \\ 0 & 3 & -1 & 0 & 1 & 0 \\ 0 & -12 & 3 & 1 & 0 & 4 \end{array}\right]
4. Make the second pivot 1 by dividing Row 2 by 3:
[14−100−101−1301300−123104]\left[\begin{array}{ccc|ccc} 1 & 4 & -1 & 0 & 0 & -1 \\ 0 & 1 & -\frac{1}{3} & 0 & \frac{1}{3} & 0 \\ 0 & -12 & 3 & 1 & 0 & 4 \end{array}\right]
5. Eliminate the second column above and below the pivot by replacing Row 1 with Row 1 minus 4 times Row 2 and Row 3 with Row 3 plus 12 times Row 2:
[10130−43−101−13013000−1148]\left[\begin{array}{ccc|ccc} 1 & 0 & \frac{1}{3} & 0 & -\frac{4}{3} & -1 \\ 0 & 1 & -\frac{1}{3} & 0 & \frac{1}{3} & 0 \\ 0 & 0 & -1 & 1 & 4 & 8 \end{array}\right]
6. Make the third pivot 1 by multiplying Row 3 by −1-1:
[10130−43−101−130130001−1−4−8]\left[\begin{array}{ccc|ccc} 1 & 0 & \frac{1}{3} & 0 & -\frac{4}{3} & -1 \\ 0 & 1 & -\frac{1}{3} & 0 & \frac{1}{3} & 0 \\ 0 & 0 & 1 & -1 & -4 & -8 \end{array}\right]
7. Eliminate the third column above the pivot by replacing Row 1 with Row 1 minus 13\frac{1}{3} times Row 3 and Row 2 with Row 2 plus 13\frac{1}{3} times Row 3:
[10013013010−13−1−83001−1−4−8]\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{1}{3} & 0 & \frac{1}{3} \\ 0 & 1 & 0 & -\frac{1}{3} & -1 & -\frac{8}{3} \\ 0 & 0 & 1 & -1 & -4 & -8 \end{array}\right]

STEP 5

The matrix B B is the inverse A−1 A^{-1} :
A−1=[13013−13−1−83−1−4−8]A^{-1} = \left[\begin{array}{ccc} \frac{1}{3} & 0 & \frac{1}{3} \\ -\frac{1}{3} & -1 & -\frac{8}{3} \\ -1 & -4 & -8 \end{array}\right]
Verify that AA−1=I A A^{-1} = I and A−1A=I A^{-1} A = I by performing matrix multiplication.
The inverse matrix A−1 A^{-1} is:
[13013−13−1−83−1−4−8]\boxed{\left[\begin{array}{ccc} \frac{1}{3} & 0 & \frac{1}{3} \\ -\frac{1}{3} & -1 & -\frac{8}{3} \\ -1 & -4 & -8 \end{array}\right]}

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