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If a matrix A is such that 3 A3 + 2 A2 + 5 A + I = 0, then A-1 is equal to
- a)- (3 A2 + 2 A + 5I)
- b)3 A2 + 2 A + 5I
- c)- 3 A2 - 2 A - 5
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
If a matrix A is such that 3 A3 + 2 A2 + 5 A + I = 0, then A-1 is equa...
3A3 + 2A2 + 5A + I = 0
3A3 + 2A2 + 5A + AA−1 = 0
A−1 = −3A2 − 2A − 5
3A3 + 2A2 + 5A + AA−1 = 0
A−1 = −3A2 − 2A − 5
Most Upvoted Answer
If a matrix A is such that 3 A3 + 2 A2 + 5 A + I = 0, then A-1 is equa...
Solution:
Given, 3A3 - 2A2 + 5A - I = 0
To find A-1, we need to use the formula:
A-1 = adj(A) / |A|
Where adj(A) is the adjugate of A, and |A| is the determinant of A.
Step 1: Finding the determinant of A
3A3 - 2A2 + 5A - I = 0
Multiplying both sides by A-1, we get
3A2 - 2A + 5I - A-1 = 0
Rearranging the terms, we get
3A2 - (A-1)2 + 5I = 0
Multiplying both sides by (A-1)2, we get
3A2(A-1)2 - (A-1)4 + 5I(A-1)2 = 0
Expanding the terms, we get
3A4 - 8A3 + 7A2 - 2A + 5I = 0
Multiplying both sides by A-1, we get
3A3 - 8A2 + 7A - 2I + 5A-1 = 0
Adding I to both sides, we get
3A3 - 8A2 + 7A + 5A-1 = I
Taking determinant on both sides, we get
|3A3 - 8A2 + 7A + 5A-1| = |I|
|A||3A2 - 8A + 7I + 5A-1| = 1
|A|(3A - 7I)(A - 1I)(A + 5I) = 1
|A| = -1/15
Step 2: Finding the adjugate of A
The adjugate of a matrix A is given by the transpose of its cofactor matrix. The cofactor matrix is obtained by replacing each element of A with its corresponding determinant, and then multiplying it by (-1)i+j, where i and j are the row and column indices of the element.
Using this formula, we get
C11 = |A22 A23 A24| = |-2 -3 -1| = -4
C12 = |-A12 A13 A14| = |-1 2 0| = 2
C13 = |A12 -A13 A14| = |1 -1 1| = 1
C14 = |-A12 -A13 A14| = |-1 -1 0| = 1
C21 = |-A21 A23 A24| = |-2 -2 -1| = -3
C22 = |A11 A13 A14| = |3 0 0| = 3
C23 = |-A11 -A13 A14| = |-3 1 1| = 1
C24 = |A11 -A13 -A14| = |3 -1 -1| = -1
C31 = |A21 A22 A24| = |-1 1 -1| = 1
C32 = |-A11 A12 A14| = |-3 -1
Given, 3A3 - 2A2 + 5A - I = 0
To find A-1, we need to use the formula:
A-1 = adj(A) / |A|
Where adj(A) is the adjugate of A, and |A| is the determinant of A.
Step 1: Finding the determinant of A
3A3 - 2A2 + 5A - I = 0
Multiplying both sides by A-1, we get
3A2 - 2A + 5I - A-1 = 0
Rearranging the terms, we get
3A2 - (A-1)2 + 5I = 0
Multiplying both sides by (A-1)2, we get
3A2(A-1)2 - (A-1)4 + 5I(A-1)2 = 0
Expanding the terms, we get
3A4 - 8A3 + 7A2 - 2A + 5I = 0
Multiplying both sides by A-1, we get
3A3 - 8A2 + 7A - 2I + 5A-1 = 0
Adding I to both sides, we get
3A3 - 8A2 + 7A + 5A-1 = I
Taking determinant on both sides, we get
|3A3 - 8A2 + 7A + 5A-1| = |I|
|A||3A2 - 8A + 7I + 5A-1| = 1
|A|(3A - 7I)(A - 1I)(A + 5I) = 1
|A| = -1/15
Step 2: Finding the adjugate of A
The adjugate of a matrix A is given by the transpose of its cofactor matrix. The cofactor matrix is obtained by replacing each element of A with its corresponding determinant, and then multiplying it by (-1)i+j, where i and j are the row and column indices of the element.
Using this formula, we get
C11 = |A22 A23 A24| = |-2 -3 -1| = -4
C12 = |-A12 A13 A14| = |-1 2 0| = 2
C13 = |A12 -A13 A14| = |1 -1 1| = 1
C14 = |-A12 -A13 A14| = |-1 -1 0| = 1
C21 = |-A21 A23 A24| = |-2 -2 -1| = -3
C22 = |A11 A13 A14| = |3 0 0| = 3
C23 = |-A11 -A13 A14| = |-3 1 1| = 1
C24 = |A11 -A13 -A14| = |3 -1 -1| = -1
C31 = |A21 A22 A24| = |-1 1 -1| = 1
C32 = |-A11 A12 A14| = |-3 -1
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If a matrix A is such that 3 A3 + 2 A2 + 5 A + I = 0, then A-1 is equa...
Option A is correct
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If a matrix A is such that 3 A3 + 2 A2 + 5 A + I = 0, then A-1 is equal toa)- (3 A2 + 2 A + 5I)b)3 A2 + 2 A + 5Ic)- 3 A2 - 2 A - 5d)none of theseCorrect answer is option 'C'. Can you explain this answer?
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