a) just checking if I got the correct answer
[6 -2 2]
Matrix A=[ 3 -1 1]
[-9 3 -3]
Compute determinant of A. My answer was Det(A) = 0
B) also need help with this problem
Let U be a square matrix such that U^T*U = I (identity matrix). What are the possible values for Det(U)?
thanks
[6 -2 2]
Matrix A=[ 3 -1 1]
[-9 3 -3]
Compute determinant of A. My answer was Det(A) = 0
B) also need help with this problem
Let U be a square matrix such that U^T*U = I (identity matrix). What are the possible values for Det(U)?
thanks
-
A =
[6 -2 2]
[3 -1 1]
[-9 3 -3]
To find the determinant, keep this in mind:
[+ - +]
[- + -]
[+ - +]
You can choose to evaluate the determinant along any row or any column. Since the second row has the smallest numbers, I'll do so by the second row. Since the entry begins at (2, 1), as per our sign matrix, we start with a minus.
-3 * det[-2 2] + (-1) det [6 2] - (1) det [6 -2]
. . . . . [3 -3] . . . . . . . .[-9 -3] . . . . . [-9 3]
(-3)( (-2)(-3) - (2)(3) ) + (-1)( -18 - (-18)) - (1) (18 - (-9)(-2))
(-3) (6 - 6) + (-1)(-18 + 18) - (1)(18 - 18)
(-3) (0) + (-1) (0) - (1)(0)
Which is 0. You are correct.
B) Let U be a square matrix such that U^T * U = I.
What are the possible values for det(U)?
First off, there is a theorem that states that a matrix's determinant is the same as the determinant of its transpose.
That means it is a given that
det(U^T) = det(U)
Also, the determinant of a product of matrices is equal to the product of the determinants, i.e.
det(AB) = det(A)det(B)
That means if we have
U^T * U = I
it follows that
det( U^T * U ) = det(I)
so
det(U^T) det(U) = det(I)
and since det(U^T) = det(U), it follows that
det(U) * det(U) = det(I)
and the determinant of the identity matrix is 1, so
det(U) * det(U) = 1
[det(U)]^2 = 1
So if we take the square root of both sides,
det(U) = +/- 1
The possible values for det(U) are: -1 and 1.
[6 -2 2]
[3 -1 1]
[-9 3 -3]
To find the determinant, keep this in mind:
[+ - +]
[- + -]
[+ - +]
You can choose to evaluate the determinant along any row or any column. Since the second row has the smallest numbers, I'll do so by the second row. Since the entry begins at (2, 1), as per our sign matrix, we start with a minus.
-3 * det[-2 2] + (-1) det [6 2] - (1) det [6 -2]
. . . . . [3 -3] . . . . . . . .[-9 -3] . . . . . [-9 3]
(-3)( (-2)(-3) - (2)(3) ) + (-1)( -18 - (-18)) - (1) (18 - (-9)(-2))
(-3) (6 - 6) + (-1)(-18 + 18) - (1)(18 - 18)
(-3) (0) + (-1) (0) - (1)(0)
Which is 0. You are correct.
B) Let U be a square matrix such that U^T * U = I.
What are the possible values for det(U)?
First off, there is a theorem that states that a matrix's determinant is the same as the determinant of its transpose.
That means it is a given that
det(U^T) = det(U)
Also, the determinant of a product of matrices is equal to the product of the determinants, i.e.
det(AB) = det(A)det(B)
That means if we have
U^T * U = I
it follows that
det( U^T * U ) = det(I)
so
det(U^T) det(U) = det(I)
and since det(U^T) = det(U), it follows that
det(U) * det(U) = det(I)
and the determinant of the identity matrix is 1, so
det(U) * det(U) = 1
[det(U)]^2 = 1
So if we take the square root of both sides,
det(U) = +/- 1
The possible values for det(U) are: -1 and 1.
-
a) Use the formula for determinants of order 3: |A| = a11 a22 a33 + a12 a23 a31 + a13 a21 a32 - a13 a22 a31 - a12 a21 a33 - a11 a23 a32.
|A| = 6•(-1)•(-3) + (-2)•1•(-9) + 3•3•2 - 2•(-1)•(-9) - (-2)•3•(-3) - 1•3•6 = 0
b) Using these properties:
det(A) = det(A^T)
det(A*B) = det(A)*det(B)
det(U^T*U) = det(U^T)*det(U) = det(U)*det(U) = det(U)^2
as U^T*U = I
det(U^T*U) = det(I) = 1
hence
det(U)^2 = 1
=> det(U) = 1 or det(U) = -1
|A| = 6•(-1)•(-3) + (-2)•1•(-9) + 3•3•2 - 2•(-1)•(-9) - (-2)•3•(-3) - 1•3•6 = 0
b) Using these properties:
det(A) = det(A^T)
det(A*B) = det(A)*det(B)
det(U^T*U) = det(U^T)*det(U) = det(U)*det(U) = det(U)^2
as U^T*U = I
det(U^T*U) = det(I) = 1
hence
det(U)^2 = 1
=> det(U) = 1 or det(U) = -1
-
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