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If |A| = |kA|, where A is a square matrix of order 2, then sum of all possible values of k is
We are given that A is a square matrix of order 2, and |A| = |kA|. We have to compute the sum of all possible values of k. First recall that if A is any square matrix of order 2, then the determinant of kA where k is any scalar is given by |kA| = k² |A| This is because for a 2x2 matrix, multiplicatiRead more
We are given that A is a square matrix of order 2, and |A| = |kA|. We have to compute the sum of all possible values of k.
First recall that if A is any square matrix of order 2, then the determinant of kA where k is any scalar is given by
|kA| = k² |A|
This is because for a 2×2 matrix, multiplication of the matrix A by scalar k scales the determinant by k².
We know that |A| = |kA|. Using the above formula, we substitute for the value of |kA|,
|A| = k² |A|
If |A| ≠ 0, we can divide both sides by |A| to get,
1 = k²
This gives two values for k
k = 1 or k = -1
Therefore, the sum of all possible values of k is:
1 + (-1) = 0
So, the correct answer is 0.
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If for a square matrix A, A² – A + I = O, then A⁻¹ equals
We have the equation of a square matrix A as A² - A + I = O Rearrange the equation to get A² - A = -I Now we factor the left-hand side, A(A - I) = -I To get the inverse of A (A⁻¹), multiply both sides of the equation by A⁻¹, A⁻¹ * A(A - I) = A⁻¹ * (-I) This gives us, (A - I) = -A⁻¹ Thus we can writeRead more
We have the equation of a square matrix A as
A² – A + I = O
Rearrange the equation to get
A² – A = -I
Now we factor the left-hand side,
A(A – I) = -I
To get the inverse of A (A⁻¹), multiply both sides of the equation by A⁻¹,
A⁻¹ * A(A – I) = A⁻¹ * (-I)
This gives us,
(A – I) = -A⁻¹
Thus we can write
A⁻¹ = I – A
Hence, the correct answer is option (c) I – A.
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If A is any square matrix of order 3 x 3 such that |adj. A| = 25 and |A| is non-positive, then the value of |A| is
We are given that A is a square matrix of order 3x3 and |adj(A)| = 25. We need to find the value of |A|. The formula for the relation between the determinant of a matrix A and its adjoint is given as under: |adj(A)| = |A|(n -1), where n is the order of the given matrix. When n = 3 for 3x3 matrix, itRead more
We are given that A is a square matrix of order 3×3 and |adj(A)| = 25. We need to find the value of |A|.
The formula for the relation between the determinant of a matrix A and its adjoint is given as under:
|adj(A)| = |A|(n -1), where n is the order of the given matrix.
When n = 3 for 3×3 matrix, it becomes:
|adj(A)| = |A|²
We are given that |adj(A)| = 25.
|A|² = 25
We find the square root of both sides of the equation
|A| = ±5
Since we know that |A| is non-positive we choose negative sign
|A| = -5
So, the correct answer is -5.
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If A is a square matrix such that |A| = 5, the value of |AAᵀ| is
A square matrix A |A| = 5 find |AA T| Theorem on Determinants for Matrix multiplication: |AB| = |A||B| AAᵀ For matrix multiplication. Then, using this theorem. |AA ᵀ |= |A|. |A|ᵀ| AṀ A being a square Matrix | AṀ ṀT|= |AT | Hence, |Aᵀ| = |A|. And we get; |AA T |= | A||A | = | A|² Since, the value ofRead more
A square matrix A |A| = 5 find |AA T|
Theorem on Determinants for Matrix multiplication:
|AB| = |A||B|
AAᵀ For matrix multiplication. Then, using this theorem. |AA ᵀ |= |A|. |A|ᵀ|
AṀ A being a square Matrix
| AṀ ṀT|= |AT |
Hence, |Aᵀ| = |A|.
And we get;
|AA T |= | A||A | = | A|²
Since, the value of A = 5
|AAᵀ| = 5² = 25
Therefore, the correct value of |AAᵀ| is 25.
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If A is any square matrix of order 3 x 3 such that |A| = 3, then the value of |adj. A| is
We are given that A is a square matrix of order 3x3 and |A| = 3. We have to find the value of |adj(A)|. The relation between the determinant of a matrix A and the determinant of its adjoint adj(A) is given by the formula: |adj(A)| = |A|^(n-1) Here, n = 3 because A is a 3x3 matrix. So the formula becRead more
We are given that A is a square matrix of order 3×3 and |A| = 3. We have to find the value of |adj(A)|.
The relation between the determinant of a matrix A and the determinant of its adjoint adj(A) is given by the formula:
|adj(A)| = |A|^(n-1)
Here, n = 3 because A is a 3×3 matrix. So the formula becomes:
|adj(A)| = |A|^(3-1) = |A|²
Since |A| = 3, we calculate:
|adj(A)| = 3² = 9
Hence, the correct value of |adj(A)| is 9.
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