If A, B and C are square matrices of order 3 and det(BC) = 2 det(A), then what is the value of det(2A-1BC)?
16
Solving the Matrix Determinant Problem
We are given three square matrices A, B, and C, all of order 3. We are also given the relationship between their determinants: \( \det(BC) = 2 \det(A) \). Our goal is to find the value of \( \det(2A^{-1}BC) \).
To solve this, we will use the fundamental properties of determinants for square matrices.
Key Determinant Properties for Square Matrices
- Determinant of a product: For square matrices X and Y of the same order, \( \det(XY) = \det(X) \det(Y) \).
- Determinant of a scalar multiple: For a scalar \( k \) and a square matrix X of order \( n \), \( \det(kX) = k^n \det(X) \).
- Determinant of an inverse: If a square matrix X is invertible (i.e., \( \det(X) \neq 0 \)), then \( \det(X^{-1}) = \frac{1}{\det(X)} \).
Step-by-Step Calculation of det(2A-1BC)
Let's break down the calculation of \( \det(2A^{-1}BC) \) using the properties listed above:
- Applying the scalar multiplication property to \( \det(2A^{-1}BC) \). The matrix is \( A^{-1}BC \) and the scalar is 2. The order of the matrix is \( n=3 \). \( \det(2A^{-1}BC) = 2^3 \det(A^{-1}BC) = 8 \det(A^{-1}BC) \)
- Now, apply the determinant of a product property to \( \det(A^{-1}BC) \). We can group the matrices as \( (A^{-1})(BC) \) or \( (A^{-1}B)(C) \). Let's use the first grouping: \( \det(A^{-1}BC) = \det(A^{-1}(BC)) = \det(A^{-1}) \det(BC) \)
- Apply the determinant of an inverse property to \( \det(A^{-1}) \). Assuming matrix A is invertible (which must be true for \( A^{-1} \) to exist), we have: \( \det(A^{-1}) = \frac{1}{\det(A)} \)
- Substitute the result from step 3 into the equation from step 2: \( \det(A^{-1}BC) = \left(\frac{1}{\det(A)}\right) \det(BC) \)
- Substitute the result from step 4 back into the equation from step 1: \( \det(2A^{-1}BC) = 8 \times \left(\frac{1}{\det(A)}\right) \det(BC) \) \( \det(2A^{-1}BC) = \frac{8 \det(BC)}{\det(A)} \)
- Now, use the given condition: \( \det(BC) = 2 \det(A) \). Substitute this into the equation from step 5: \( \det(2A^{-1}BC) = \frac{8 \times (2 \det(A))}{\det(A)} \)
- Assuming \( \det(A) \neq 0 \) (as \( A^{-1} \) exists), we can cancel \( \det(A) \) from the numerator and the denominator: \( \det(2A^{-1}BC) = 8 \times 2 \) \( \det(2A^{-1}BC) = 16 \)
Result of det(2A-1BC) Calculation
Based on the properties of determinants and the given condition, the value of \( \det(2A^{-1}BC) \) is 16.
Revision Table: Matrix Determinant Properties
| Property | Formula | Description | Condition |
|---|---|---|---|
| Scalar Multiple | \( \det(kX) = k^n \det(X) \) | Determinant of a matrix multiplied by a scalar \( k \). \( n \) is the order of matrix X. | X is a square matrix of order \( n \). |
| Product | \( \det(XY) = \det(X) \det(Y) \) | Determinant of the product of two matrices. | X and Y are square matrices of the same order. |
| Inverse | \( \det(X^{-1}) = \frac{1}{\det(X)} \) | Determinant of the inverse of a matrix. | X is an invertible square matrix. |
| Transpose | \( \det(X^T) = \det(X) \) | Determinant of the transpose of a matrix. | X is a square matrix. |
Additional Information on Matrix Invertibility
For a square matrix \( A \) to have an inverse, denoted as \( A^{-1} \), its determinant \( \det(A) \) must be non-zero. If \( \det(A) = 0 \), the matrix \( A \) is called a singular matrix, and its inverse does not exist. In this problem, since \( A^{-1} \) is part of the expression \( 2A^{-1}BC \), it is implicitly assumed that \( A \) is invertible, meaning \( \det(A) \neq 0 \). This allows us to use the property \( \det(A^{-1}) = 1/\det(A) \) and also to cancel \( \det(A) \) during the calculation.
Important Questions from Determinants
If \(A=\left[\begin{array}{rrr} 2 & -1 & 0 \\ -1 & 3 & 0 \\ 1 & 0 & 1 \end{array}\right]\), then what is the value of det[adj(adjA)] ?
If \(A=\left[\begin{array}{rrr} 0 & 3 & 4 \\ -3 & 0 & 5 \\ -4 & -5 & 0 \end{array}\right]\), then which one of the following statements is correct?
If \(\left|\begin{array}{ccc} x^2+3 x & x-1 & x+3 \\ x+1 & -2 x & x-4 \\ x-3 & x+4 & 3 x \end{array}\right|\) = ax4 + bx3 + cx2 + dx + e, then what is the value of e?"
If all elements of a third order determinant are equal to 1 or -1, then the value of the determinant is:
If a, b, c are in AP, then what is \(\left|\begin{array}{lll} x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \\ x+a & x+b & x+c \end{array}\right|\) equal to ?
