If A is a square matrix of order 3 and det A = 5, then what is det [(2A) -1 ] equal to ?
1/40
Calculating Determinant of Inverse of Scaled Matrix
This problem asks us to find the determinant of the inverse of a scaled matrix, given the determinant of the original matrix and its order. We need to use fundamental properties of determinants to solve this.
Let A be a square matrix of order n. The key properties of determinants that are relevant here are:
- The determinant of a scalar multiple of a matrix: $det(kA) = k^n \cdot det(A)$, where k is a scalar and n is the order of matrix A.
- The determinant of the inverse of a matrix: $det(A^{-1}) = \frac{1}{det(A)}$, provided $det(A) \neq 0$.
In this question, we are given that A is a square matrix of order 3, so $n=3$. We are also given that $det(A) = 5$. We need to find $det[(2A)^{-1}]$.
Step-by-Step Calculation of det[(2A)$^{-1}$]
First, let's find the determinant of the matrix (2A). Using the property $det(kA) = k^n \cdot det(A)$, with $k=2$ and $n=3$, we have:
$$det(2A) = 2^3 \cdot det(A)$$
Substitute the given value of $det(A) = 5$ into the equation:
$$det(2A) = 8 \cdot 5$$
$$det(2A) = 40$$
Now we need to find the determinant of the inverse of the matrix (2A). Let B = 2A. We want to find $det(B^{-1})$. Using the property $det(B^{-1}) = \frac{1}{det(B)}$, we can write:
$$det[(2A)^{-1}] = \frac{1}{det(2A)}$$
Substitute the value $det(2A) = 40$ we calculated:
$$det[(2A)^{-1}] = \frac{1}{40}$$
Thus, the determinant of $(2A)^{-1}$ is $\frac{1}{40}$.
Let's compare this result with the given options:
- Option 1: 1/10
- Option 2: 2/5
- Option 3: 8/5
- Option 4: 1/40
Our calculated value, $\frac{1}{40}$, matches Option 4.
Revision Table: Key Determinant Properties
| Property | Formula (for matrix A of order n) | Explanation |
|---|---|---|
| Determinant of Scalar Multiple | $det(kA) = k^n \cdot det(A)$ | Scaling a matrix by k scales its determinant by $k^n$. |
| Determinant of Inverse | $det(A^{-1}) = \frac{1}{det(A)}$ | The determinant of the inverse is the reciprocal of the original determinant (if det(A) ≠ 0). |
| Determinant of Product | $det(AB) = det(A) \cdot det(B)$ | The determinant of a matrix product is the product of their determinants. |
| Determinant of Transpose | $det(A^T) = det(A)$ | The determinant of the transpose of a matrix is the same as the determinant of the original matrix. |
Additional Information: Understanding Matrix Determinants
The determinant is a scalar value that can be computed from the elements of a square matrix. It has several important properties and uses in linear algebra.
- Geometric Interpretation: For a 2x2 matrix, the absolute value of the determinant represents the area of the parallelogram formed by the column vectors (or row vectors). For a 3x3 matrix, it represents the volume of the parallelepiped formed by the column vectors.
- Invertibility: A square matrix A is invertible (has an inverse $A^{-1}$) if and only if its determinant, $det(A)$, is non-zero. A matrix with a determinant of zero is called a singular matrix.
- Solving Linear Systems: Determinants are used in Cramer's rule to solve systems of linear equations.
- Eigenvalues: Determinants are used in finding the eigenvalues of a matrix, which are crucial in many areas of science and engineering.
Understanding how determinants behave under operations like scalar multiplication and inversion is fundamental to working with matrices.
Important Questions from Determinants
Let $A$ and $B$ be two invertible matrices of order $3 \times 3$. If $\det(A^2 B (A^T)^3) = 16$ and $\det(A^3 B^{-2}) = 32$, then $\det(B^2 A^{-1} (B^T)^3)$ is equal to:
The value of determinant \(\left| {\begin{array}{*{20}{c}} {a - b - c}&{2a}&{2a}\\ {2b}&{b - c - a}&{2b}\\ {2c}&{2c}&{c - a - b} \end{array}} \right|\) is:
If \(\left| {\begin{array}{*{20}{c}} 5&a\\ a&2 \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 2&1\\ 3&2 \end{array}} \right|\), then the values of a are:
The value of the determinant \(\left| {\begin{array}{*{20}{c}} {\sqrt {13} + \sqrt 3 }&{2\sqrt 5 }&{\sqrt 5 }\\ {\sqrt {15} + \sqrt {26} }&5&{\sqrt {10} }\\ {3 + \sqrt {65} }&{\sqrt {15} }&5 \end{array}} \right|\) is
The determinant \(\left| {\begin{array}{*{20}{c}} {xp + y}&x&y\\ {yp + z}&y&z\\ 0&{xp + y}&{yp + z} \end{array}} \right| = 0,\) if
