If A is an invertible matrix of order n and k is any positive real number, then the value of [det(kA)] -1 det A is
k -n
Understanding Determinants and Scalar Multiples
This question asks us to evaluate an expression involving the determinant of an invertible matrix A and a scalar multiple of A. The expression is $[det(kA)]^{-1} det A$. To solve this, we need to use properties of determinants, specifically how the determinant changes when a matrix is multiplied by a scalar.
Key Property of Determinants
For an n x n matrix A and a scalar k, the determinant of the scalar multiple kA is given by:
\(det(kA) = k^n det A\)
Here, A is an invertible matrix of order n, and k is a positive real number.
Step-by-Step Calculation
Let's break down the given expression step by step:
We are given the expression: \([det(kA)]^{-1} det A\)
- First, let's find the value of
\(det(kA)\). Using the property mentioned above, for an n x n matrix A and scalar k, we have:\(det(kA) = k^n det A\) - Now, substitute this into the original expression:
\([k^n det A]^{-1} det A\) - Recall that for any non-zero number x,
\(x^{-1} = \frac{1}{x}\). Also, for any numbers a and b,\((ab)^{-1} = a^{-1} b^{-1}\). Applying this to\([k^n det A]^{-1}\):\([k^n det A]^{-1} = (k^n)^{-1} (det A)^{-1}\)Also,
\((k^n)^{-1} = k^{-n}\).So,
\([k^n det A]^{-1} = k^{-n} (det A)^{-1}\) - Substitute this result back into the expression from step 2:
\(k^{-n} (det A)^{-1} det A\) - Since
\(det A\)is a non-zero number (because A is invertible), the product of\((det A)^{-1}\)and\(det A\)is 1:\((det A)^{-1} det A = 1\) - Finally, substitute 1 into the expression from step 4:
\(k^{-n} \times 1 = k^{-n}\)
Thus, the value of the expression \([det(kA)]^{-1} det A\) is \(k^{-n}\).
Analysis of Options
Let's compare our result with the given options:
- Option 1:
\(k^{-n}\) - Option 2:
\(k^{-1}\) - Option 3:
\(k^n\) - Option 4:
\(nk\)
Our calculated value, \(k^{-n}\), matches Option 1.
| Expression Part | Property Used | Result |
|---|---|---|
\(det(kA)\) |
\(det(kA) = k^n det A\) |
\(k^n det A\) |
\([det(kA)]^{-1}\) |
\((xy)^{-1} = x^{-1}y^{-1}\) and \((a^m)^n = a^{mn}\) |
\((k^n det A)^{-1} = k^{-n} (det A)^{-1}\) |
\([det(kA)]^{-1} det A\) |
\(y^{-1} y = 1\) for \(y \neq 0\) |
\(k^{-n} (det A)^{-1} det A = k^{-n} \times 1 = k^{-n}\) |
This confirms our result that the value of \([det(kA)]^{-1} det A\) is \(k^{-n}\).
Revision Table: Determinant Properties
| Property | Description | Formula |
|---|---|---|
| Determinant of Scalar Multiple | If A is an n x n matrix and k is a scalar. | \(det(kA) = k^n det A\) |
| Determinant of Inverse | If A is an invertible matrix. | \(det(A^{-1}) = (det A)^{-1} = \frac{1}{det A}\) |
| Determinant of Product | If A and B are n x n matrices. | \(det(AB) = det A \times det B\) |
| Determinant of Identity | The identity matrix \(I_n\) of order n. |
\(det(I_n) = 1\) |
Additional Information: Invertible Matrices and Determinants
A matrix A is considered invertible (or non-singular) if and only if its determinant is non-zero. That is, \(det A \neq 0\).
The determinant of a matrix is a scalar value that can be computed from the elements of a square matrix. It has many uses, including determining if a matrix is invertible, solving systems of linear equations, and finding eigenvalues.
In this problem, the fact that A is invertible is crucial because it guarantees that \(det A \neq 0\), which allows us to use the property \((det A)^{-1} det A = 1\).
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