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Chapter 19: Problem 55

Let \(\mathrm{A}\) and \(\mathrm{B}\) be two invertible matrices of order \(3 \times 3\). If \(\operatorname{det}\left(\mathrm{ABA}^{\mathrm{T}}\right)=8\) and \(\operatorname{det}\left(\mathrm{AB}^{-1}\right)=8\), then \(\operatorname{det}\left(\mathrm{BA}^{-1} \mathrm{~B}^{\mathrm{T}}\right)\) is equal to: [Jan. 11, 2019 (II)] (a) \(\frac{1}{4}\) (b) 1 (c) \(\frac{1}{16}\) (d) 16

Short Answer

Expert verified
The determinant \(\det(BA^{-1}B^T)\) is \(\frac{1}{16}\).

Step by step solution

01

Understanding the Problem

We need to find \(\det\left(\mathrm{BA}^{-1} \mathrm{~B}^{\mathrm{T}}\right)\) given that \(\det\left(\mathrm{ABA}^{\mathrm{T}}\right)=8\) and \(\det\left(\mathrm{AB}^{-1}\right)=8\). The matrices \(\mathrm{A}\) and \(\mathrm{B}\) are invertible matrices of order \(3 \times 3\).
02

Determinant Properties

Recall that the determinant of the product of matrices is the product of their determinants. Also, the determinant of the transpose of a matrix is equal to the determinant of the matrix itself. Using these properties we can simplify the expressions for the determinants.
03

Expressing \(\det(ABA^{T})\)

We have \(\det(ABA^{T}) = \det(A) \cdot \det(B) \cdot \det(A^{T})\). Since \(\det(A^{T}) = \det(A)\), this simplifies to \(\det(A)^2 \cdot \det(B) = 8\).
04

Expressing \(\det(AB^{-1})\)

Similarly, for \(\det(AB^{-1}) = \det(A) \cdot \det(B^{-1})\). Since \(\det(B^{-1}) = \det(B)^{-1}\), it can be simplified to \(\det(A) \cdot \frac{1}{\det(B)} = 8\).
05

Solving for \(\det(A)\) and \(\det(B)\)

From \(\det(A)^2 \cdot \det(B) = 8\) and \(\det(A) \cdot \frac{1}{\det(B)} = 8\), set \(\det(A) = x\) and \(\det(B) = y\). This gives two equations: \(x^2y = 8\) and \(\frac{x}{y} = 8\).
06

Finding \(\det(A)\)

From \(\frac{x}{y} = 8\), we can express \(y = \frac{x}{8}\). Substituting \(y\) in \(x^2y = 8\), we have \(x^2 \cdot \frac{x}{8} = 8\).
07

Solve for \(x\)

This equation becomes \(\frac{x^3}{8} = 8\). Solving for \(x\), we get \(x^3 = 64\) hence \(x = 4\). Thus \(\det(A) = 4\).
08

Find \(\det(B)\)

Substitute \(x = 4\) back into \(y = \frac{x}{8}\) to get \(y = \frac{4}{8} = \frac{1}{2}\). Hence \(\det(B) = \frac{1}{2}\).
09

Calculate \(\det(BA^{-1}B^T)\)

We need \(\det(BA^{-1}B^T) = \det(B) \cdot \det(A^{-1}) \cdot \det(B^T)\). We know \(\det(B^T) = \det(B)\) and \(\det(A^{-1}) = \frac{1}{\det(A)}\). This gives us \(\det(B)^2 \cdot \frac{1}{\det(A)} = \left(\frac{1}{2}\right)^2 \cdot \frac{1}{4}\).
10

Final Calculation

\(\det(BA^{-1}B^T) = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Invertible Matrices
Invertible matrices, also known as non-singular or non-degenerate matrices, are matrices that can be reversed or inverted. For a square matrix to be invertible, it must satisfy two main conditions. First, its determinant must be non-zero. Second, there exists another matrix such that when multiplied by the original matrix, results in the identity matrix.
  • The identity matrix is a square matrix with ones on the diagonal and zeros elsewhere.
  • If a matrix has an inverse, denoted as \(A^{-1}\), then \(A \, A^{-1} = I\), where \(I\) is the identity matrix.
  • Invertible matrices play a crucial role in solving linear equations where matrix techniques are used.
Understanding the invertibility of matrices helps in determining the feasibility and solutions of the problems involving calculations over matrices.
Properties of Determinants
Determinants are scalar values that can be computed from a square matrix and they offer insights into the matrix鈥檚 properties. The determinant provides useful information such as whether a matrix is invertible or not, because a non-zero determinant indicates that a matrix is invertible.
  • The determinant of a product of matrices is equal to the product of their determinants: \(\det(AB) = \det(A) \cdot \det(B)\).
  • The determinant also remains unchanged when dealing with the transpose of a matrix, meaning \(\det(A) = \det(A^T)\).
  • For an inverse matrix, \(\det(A^{-1}) = \frac{1}{\det(A)}\).
These properties are useful when solving matrix equations, as demonstrated in calculations where they allow simplification and make problem-solving more manageable.
Matrix Transpose
The transpose of a matrix involves flipping the matrix over its diagonal, which switches the row and column indices of each element. This operation is quite straightforward but has significant implications in matrix algebra.
  • If a matrix \(A\) is transposed to form \(A^T\), then each element \(a_{ij}\) becomes \(a_{ji}\).
  • The operation doesn鈥檛 change the determinant of the matrix, reinforcing the property that \(\det(A) = \det(A^T)\).
  • Transposing twice returns the matrix to its original form: \((A^T)^T = A\).
Understanding the transpose operation is essential in various matrix computations and in understanding symmetries in mathematical expressions and equations.
Matrix Inverse
Finding the inverse of a matrix is like finding a key that "unlocks" the original matrix back into the identity matrix through multiplication. Only invertible matrices have inverses, and this concept is crucial in various applications, from solving systems of equations to computer graphics.
  • The inverse of a matrix \(A\) is denoted as \(A^{-1}\), and it satisfies \(A \, A^{-1} = I\).
  • To find the inverse, one can use methods like row reduction, the adjugate method, or numerical algorithms for larger matrices.
  • If the determinant \(\det(A) = 0\), then the matrix is not invertible and does not have an inverse.
Mastering the concept of matrix inverses allows for the seamless manipulation of matrix equations and enhances comprehension in higher-level mathematical analysis.

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Most popular questions from this chapter

Let \(A\) be a \(3 \times 3\) matrix such that adj \(A=\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 0 & 2 \\ 1 & -2 & -1\end{array}\right]\) and \(B=\operatorname{adj}(\operatorname{adj} A)\). If \(|A|=\lambda\) and \(\left|\left(B^{-1}\right)^{T}\right|=\mu\), then the ordered pair, \((|\lambda|, \mu)\) is equal to: \(\quad\) [Sep. 03, 2020 (II)] (a) \(\left(3, \frac{1}{81}\right)\) (b) \(\left(9, \frac{1}{9}\right)\) (c) \((3,81)\) (d) \(\left(9, \frac{1}{81}\right)\)

If the system of equations \(2 x+3 y-z=0, x+k y-2 z=0\) and \(2 x-y+z=0\) has a non-trivial solution \((x, y, z)\), then \(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k\) is equal to: \(\quad\) [April 09, 2019 (II)] (a) \(\frac{3}{4}\) (b) \(\frac{1}{2}\) (c) \(-\frac{1}{4}\) (d) \(-4\)

The area of triangle formed by the lines joining the vertex of the parabola, \(x^{2}=8 y\), to the extremities of its latus rectum is \mathrm{\\{} D n l i n e ~ M a y ~ 1 2 , ~ 2 0 1 2 ] ~ (a) 2 (b) 8 (c) 1 (d) 4

Let \(A=\left(\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right)\) and \(B=\left(\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right)\). If \(B\) is the inverse of matrix \(A\), then \(\alpha\) is (a) 5 (b) \(-1\) (c) 2 (d) \(-2\)

Let \(A\) and \(B\) be real matrices of the form \(\left[\begin{array}{ll}\alpha & 0 \\ 0 & \beta\end{array}\right]\) and \(\left[\begin{array}{ll}0 & \gamma \\\ \delta & 0\end{array}\right]\), respectively. \(\quad\) Online May 12, 2012] Statement 1: \(A B-B A\) is always an invertible matrix. Statement 2: \(A B-B A\) is never an identity matrix. (a) Statement 1 is true, Statement 2 is false. (b) Statement 1 is false, Statement 2 is true. (c) Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation of Statement 1 . (d) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1 .
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