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Chapter 3: Problem 50

Let \(A\) and \(B\) be \(n \times n\) matrices such that \(A B\) is singular. Prove that either \(A\) or \(B\) is singular.

Short Answer

Expert verified
If the product of two matrices A and B is singular, it has been proven that either matrix A or matrix B, or both, must be singular. This statement holds due to the properties of determinants and multiplication.

Step by step solution

01

Preliminary Definitions

Start by defining a singular matrix and a determinant. A singular matrix is a square matrix that does not have an inverse, meaning its determinant is equal to zero. The determinant, for a square matrix, is a value that is calculated from the elements of the matrix. It gives significant information about the matrix it is derived from.
02

Apply the Determinant Property

Next, the determinant has a property where the determinant of a product of matrices equals the product of their determinants. Formally, \(det(AB) = det(A) * det(B)\). Apply this property to matrices A and B.
03

Consider the Result of the Product

It is given that the product \(AB\) is singular. Hence, \(det(AB) = 0\). Substituting this into the equation \(det(AB) = det(A) * det(B)\) from Step 2, it is clear that \(0 = det(A) * det(B)\).
04

Conclude the Result

Because of the properties of multiplication, it's known that if the product of two real numbers is zero, then at least one of the numbers has to be zero too. Hence, it must follow that either \(det(A) = 0\) or \(det(B) = 0\). If a determinant of a matrix is zero, the matrix is singular. Therefore, either A or B is singular.

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Most popular questions from this chapter

Find the volume of the tetrahedron having the given vertices. $$(1,1,1),(0,0,0),(2,1,-1),(-1,1,2)$$

Find the adjoint of the matrix \(A .\) Then use the adjoint to find the inverse of \(A,\) if possible. $$A=\left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & -1 \\ 2 & 2 & 2 \end{array}\right]$$

Use a graphing utility with matrix capabilities to determine whether \(A\) is orthogonal. To test for orthogonality, find (a) \(A^{-1},\) (b) \(A^{T},\) and (c) \(|A|,\) and verify that \(A^{-1}=A^{T}\) and \(|A|=\pm 1\) $$A=\left[\begin{array}{llr} \frac{3}{5} & 0 & -\frac{4}{5} \\ 0 & 1 & 0 \\ \frac{4}{5} & 0 & \frac{3}{5} \end{array}\right]$$

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Verify that \(\lambda_{i}\) is an eigenvalue of \(A\) and that \(\mathbf{x}_{i}\) is a corresponding eigenvector. $$\begin{array}{l} A=\left[\begin{array}{lll} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{array}\right] ; \quad \lambda_{1}=2, \quad \mathbf{x}_{1}=\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right] \\ \lambda_{2}=0, \quad \mathbf{x}_{2}=\left[\begin{array}{r} -1 \\ 0 \\ 1 \end{array}\right] ; \quad \lambda_{3}=1, \quad \mathbf{x}_{3}=\left[\begin{array}{r} -1 \\ 1 \\ -1 \end{array}\right] \end{array}$$
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