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Chapter 2: Problem 6

If \(A\) is singular, what can you say about the product \(A\) adj \(A ?\)

Short Answer

Expert verified
For a singular matrix A, its determinant is zero and the product A * adj(A) is equal to the zero matrix: A * adj(A) = 0

Step by step solution

01

Recall the definition of a singular matrix and its adjugate matrix and the determinant property#

A singular matrix is a square matrix A with a determinant equal to zero, i.e., det(A) = 0. The adjugate of a matrix A, denoted adj(A) is the transpose of the cofactor matrix of A. The relationship between the product of a matrix and its adjugate is given by: A * adj(A) = det(A) * I where I is the identity matrix.#tag_end#
02

Calculate the product A * adj(A) for a singular matrix A#

Since we know that A is a singular matrix, its determinant is zero: det(A) = 0 Using the relationship between the product of a matrix and its adjugate, we can calculate the product A * adj(A): A * adj(A) = det(A) * I Substitute det(A) = 0: A * adj(A) = 0 * I Therefore, the product simplifies to: A * adj(A) = 0 #tag_end#
03

Conclusion#

For a singular matrix A, its determinant is zero and the product A * adj(A) is equal to the zero matrix: A * adj(A) = 0#tag_end#

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Key Concepts

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

Adjugate Matrix
The adjugate matrix, often denoted as adj(A), is a key concept in linear algebra. To form the adjugate of a matrix, we first find the cofactor matrix of the original matrix. This involves calculating the cofactor for each element of the matrix, treating each one as if it's separate. For each element of the matrix, its cofactor is determined by removing the row and column to which the element belongs, computing the determinant of the remaining submatrix, and multiplying by \ (-1)^{i+j} \, where \(i\) and \(j\) are the indices of the element. Using these cofactors, we construct the cofactor matrix and then transpose it to get the adjugate.
  • The adjugate matrix is useful in various calculations, including finding the inverse of a matrix.
  • For any square matrix A, the relationship \(A \, \text{adj}(A) = \text{det}(A) \, I\) holds, where \(I\) is the identity matrix.
Determinant
The determinant is a special number calculated from a square matrix. Understanding this concept is crucial, as it reveals several properties of the matrix.
The process of finding a determinant varies depending on the size of the matrix. For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant is calculated as \(ad - bc\). For larger matrices, the calculation is more complex, involving recursive expansion along a row or column using smaller determinants.
  • The determinant of a matrix provides insights into the matrix properties such as invertibility. If the determinant is zero, the matrix is called singular and does not have an inverse.
  • A non-zero determinant indicates that the matrix is invertible, which is a key condition for many linear algebra applications.
Identity Matrix
The identity matrix, often denoted by \(I\), plays a central role in matrix operations. This special matrix serves as the multiplicative identity in matrix algebra, much like the number 1 operates in arithmetic. It is square and contains ones along its main diagonal, from the top left to the bottom right, with all other entries being zero. For example, a 3x3 identity matrix looks like this:
\[ I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \]
  • The identity matrix has the property that for any square matrix A, the product \(AI = IA = A\).
  • In the context of singular matrices, the identity matrix helps demonstrate that \(A \, \text{adj}(A) = \text{det}(A) \, I.\) For a singular matrix with \(\text{det}(A) = 0\), this relationship confirms that \(A \, \text{adj}(A)\) results in the zero matrix.

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Most popular questions from this chapter

Let \(A\) be a \(k \times k\) matrix and let \(B\) be an \((n-k) \times(n-k)\) matrix. Let \\[ \begin{array}{c} E=\left(\begin{array}{cc} I_{k} & O \\ O & B \end{array}\right), \quad F=\left(\begin{array}{cc} A & O \\ O & I_{n-k} \end{array}\right) \\ C=\left(\begin{array}{cc} A & O \\ O & B \end{array}\right) \end{array} \\] where \(I_{k}\) and \(I_{n-k}\) are the \(k \times k\) and \((n-k) \times(n-k)\) identity matrices. (a) Show that det( \(E)=\operatorname{det}(B)\) (b) Show that \(\operatorname{det}(F)=\operatorname{det}(A)\) (c) Show that \(\operatorname{det}(C)=\operatorname{det}(A) \operatorname{det}(B)\).

Let \(A\) and \(B\) be \(2 \times 2\) matrices (a) \(\operatorname{Does} \operatorname{det}(A+B)=\operatorname{det}(A)+\operatorname{det}(B) ?\) (b) \(\operatorname{Does} \operatorname{det}(A B)=\operatorname{det}(A) \operatorname{det}(B) ?\) (c) \(\operatorname{Does} \operatorname{det}(A B)=\operatorname{det}(B A) ?\) Justify your answers.

Given \\[ A=\left(\begin{array}{lll} 1 & 2 & 1 \\ 0 & 4 & 3 \\ 1 & 2 & 2 \end{array}\right) \\] determine the (2,3) entry of \(A^{-1}\) by computing a quotient of two determinants.

Let \(A\) be an \(n \times n\) matrix and \(\alpha\) a scalar. Show that \\[ \operatorname{det}(\alpha A)=\alpha^{n} \operatorname{det}(A) \\]

Let \(A\) and \(B\) be \(3 \times 3\) matrices with \(\operatorname{det}(A)=4\) and \(\operatorname{det}(B)=5 .\) Find the value of (a) \(\operatorname{det}(A B)\) (b) \(\operatorname{det}(3 A)\) (c) \(\operatorname{det}(2 A B)\) (d) \(\operatorname{det}\left(A^{-1} B\right)\)
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