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Chapter 8: Problem 34

Prove Theorem \(8.9: A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I\)

Short Answer

Expert verified
To prove Theorem 8.9, we first defined matrix A and its adjugate adj(A). We then computed the products A(adj A) and (adj A)A and applied properties of determinants to show that their diagonal elements are equal to the determinant of A and their off-diagonal elements are 0. Therefore, we proved that \( A(\operatorname{adj} A) = (\operatorname{adj} A) A = |A| I \).

Step by step solution

01

Define matrix A and its adjugate

Let A be an \( n \times n \) square matrix with elements \( a_{ij} \), and let adj(A) be the matrix whose elements are the cofactors of A. The adjugate matrix is calculated by first finding the matrix of minors for A, then getting the matrix of cofactors by applying the checkerboard pattern of signs, and finally transposing the matrix of cofactors.
02

Compute A(adj A)

Calculate the product of A and its adjugate: \( A(\operatorname{adj} A) \). The element in the i-th row and j-th column of this product matrix is given by: \( [A(\operatorname{adj} A)]_{ij} = \sum_{k=1}^{n} a_{ik} C_{kj} \) Here, \( C_{kj} \) represents the cofactor of the element in the k-th row and j-th column of A. Now, note that the cofactor \( C_{kj} \) can also be written as \( (-1)^{k+j} M_{kj} \), where \( M_{kj} \) is the minor of the element in the k-th row and j-th column of A. So, \( [A(\operatorname{adj} A)]_{ij} = \sum_{k=1}^{n} a_{ik} (-1)^{k+j} M_{kj} \)
03

Use the properties of determinants

When i = j, the sum becomes the determinant of A: \( [A(\operatorname{adj} A)]_{ii} = |A| \) When i 鈮� j, we have a sum of products that can be considered as expanding a determinant along a row different from the i-th row. This sum is equal to 0: \( [A(\operatorname{adj} A)]_{ij} = 0 \) Therefore, the diagonal elements of the product matrix are equal to the determinant of A, and the off-diagonal elements are 0: \( A(\operatorname{adj} A) = |A| I \)
04

Compute (adj A) A

Calculate the product of the adjugate of A and A: \( (\operatorname{adj} A) A \). The element in the i-th row and j-th column of this product matrix is given by: \( [(\operatorname{adj} A) A]_{ij} = \sum_{k=1}^{n} C_{ik} a_{kj} \) Here, C_{ik} represents the cofactor of the element in the i-th row and k-th column of A. Now, note that the cofactor C_{ik} can also be written as \( (-1)^{i+k} M_{ik} \), where M_{ik} is the minor of the element in the i-th row and k-th column of A. So, \( [(\operatorname{adj} A) A]_{ij} = \sum_{k=1}^{n} (-1)^{i+k} M_{ik} a_{kj} \)
05

Use the properties of determinants again

When i = j, the sum becomes the determinant of A: \( [(\operatorname{adj} A) A]_{ii} = |A| \) When i 鈮� j, we have a sum of products that can be considered as expanding a determinant along a column different from the j-th column. This sum is equal to 0: \( [(\operatorname{adj} A) A]_{ij} = 0 \) Therefore, the diagonal elements of the product matrix are equal to the determinant of A, and the off-diagonal elements are 0: \( (\operatorname{adj} A) A = |A| I \)
06

Conclusion

From Steps 3 and 5, we have shown that both A(adj A) and (adj A) A are equal to |A|I. Thus, Theorem 8.9 is proven: \( A(\operatorname{adj} A) = (\operatorname{adj} A) A = |A| I \)

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Most popular questions from this chapter

Solve the following systems by determinants: (a) \(\left\\{\begin{array}{l}3 x+5 y=8 \\ 4 x-2 y=1\end{array}\right.\) (b) \(\left\\{\begin{array}{l}2 x-3 y=-1 \\ 4 x+7 y=-1\end{array}\right.\) (c) \(\left\\{\begin{array}{c}a x-2 b y=c \\ 3 a x-5 b y=2 c\end{array} \quad(a b \neq 0)\right.\)

Show that \(g=g\left(x_{1}, \ldots, x_{n}\right)=(-1)^{n} V_{n-1}(x)\) where \(g=g\left(x_{i}\right)\) is the difference product in Problem 8.19 \(x=x_{n},\) and \(V_{n-1}\) is the Vandermonde determinant defined by

Evaluate the determinant of each of the following matrices: (a) $A=\left[\begin{array}{lll}2 & 3 & 4 \\ 5 & 4 & 3 \\ 1 & 2 & 1\end{array}\right]$ (b) $B=\left[\begin{array}{rrr}1 & -2 & 3 \\ 2 & 4 & -1 \\ 1 & 5 & -2\end{array}\right]$ (c) $C=\left[\begin{array}{rrr}1 & 3 & -5 \\ 3 & -1 & 2 \\ 1 & -2 & 1\end{array}\right]\(Use the diagram in Fig. \)8-1$ to obtain the six products: (a) \(|A|=2(4)(1)+3(3)(1)+4(2)(5)-1(4)(4)-2(3)(2)-1(3)(5)=8+9+40-16-12-15=14\) (b) \(|B|=-8+2+30-12+5-8=9\) (c) \(|C|=-1+6+30-5+4-9=25\)

Show that if \(A\) is diagonal (triangular) then adj \(A\) is diagonal (triangular).

Suppose \(B\) is row equivalent to a square matrix \(A\). Prove that \(|B|=0\) if and only if \(|A|=0\).
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