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Question: If A=$\begin{bmatrix} 3 & 0 & 0 \\ 2 & 2 & 0 \\ 4 & 5 & 3 \end{bmatrix}$ then which of the following...

If A=[300220453]\begin{bmatrix} 3 & 0 & 0 \\ 2 & 2 & 0 \\ 4 & 5 & 3 \end{bmatrix} then which of the following is (are) true?

A

A is invertible.

B

trace (adj (adj A))=144.

C

trace (adj (adj A))= 108.

D

|adj A| is less than 400.

Answer

(A), (B), (D)

Explanation

Solution

Let the given matrix be A=[300220453]A = \begin{bmatrix} 3 & 0 & 0 \\ 2 & 2 & 0 \\ 4 & 5 & 3 \end{bmatrix}. The order of the matrix A is n=3n=3.

We analyze each statement:

(A) A is invertible.

A matrix A is invertible if and only if its determinant is non-zero. The given matrix A is a lower triangular matrix. The determinant of a triangular matrix is the product of its diagonal elements. A=det(A)=3×2×3=18|A| = \det(A) = 3 \times 2 \times 3 = 18. Since A=180|A| = 18 \neq 0, the matrix A is invertible. So, statement (A) is true.

(B) trace (adj (adj A))=144.

(C) trace (adj (adj A))= 108.

For a non-singular matrix A of order n, the adjugate of the adjugate of A is given by the property: adj(adj A)=An2A\text{adj}(\text{adj } A) = |A|^{n-2} A. In this case, n=3n=3 and A=18|A|=18. Since A0|A| \neq 0, A is non-singular. adj(adj A)=A32A=A1A=AA=18A\text{adj}(\text{adj } A) = |A|^{3-2} A = |A|^1 A = |A| A = 18A. The trace of a matrix is the sum of its diagonal elements. The trace of a scalar multiple of a matrix is the scalar multiple of the trace. trace(kA)=k trace(A)\text{trace}(kA) = k \text{ trace}(A). trace(adj(adj A))=trace(18A)=18 trace(A)\text{trace}(\text{adj}(\text{adj } A)) = \text{trace}(18A) = 18 \text{ trace}(A). The trace of A is the sum of its diagonal elements: trace(A)=3+2+3=8\text{trace}(A) = 3 + 2 + 3 = 8. So, trace(adj(adj A))=18×8=144\text{trace}(\text{adj}(\text{adj } A)) = 18 \times 8 = 144. Statement (B) says trace (adj (adj A))=144, which is true. Statement (C) says trace (adj (adj A))= 108, which is false.

(D) |adj A| is less than 400.

For a non-singular matrix A of order n, the determinant of the adjugate of A is given by the property: adj A=An1|\text{adj } A| = |A|^{n-1}. In this case, n=3n=3 and A=18|A|=18. adj A=A31=A2=182|\text{adj } A| = |A|^{3-1} = |A|^2 = 18^2. 182=32418^2 = 324. The statement says |adj A| is less than 400. 324<400324 < 400. This inequality is true. So, statement (D) is true.

Based on the analysis, statements (A), (B), and (D) are true.