Trace, Transpose of a Matrix – Problems
1. If the trace of the Matrix \(A=\left[ \begin{matrix} x-1 & 0 & 2 & 5 \\ 3 & {{x}^{2}}-2 & 4 & 1 \\ -1 & -2 & x-3 & 1 \\ 2 & 0 & 4 & {{x}^{2}}-6 \\\end{matrix} \right]\).
Solution: Given \(A=\left[ \begin{matrix} x-1 & 0 & 2 & 5 \\ 3 & {{x}^{2}}-2 & 4 & 1 \\ -1 & -2 & x-3 & 1 \\ 2 & 0 & 4 & {{x}^{2}}-6 \\\end{matrix} \right]\).
Trace of matrices is defined as Tr(D) = \(\sum\limits_{i=1}^{n}{{{a}_{ij}}}\) = (x – 1) + (x² – 2) + (x – 3) + (x² – 6) = 0
x – 1 + x² – 2 + x- 3 + x² – 6 = 0
2x² + 2x – 12 = 0
x² + x – 6 = 0
x² + 3x – 2x – 6 = 0
x (x + 3) – 2(x + 3) = 0
(x + 3) (x – 2) = 0
x + 3 = 0
x = – 3
x -2 = 0
x = 2
x = – 3, 2
2. If \(A+2B=\left[ \begin{matrix} 2 & -4 \\ 1 & 6 \\\end{matrix} \right]\), \(A’+B’=\left[ \begin{matrix} 1 & 2 \\ 0 & -1 \\\end{matrix} \right]\), then A is equal to
Solution: Given
\(A+2B=\left[ \begin{matrix} 2 & -4 \\ 1 & 6 \\\end{matrix} \right]\) and \(A’+B’=\left[ \begin{matrix} 1 & 2 \\ 0 & -1 \\\end{matrix} \right]\).
(A’ + B’)T = (A + B)
\({{({{A}^{‘}}+{{B}^{‘}})}^{T}}={{\left[ \begin{matrix} 1 & 2 \\ 0 & -1 \\\end{matrix} \right]}^{T}}\) .
\({{({{A}^{‘}}+{{B}^{‘}})}^{T}}=\left[ \begin{matrix} 1 & 0 \\ 2 & -1 \\\end{matrix} \right]\) .
A = 2 (A + B) – (A + 2B)
\(=2\left[ \begin{matrix} 1 & 0 \\ 2 & -1 \\\end{matrix} \right]-\left[ \begin{matrix} 2 & -4 \\ 1 & 6 \\\end{matrix} \right]\) .
\(=\left[ \begin{matrix} 0 & 4 \\ 3 & -8 \\\end{matrix} \right]\).
3. If a square matrix A is such that AAT = I = ATA, then |A| is equal to
Solution: Given matrix A is a square matrix
AAT = I = ATA
|AA’| = I = |A’A|
|A| |A’| = I = |A’| |A|
(Since |A| |A’| =| A|²)
|A| = ± 1.