# Trace, Transpose of a Matrix – Problems

## 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.