Question

a)Show that, if A is any n×n matrix with real entries, then there is a n×n symmetric matrix S and a n×n skew symmetric matrix S' such that A=S+S'.(3)b)Find the solutions to the following system of equations by reducing the corresponding augmented matrix to row-reduced echelon form.
2a+3b+4c+d=8
a+2b+2c+2d=3
a−b+c+3d=3

Answer 1

Answer on Question #52683 – Math – Linear Algebra

a) Show that, if $A$ is any $n \times n$ matrix with real entries, then there is a $n \times n$ symmetric matrix $S$ and a $n \times n$ skew symmetric matrix $S'$ such that $A = S + S'$.

b) Find the solutions to the following system of equations by reducing the corresponding augmented matrix to row-reduced echelon form.

$$\begin{array}{l} 2a + 3b + 4c + d = 8 \\ a + 2b + 2c + 2d = 3 \\ a - b + c + 3d = 3 \\ \end{array}$$

Solution

a) Let's rewrite $A$ in the following way

$$A = \frac{A + A^T}{2} + \frac{A - A^T}{2}$$

By properties of transpose matrices, the following identities hold true:

$$(A + B)^T = A^T + B^T, \quad (A - B)^T = A^T - B^T, \quad (A^T)^T = A.$$

It's easy to verify that $S = \frac{A + A^T}{2}$ is symmetric and $S' = \frac{A - A^T}{2}$ is skew symmetric:

$$S^T = \left(\frac{A + A^T}{2}\right)^T = \frac{A^T + A}{2} = S$$ $$(S')^T = \left(\frac{A - A^T}{2}\right)^T = \frac{A^T - A}{2} = -\frac{A - A^T}{2} = -S'$$

b) The given system can be written in the matrix form as follows

$$\left( \begin{array}{ccccc} 2 & 3 & 4 & 1 \\ 1 & 2 & 2 & 2 \\ 1 & -1 & 1 & 3 \end{array} \right) \left( \begin{array}{c} a \\ b \\ c \\ d \end{array} \right) = \left( \begin{array}{c} 8 \\ 3 \\ 3 \end{array} \right)$$

The augmented matrix

$$\left( \begin{array}{ccccc} 2 & 3 & 4 & 1 \\ 1 & 2 & 2 & 2 \\ 1 & -1 & 1 & 3 \end{array} \right) \left( \begin{array}{c} 8 \\ 3 \\ 3 \\ 3 \end{array} \right)$$

Transformation to row echelon form:

$$\left( \begin{array}{ccccc} 2 & 3 & 4 & 1 & 8 \\ 1 & 2 & 2 & 2 & 3 \\ 1 & -1 & 1 & 3 & 3 \end{array} \right)$$ $$\xrightarrow{\text{subtract row 1 from doubled row 2 and from doubled row 3}} \left( \begin{array}{ccccc} 2 & 3 & 4 & 1 & 8 \\ 0 & 1 & 0 & 3 & -2 \\ 0 & -5 & -2 & 5 & -2 \end{array} \right)$$ $$\left( \begin{array}{ccccc} 2 & 3 & 4 & 1 & 8 \\ 0 & 1 & 0 & 3 & -2 \\ 0 & -5 & -2 & 5 & -2 \end{array} \right) \xrightarrow{\text{add 5 times row 2 to row 3}} \left( \begin{array}{ccccc} 2 & 3 & 4 & 1 & 8 \\ 0 & 1 & 0 & 3 & -2 \\ 0 & 0 & -2 & 20 & -12 \end{array} \right) \xrightarrow{\text{divide row 3 by } (-2)}$$ $$\left( \begin{array}{ccccc} 2 & 3 & 4 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -10 \\ \end{array} \right) \xrightarrow{\text{subtract 3 times row 2 and 4 times row 3 from row 1}} \xrightarrow{\text{divide row 1 by 2}}$$ $$\left( \begin{array}{ccccc} 2 & 0 & 0 & 32 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -10 \end{array} \right) \xrightarrow{-10} \text{divide row 1 by 2}$$ $$\left( \begin{array}{cccc} 1 & 0 & 0 & 16 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -10 \end{array} \right)$$

Therefore,

$$a + 16d = -5 \quad a = -16d - 5$$ $$b + 3d = -2 \quad \Rightarrow \quad b = -3d - 2$$ $$c - 10d = 6 \quad c = 10d + 6$$

$d$ is an arbitrary real constant.

Thus,

$$\left( \begin{array}{c} a \\ b \\ c \\ d \end{array} \right) = \left( \begin{array}{c} -16d - 5 \\ -3d - 2 \\ 10d + 6 \\ d \end{array} \right) = \left( \begin{array}{c} -16 \\ -3 \\ 10 \\ 1 \end{array} \right) d + \left( \begin{array}{c} -5 \\ -2 \\ 6 \\ 0 \end{array} \right),$$

$d$ is an arbitrary real constant.

**Answer:**

a) $S = \frac{A + A^T}{2}, S' = \frac{A - A^T}{2}$

b) $\left( \begin{array}{c} a \\ b \\ c \\ d \end{array} \right) = \left( \begin{array}{c} -16d - 5 \\ -3d - 2 \\ 10d + 6 \\ d \end{array} \right) = \left( \begin{array}{c} -16 \\ -3 \\ 10 \\ 1 \end{array} \right) d + \left( \begin{array}{c} -5 \\ -2 \\ 6 \\ 0 \end{array} \right)$

www.AssignmentExpert.com