Question

4)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'.
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 #52679 – 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

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

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:

$$\left( \begin{array}{ccccc} 2 & 3 & 4 & 1 & 8 \\ 1 & 2 & 2 & 2 & 3 \\ 1 & -1 & 1 & 3 & 3 \end{array} \right) \xrightarrow{\text{subtract row 1 from doubled row 2 and row 3}} \left( \begin{array}{ccccc} 2 & 3 & 4 & 1 & 8 \\ 0 & 1 & 0 & 3 & -2 \\ 0 & -5 & -2 & 5 & -2 \end{array} \right)$$ $$\left( \begin{array}{ccccc} 2 & 3 & 4 & 1 & 8 \\ 0 & 1 & 0 & 3 & -2 \\ 0 & -5 & -2 & 5 & -2 \end{array} \right) \xrightarrow{\text{subtract 5 times row 2 from row 3}} \left( \begin{array}{ccccc} 2 & 3 & 4 & 1 & 8 \\ 0 & 1 & 0 & 3 & -2 \\ 0 & 0 & -2 & 20 & -12 \end{array} \right)$$

Therefore,

$$\begin{array}{l} 2a + 3b + 4c + d = 8 \quad a = -16d - 5 \\ 1b + 3d = -2 \quad \Rightarrow \quad b = -3d - 2 \\ -2c + 20d = -12 \quad c = 10d + 6 \\ \end{array}$$

Answer:

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

b) $\begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} = \begin{pmatrix} -16d - 5 \\ -3d - 2 \\ 10d + 6 \\ d \end{pmatrix} = \begin{pmatrix} -5 \\ -2 \\ 6 \\ 0 \end{pmatrix} + d \begin{pmatrix} -16 \\ -3 \\ 10 \\ 1 \end{pmatrix}$, $d$ is arbitrary, real.

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