Chapter 9, Problem 1P

(a) Write the following set of equations in matrix form:

$\begin{array}{l}8=6x_3+2x_2\\ 2-x_1=x_3\\ 5x_2+8x_1=13\end{array}$

(b) Multiply the matrix of coefficients by its transpose; i.e., $\left[A\right]{\left[A\right]}^T$.

To determine

The matrix form of the following set of equations,

$\begin{array}{c}8=6x_3+2x_2\\ 2-x_1=x_3\\ 5x_2+8x_1=13\end{array}$

Answer to Problem 1P

Solution:

The matrix form of the set of equations is, $\left[\begin{array}{ccc}0& 2& 6\\ 1& 0& 1\\ 8& 5& 0\end{array}\right]\left\{\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right\}=\left\{\begin{array}{c}8\\ 2\\ 13\end{array}\right\}$.

Explanation of Solution

Given:

The set of equations,

$\begin{array}{c}8=6x_3+2x_2\\ 2-x_1=x_3\\ 5x_2+8x_1=13\end{array}$

Consider the set of equations,

$\begin{array}{c}8=6x_3+2x_2\\ 2-x_1=x_3\\ 5x_2+8x_1=13\end{array}$

The set of equations can be written as,

$\begin{array}{c}0x_1+2x_2+6x_3=8\\ x_1+0x_2+x_3=2\\ 8x_1+5x_2+0x_3=13\end{array}$

These equations have three unknowns $x_1,x_2\text{ and }{x}_3$.

The matrix form of the above set of equations can be written as,

$\left[A\right]\left\{X\right\}=\left\{B\right\}$

Where, A is the coefficient matrix.

Therefore, the matrix form of the set of equations is,

$\left[\begin{array}{ccc}0& 2& 6\\ 1& 0& 1\\ 8& 5& 0\end{array}\right]\left\{\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right\}=\left\{\begin{array}{c}8\\ 2\\ 13\end{array}\right\}$.

To determine

To calculate: The matrix multiplication of the coefficient matrix with its transpose, that is $\left[A\right].{\left[A\right]}^T$. Where the coefficient matrix is, $\left[A\right]=\left[\begin{array}{ccc}0& 2& 6\\ 1& 0& 1\\ 8& 5& 0\end{array}\right]$.

Answer to Problem 1P

Solution:

The matrix multiplication of the coefficient matrix with its transpose is,

$\left[A\right].{\left[A\right]}^T=\left[\begin{array}{ccc}40& 6& 10\\ 6& 2& 8\\ 10& 8& 89\end{array}\right]$

Explanation of Solution

Given:

The coefficient matrix, $\left[A\right]=\left[\begin{array}{ccc}0& 2& 6\\ 1& 0& 1\\ 8& 5& 0\end{array}\right]$.

Formula used:

Transpose of the matrix $A$ is denoted by $A^T$. If A is any matrix $\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$, then its transpose will be $\left[\begin{array}{ccc}a& d& g\\ b& e& h\\ c& f& i\end{array}\right]$.

Calculation:

Consider the coefficient matrix,

$\left[A\right]=\left[\begin{array}{ccc}0& 2& 6\\ 1& 0& 1\\ 8& 5& 0\end{array}\right]$

The transpose of the matrix A is denoted by ${\left[A\right]}^T$ and is equal to,

${\left[A\right]}^T=\left[\begin{array}{ccc}0& 1& 8\\ 2& 0& 5\\ 6& 1& 0\end{array}\right]$

Multiplication of the coefficient matrix and its transpose is,

$\begin{array}{c}\left[A\right].{\left[A\right]}^T=\left[\begin{array}{ccc}0& 2& 6\\ 1& 0& 1\\ 8& 5& 0\end{array}\right].\left[\begin{array}{ccc}0& 1& 8\\ 2& 0& 5\\ 6& 1& 0\end{array}\right]\\ =\left[\begin{array}{ccc}0\times 0+2\times 2+6\times 6& 0\times 1+2\times 0+6\times 1& 0\times 8+2\times 5+6\times 0\\ 1\times 0+0\times 2+1\times 6& 1\times 1+0\times 0+1\times 1& 1\times 8+0\times 5+1\times 0\\ 8\times 0+5\times 2+0\times 6& 8\times 1+5\times 0+0\times 1& 8\times 8+5\times 5+0\times 0\end{array}\right]\\ =\left[\begin{array}{ccc}4+36& 6& 10\\ 6& 2& 8\\ 10& 8& 64+25\end{array}\right]\\ =\left[\begin{array}{ccc}40& 6& 10\\ 6& 2& 8\\ 10& 8& 89\end{array}\right]\end{array}$

Therefore, the matrix multiplication of the coefficient matrix with its transpose is,

$\left[A\right].{\left[A\right]}^T=\left[\begin{array}{ccc}40& 6& 10\\ 6& 2& 8\\ 10& 8& 89\end{array}\right]$.