Chapter 2.3, Problem 51E

a.

To determine

To find: the values of a, b, c, and d to make the statement true.

Answer to Problem 51E

  $$

  $\begin{array}{l}a=1\\ b=0\\ c=0\\ d=1\end{array}$

Explanation of Solution

Given information: The given matrix equation is as follows:

  $\left[\begin{array}{cc}\begin{array}{l}-2\\ 4\end{array}& \begin{array}{l}3\\ -5\end{array}\end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{l}a\\ c\end{array}& \begin{array}{l}b\\ d\end{array}\end{array}\right]=\left[\begin{array}{cc}\begin{array}{l}-2\\ 4\end{array}& \begin{array}{l}3\\ -5\end{array}\end{array}\right]$

Calculation:

First evaluate the determinant of first matrix as follows:

  $\begin{array}{l}\left|\begin{array}{cc}\begin{array}{l}-2\\ 4\end{array}& \begin{array}{l}3\\ -5\end{array}\end{array}\right|=\left[\left(-2\right)\left(-5\right)\right]-\left[4\left(3\right)\right]\\ =10-12\\ =-2\end{array}$

This means that this matrix has inverse. Multiply the whole equation with the inverse from the left side to calculate the inverse as shown below.

  $\begin{array}{l}\left[\begin{array}{cc}\begin{array}{l}-2\\ 4\end{array}& \begin{array}{l}3\\ -5\end{array}\end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{l}a\\ c\end{array}& \begin{array}{l}b\\ d\end{array}\end{array}\right]=\left[\begin{array}{cc}\begin{array}{l}-2\\ 4\end{array}& \begin{array}{l}3\\ -5\end{array}\end{array}\right]\\ {\left[\begin{array}{cc}\begin{array}{l}-2\\ 4\end{array}& \begin{array}{l}3\\ -5\end{array}\end{array}\right]}^{-1}\cdot \left[\begin{array}{cc}\begin{array}{l}-2\\ 4\end{array}& \begin{array}{l}3\\ -5\end{array}\end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{l}a\\ c\end{array}& \begin{array}{l}b\\ d\end{array}\end{array}\right]=\left[\begin{array}{cc}\begin{array}{l}-2\\ 4\end{array}& \begin{array}{l}3\\ -5\end{array}\end{array}\right]\cdot {\left[\begin{array}{cc}\begin{array}{l}-2\\ 4\end{array}& \begin{array}{l}3\\ -5\end{array}\end{array}\right]}^{-1}\end{array}$

Matrix multiplied with its inverse is identity matrix. So,

  $\begin{array}{c}I\cdot \left[\begin{array}{cc}\begin{array}{l}a\\ c\end{array}& \begin{array}{l}b\\ d\end{array}\end{array}\right]=I\\ \left[\begin{array}{cc}\begin{array}{l}1\\ 0\end{array}& \begin{array}{l}\text{ 0}\\ \text{ 1}\end{array}\end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{l}a\\ c\end{array}& \begin{array}{l}b\\ d\end{array}\end{array}\right]=\left[\begin{array}{cc}\begin{array}{l}1\\ 0\end{array}& \begin{array}{l}\text{ 0}\\ \text{ 1}\end{array}\end{array}\right]\\ \left[\begin{array}{cc}\begin{array}{l}a\\ c\end{array}& \begin{array}{l}b\\ d\end{array}\end{array}\right]=\left[\begin{array}{cc}\begin{array}{l}1\\ 0\end{array}& \begin{array}{l}\text{ 0}\\ \text{ 1}\end{array}\end{array}\right]\end{array}$

Therefore, the values of a, b, c, and d is 1, 0, 0, and 1 respectively.

b.

To determine

To find: and explain the result if matrix containing two column is multiplied by the matrix containing a, b, c, and d.

Answer to Problem 51E

  $\left[\begin{array}{cc}\begin{array}{l}a\\ c\end{array}& \begin{array}{l}b\\ d\end{array}\end{array}\right]=\left[\begin{array}{cc}\begin{array}{l}1\\ 0\end{array}& \begin{array}{l}\text{ 0}\\ \text{ 1}\end{array}\end{array}\right]$

Any matrix multiplied by the identity matrix is that same matrix.

Explanation of Solution

Given information: The given matrix equation is as follows:

  $\left[\begin{array}{cc}\begin{array}{l}-2\\ 4\end{array}& \begin{array}{l}3\\ -5\end{array}\end{array}\right]\cdot \left[\begin{array}{cc}\begin{array}{l}a\\ c\end{array}& \begin{array}{l}b\\ d\end{array}\end{array}\right]=\left[\begin{array}{cc}\begin{array}{l}-2\\ 4\end{array}& \begin{array}{l}3\\ -5\end{array}\end{array}\right]$

Calculation:

The matrix evaluated in part a. is as follows:

  $\begin{array}{c}\left[\begin{array}{cc}\begin{array}{l}a\\ c\end{array}& \begin{array}{l}b\\ d\end{array}\end{array}\right]=\left[\begin{array}{cc}\begin{array}{l}1\\ 0\end{array}& \begin{array}{l}\text{ 0}\\ \text{ 1}\end{array}\end{array}\right]\\ =I\end{array}$

Since matrix containing a, b, c, and d is identity matrix, and any matrix multiplied by the identity matrix is that same matrix.

Example: If A is any matrix with two columns, then $A\cdot I=A$ .