Chapter 4, Problem 33SGR

To determine

To find: the value of variable using matrix equation and its multiple.

Answer to Problem 33SGR

  $\text{a=2}$ and $\text{b=1}$ .

Explanation of Solution

Given:

  $\left[\begin{array}{cc}\text{3}& \text{-1}\\ \text{1}& \text{2}\end{array}\right]\text{.}\left[\begin{array}{c}\text{a}\\ \text{b}\end{array}\right]\text{=}\left[\begin{array}{c}\text{5}\\ \text{4}\end{array}\right]$ .

Concept used:

Dimension of the product of two matrix are outer dimension both multiplier matrix.

Addition or elimination methods is used to calculate the two-unknown variable. To multiply the two matrices column of the first matrices must equals to the raw of the second Matrices.

The product will be defined. The dimension of the product of two matrices can be known taking Raw of the first matrix and taking the column of the second matrix.

Using substitution, elimination and addition can solve the two un-known variables.

Calculation:

  $\left[\begin{array}{cc}\text{3}& \text{-1}\\ \text{1}& \text{2}\end{array}\right]\text{.}\left[\begin{array}{c}\text{a}\\ \text{b}\end{array}\right]\text{=}\left[\begin{array}{c}\text{5}\\ \text{4}\end{array}\right]$

Solving by matrix multiplication.

  $\Rightarrow \left[\begin{array}{c}\text{3a-b}\\ \text{a+2b}\end{array}\right]\text{=}\left[\begin{array}{c}5\\ 4\end{array}\right]$ .

  $\text{3a-b}$ = $5$ ………… $\left(\text{i}\right)$

  $\text{a+2b}$

  $\text{=4}$ …………. $\left(\text{ii}\right)$

By multiply the equation $\left(\text{i}\right)$ by $\text{2}$ and $\left(\text{ii}\right)$ by $1$ :

  $\text{6a-2b=10}$ ……. $\left(\text{iii}\right)$

  $\text{a+2b}$

  $\text{=4}$ ……… $\left(\text{iv}\right)$

By using addition method in equation $\left(\text{iii}\right)$and $\left(\text{iv}\right)$ .

  $\text{7a=14}$ .

  $\text{a=2}$ .

By putting value of $\text{a=2}$ in equation $\left(\text{iv}\right)$ :

  $\text{2+2b=4}$ .

  $\text{b=}\frac{\text{4-2}}{2}=1.$

  $\text{b=1}$ .

Hence $\text{a=2}$ and $\text{b=1}$ .