Chapter 7.7, Problem 42E

(a)

To determine

The determinant of matrix product.

Answer to Problem 42E

Therefore, $\left|\text{A}\right|=9$ .

Explanation of Solution

Given information: $\text{A}=\left[\begin{array}{ccc}2& 0& 1\\ 1& -4& 2\\ 3& 1& 0\end{array}\right],\text{B}=\left[\begin{array}{ccc}2& -1& 4\\ 0& 1& 0\\ 3& -2& 1\end{array}\right]$

Formula used:

  $\begin{array}{l}\left|\text{A}\right|=\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|\\ \left|\text{A}\right|=a_{11}\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|-a_{12}\left|\begin{array}{cc}{a}_{21}& a_{23}\\ a_{31}& a_{33}\end{array}\right|+a_{13}\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|\end{array}$

Calculation: To calculate $\left|\text{A}\right|$ we will use standard formula,

  $\begin{array}{l}\text{A}=\left[\begin{array}{ccc}2& 0& 1\\ 1& -4& 2\\ 3& 1& 0\end{array}\right]\\ \left|\text{A}\right|=2\left|\begin{array}{cc}-4& 2\\ 1& 0\end{array}\right|-0\left|\begin{array}{cc}1& 2\\ 3& 0\end{array}\right|+1\left|\begin{array}{cc}1& -4\\ 3& 1\end{array}\right|\\ \left|\text{A}\right|=2\times \left(\left(-4\times 0\right)-\left(2\times 1\right)\right)+1\left(\left(1\times 1\right)-\left(-4\times 3\right)\right)\\ \left|\text{A}\right|=-4+13\\ \left|\text{A}\right|=9\end{array}$

Therefore, $\left|\text{A}\right|=9$ .

(b)

To determine

The determinant of matrix product.

Answer to Problem 42E

Therefore, $\left|\text{B}\right|=-10$ .

Explanation of Solution

Given information: $\text{A}=\left[\begin{array}{ccc}2& 0& 1\\ 1& -4& 2\\ 3& 1& 0\end{array}\right],\text{B}=\left[\begin{array}{ccc}2& -1& 4\\ 0& 1& 0\\ 3& -2& 1\end{array}\right]$

Formula used:

  $\begin{array}{l}\left|\text{A}\right|=\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|\\ \left|\text{A}\right|=a_{11}\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|-a_{12}\left|\begin{array}{cc}{a}_{21}& a_{23}\\ a_{31}& a_{33}\end{array}\right|+a_{13}\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|\end{array}$

Calculation: To calculate $\left|\text{B}\right|$ we will use standard formula,

  $\begin{array}{l}\text{B}=\left[\begin{array}{ccc}2& -1& 4\\ 0& 1& 0\\ 3& -2& 1\end{array}\right]\\ \left|\text{B}\right|=2\left|\begin{array}{cc}1& 0\\ -2& 1\end{array}\right|-\left(-1\right)\left|\begin{array}{cc}0& 0\\ 3& 1\end{array}\right|+4\left|\begin{array}{cc}0& 1\\ 3& -2\end{array}\right|\\ \left|\text{B}\right|=2\times \left(\left(1\times 1\right)-\left(-2\times 0\right)\right)+1\left(\left(0\times 1\right)-\left(0\times 3\right)\right)+4\left(\left(0\times -2\right)-\left(1\times 3\right)\right)\\ \left|\text{B}\right|=2-12\\ \left|\text{B}\right|=-10\end{array}$

Therefore, $\left|\text{B}\right|=-10$ .

(c)

To determine

The determinant of matrix product.

Answer to Problem 42E

Therefore, $\text{AB}=\left[\begin{array}{ccc}7& 8& 6\\ -2& -5& -2\\ 9& -2& 10\end{array}\right]$ .

Explanation of Solution

Given information: $\text{A}=\left[\begin{array}{ccc}2& 0& 1\\ 1& -4& 2\\ 3& 1& 0\end{array}\right],\text{B}=\left[\begin{array}{ccc}2& -1& 4\\ 0& 1& 0\\ 3& -2& 1\end{array}\right]$

Formula used:

  $\text{AB}=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]\times \left[\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right]$

  

Calculation: To calculate $\text{AB}$ we will use standard formula,

  $\begin{array}{l}\text{AB}=\left[\begin{array}{ccc}2& 0& 1\\ 1& -4& 2\\ 3& 1& 0\end{array}\right]\times \left[\begin{array}{ccc}2& -1& 4\\ 0& 1& 0\\ 3& -2& 1\end{array}\right]\\ \text{AB}=\left[\begin{array}{ccc}\left(2\times 2\right)+\left(0\times 0\right)+\left(1\times 3\right)& \left(1\times 2\right)+\left(-4\times 0\right)+\left(2\times 3\right)& \left(3\times 2\right)+\left(1\times 0\right)+\left(0\times 3\right)\\ \left(2\times -1\right)+\left(0\times 1\right)+\left(1\times 0\right)& \left(1\times -1\right)+\left(-4\times 1\right)+\left(2\times 0\right)& \left(3\times -1\right)+\left(1\times 1\right)+\left(0\times 0\right)\\ \left(2\times 4\right)+\left(0\times -2\right)+\left(1\times 1\right)& \left(1\times 4\right)+\left(-4\times -2\right)+\left(2\times 1\right)& \left(3\times 4\right)+\left(1\times -2\right)+\left(0\times 1\right)\end{array}\right]\\ \text{AB}=\left[\begin{array}{ccc}7& 8& 6\\ -2& -5& -2\\ 9& -2& 10\end{array}\right]\end{array}$

Therefore, $\text{AB}=\left[\begin{array}{ccc}7& 8& 6\\ -2& -5& -2\\ 9& -2& 10\end{array}\right]$ .

(d)

To determine

The determinant of matrix product.

Therefore, $\left|\text{AB}\right|=-17$ .

Given information: $\text{A}=\left[\begin{array}{ccc}2& 0& 1\\ 1& -4& 2\\ 3& 1& 0\end{array}\right],\text{B}=\left[\begin{array}{ccc}2& -1& 4\\ 0& 1& 0\\ 3& -2& 1\end{array}\right]$

Formula used:

  $\text{AB}=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]\times \left[\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right]$

  

Calculation: To calculate $\text{AB}$ we will use standard formula,

  $\text{AB}=\left[\begin{array}{ccc}2& 0& 1\\ 1& -4& 2\\ 3& 1& 0\end{array}\right]\times \left[\begin{array}{ccc}2& -1& 4\\ 0& 1& 0\\ 3& -2& 1\end{array}\right]$

  $\text{AB}=\left[\begin{array}{ccc}\left(2\times 2\right)+\left(0\times 0\right)+\left(1\times 3\right)& \left(1\times 2\right)+\left(-4\times 0\right)+\left(2\times 3\right)& \left(3\times 2\right)+\left(1\times 0\right)+\left(0\times 3\right)\\ \left(2\times -1\right)+\left(0\times 1\right)+\left(1\times 0\right)& \left(1\times -1\right)+\left(-4\times 1\right)+\left(2\times 0\right)& \left(3\times -1\right)+\left(1\times 1\right)+\left(0\times 0\right)\\ \left(2\times 4\right)+\left(0\times -2\right)+\left(1\times 1\right)& \left(1\times 4\right)+\left(-4\times -2\right)+\left(2\times 1\right)& \left(3\times 4\right)+\left(1\times -2\right)+\left(0\times 1\right)\end{array}\right]$

  $\text{AB}=\left[\begin{array}{ccc}7& 8& 6\\ -2& -5& -2\\ 9& -2& 10\end{array}\right]$

  $\begin{array}{l}\left|\text{AB}\right|=7\left|\begin{array}{cc}-5& -2\\ -2& 10\end{array}\right|-8\left|\begin{array}{cc}-2& -2\\ 9& 10\end{array}\right|+6\left|\begin{array}{cc}-2& -5\\ 9& -2\end{array}\right|\\ \left|\text{AB}\right|=7\left(-35-4\right)-8\left(-20+18\right)+6\left(-4+45\right)\\ \left|\text{AB}\right|=-279+16+246\\ \left|\text{AB}\right|=-17\end{array}$

Therefore, $\left|\text{AB}\right|=-17$ .

Answer to Problem 42E

Therefore, $\left|\text{AB}\right|=-17$ .

Explanation of Solution

Given information: $\text{A}=\left[\begin{array}{ccc}2& 0& 1\\ 1& -4& 2\\ 3& 1& 0\end{array}\right],\text{B}=\left[\begin{array}{ccc}2& -1& 4\\ 0& 1& 0\\ 3& -2& 1\end{array}\right]$

Formula used:

  $\text{AB}=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]\times \left[\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right]$

  

Calculation: To calculate $\text{AB}$ we will use standard formula,

  $\text{AB}=\left[\begin{array}{ccc}2& 0& 1\\ 1& -4& 2\\ 3& 1& 0\end{array}\right]\times \left[\begin{array}{ccc}2& -1& 4\\ 0& 1& 0\\ 3& -2& 1\end{array}\right]$

  $\text{AB}=\left[\begin{array}{ccc}\left(2\times 2\right)+\left(0\times 0\right)+\left(1\times 3\right)& \left(1\times 2\right)+\left(-4\times 0\right)+\left(2\times 3\right)& \left(3\times 2\right)+\left(1\times 0\right)+\left(0\times 3\right)\\ \left(2\times -1\right)+\left(0\times 1\right)+\left(1\times 0\right)& \left(1\times -1\right)+\left(-4\times 1\right)+\left(2\times 0\right)& \left(3\times -1\right)+\left(1\times 1\right)+\left(0\times 0\right)\\ \left(2\times 4\right)+\left(0\times -2\right)+\left(1\times 1\right)& \left(1\times 4\right)+\left(-4\times -2\right)+\left(2\times 1\right)& \left(3\times 4\right)+\left(1\times -2\right)+\left(0\times 1\right)\end{array}\right]$

  $\text{AB}=\left[\begin{array}{ccc}7& 8& 6\\ -2& -5& -2\\ 9& -2& 10\end{array}\right]$

  $\begin{array}{l}\left|\text{AB}\right|=7\left|\begin{array}{cc}-5& -2\\ -2& 10\end{array}\right|-8\left|\begin{array}{cc}-2& -2\\ 9& 10\end{array}\right|+6\left|\begin{array}{cc}-2& -5\\ 9& -2\end{array}\right|\\ \left|\text{AB}\right|=7\left(-35-4\right)-8\left(-20+18\right)+6\left(-4+45\right)\\ \left|\text{AB}\right|=-279+16+246\\ \left|\text{AB}\right|=-17\end{array}$

Therefore, $\left|\text{AB}\right|=-17$ .