Chapter 11, Problem 4CRT

To determine

To Find: The volumes of $A+B,C-D,AB,BD,\det (B),\det (C),\det (D).$

Answer to Problem 4CRT

  $A+B\to \text{ Not possible}$

  $CB=$Not possible $\det (B)=$ Not possible

  $C-D=\left(\begin{array}{ccc}0& -4& -2\\ -1& -4& -4\\ -1& -1& -1\end{array}\right)$

  $BD=\left(\begin{array}{c}-1\\ \frac{-1}{2}\end{array}\begin{array}{c}-2\\ -1\end{array}\begin{array}{c}-1\\ \frac{-1}{2}\end{array}\right)$

  $\det (C)=2$

  $AB=\left(\begin{array}{c}\frac{-9}{2}\\ -4\end{array}\begin{array}{c}1\\ 2\end{array}\begin{array}{c}5\\ 0\end{array}\right)$

  $C^{-1}=\left(\begin{array}{ccc}0& 0& -1\\ \frac{-1}{2}& \frac{1}{2}& \frac{-1}{2}\\ 1& 0& 1\end{array}\right)$

  $\det (D)=0$

Explanation of Solution

Given information:

  $A=\left(\begin{array}{cc}1& 5\\ 2& 0\end{array}\right)B=\left(\begin{array}{c}-2\\ \frac{-1}{2}\end{array}\begin{array}{c}1\\ 0\end{array}\begin{array}{c}0\\ 1\end{array}\right)$

  $C=\left(\begin{array}{ccc}1& 0& 1\\ 0& 2& 1\\ -1& 0& 0\end{array}\right)D=\left(\begin{array}{ccc}1& 4& 3\\ 1& 6& 5\\ 0& 1& 1\end{array}\right)$

Calculation:

  $A=\left(\begin{array}{ccc}a& b& c\\ d& e& f\\ q& h& i\end{array}\right)$

  $\left|A\right|=a(ei-fh)-b(di-gf)+c(dh-eg)$

The order of matrix $A=2\times 2$

The order of matrix $B=2\times 3$

The order of matrix $C=3\times 3$

The order of matrix $D=3\times 3$

The matrix sum is possible only if the order of two matrices are same.

  $\to$ Matrices sum $A+B$ is not possible because both have different orders.

  $\to C-D=\left(\begin{array}{ccc}1& 0& 1\\ 0& 2& 1\\ -1& 0& 0\end{array}\right)-\left(\begin{array}{ccc}1& 4& 3\\ 1& 6& 5\\ 0& 1& 1\end{array}\right)=\left(\begin{array}{ccc}1-1& 0-4& 1-3\\ 0-1& 2-6& 1-5\\ -1-0& 0-1& 0-1\end{array}\right)$

  $\therefore C-D=\left(\begin{array}{ccc}0& -4& -2\\ -1& -4& -4\\ -1& -1& -1\end{array}\right)$

The multiplication of matrices $A,B$ is possible if

Columns of $A=\text{ raws of }B$

  $\to AB={\left(\begin{array}{cc}1& 5\\ 2& 0\end{array}\right)}_{2\times 2}\times {\left(\begin{array}{c}-2\\ \frac{-1}{2}\end{array}\begin{array}{c}1\\ 0\end{array}\begin{array}{c}0\\ 1\end{array}\right)}_{2\times 3}={\left(\begin{array}{c}(1\times -2)+(5\times \frac{-1}{2})\\ -4+0\end{array}\begin{array}{c}1+0\\ 2+0\end{array}\begin{array}{c}0+5\\ 0+0\end{array}\right)}_{{}_{2\times 3}}$

  $AB=\left(\begin{array}{c}\frac{-9}{2}\\ -4\end{array}\begin{array}{c}1\\ 2\end{array}\begin{array}{c}5\\ 0\end{array}\right)$

  $\to CB$ order $(C)=3\times 3$ order $(B)=2\times 3$

Multiplication is not possible.

  $\to CB$ order $(B)=2\times 3$ order $(D)=3\times 3$

Multiplication is possible whose resultant order is $2\times 3$

  $BD= \left(\begin{array}{c}-2\\ -1\end{array}\begin{array}{c}1\\ 0\end{array}\begin{array}{c}0\\ 1\end{array}\right)\left(\begin{array}{ccc}1& 4& 3\\ 1& 6& 5\\ 0& 1& 1\end{array}\right)$

  $BD= \left(\begin{array}{c}\left(-2\times 1\right)+\left(1\times 1\right)+\left(0\times 0\right)\\ \left(-\frac{1}{2}\times 1\right)+\left(0\times 1\right)+\left(1\times 0\right)\end{array}\begin{array}{c}\left(-2\times 4\right)+\left(1\times 6\right)+\left(0\times 1\right)\\ \left(\frac{-1}{2}\times 4\right)+\left(0\times 6\right)+\left(1\times 1\right)\end{array}\begin{array}{c}\left(-2\times 3\right)+\left(1\times 5\right)+\left(0\times 1\right)\\ \left(\frac{-1}{2}\times 3\right)+\left(0\times 5\right)+\left(1\times 1\right)\end{array}\right)$

  $BD= \left(\begin{array}{c}-1\\ \frac{-1}{2}\end{array}\begin{array}{c}-2\\ -1\end{array}\begin{array}{c}-1\\ \frac{-1}{2}\end{array}\right)$

Determinant of a matrix is possible only if the $\text{number of rows }=\text{ number of columns}$

  $\begin{array}{l}\to \det \left(B\right)\to order\left(B\right)=2\times 3\\ \therefore \det \left(B\right)\text{ is not posssible}\text{.}\end{array}$

  $\begin{array}{l}\to \det \left(C\right)\to order\left(C\right)=3\times 3\\ \left|C\right|=\left(\begin{array}{ccc}1& 0& 1\\ 0& 2& 1\\ -1& 0& 0\end{array}\right)=1\left(0-0\right)-0\left(0-\left(-1\right)\right)+1\left(0-\left(-2\right)\right)\\ \text{ =}0-0+1\left(2\right)=2\end{array}$

  $\to \det \left(D\right)\to order\left(D\right)=3\times 3$

  $\begin{array}{l}D=\left(\begin{array}{ccc}1& 4& 3\\ 1& 6& 5\\ 0& 1& 1\end{array}\right)=1(6-5)-4(1-0)+3(1-0)\\ \text{ =}1(1)-4(1)+3(1)\\ \text{ =}1-4+3=0\end{array}$

If determinant of a matrix is zero then inverse is not possible. Therefore inverse of D is not possible. But inverse of C is possible.

If $A=\left(\begin{array}{ccc}a& b& c\\ d& e& f\\ q& h& i\end{array}\right)$ then $A^{-1}=\frac{1}{\det (A)}\left(\begin{array}{ccc}(ei-fh)& -(bi-ch)& (bf-ce)\\ -(di-gf)& (ai-cg)& -(af-cd)\\ (dh-eg)& -(ah-bg)& (ae-bd)\end{array}\right)$

  $\therefore C^{-1}=\left(\begin{array}{ccc}0& 0& -2\\ -1& 1& -1\\ 2& 0& 2\end{array}\right)=\left(\begin{array}{ccc}0& 0& -1\\ \frac{-1}{2}& \frac{1}{2}& \frac{-1}{2}\\ 1& 0& 1\end{array}\right)$