Chapter 10, Problem 11RCC

To determine

To find: the value of $A+B,A-B$ and $kA$

Answer to Problem 11RCC

The answer is

  $A+B=\left[\begin{array}{ccc}6& 8& 10\\ 12& 14& 16\end{array}\right],A-B=\left[\begin{array}{ccc}-6& -6& -6\\ -6& -6& -6\end{array}\right],kA=\left[\begin{array}{ccc}0& 3& 6\\ 9& 12& 15\end{array}\right]$

Explanation of Solution

Given that A and B is matrices with the same dimension and that k are a real number. Find $A+B,A-B$ and $kA$ . These sum, difference, and scalar product formulas of matrices can be found in Chapter 10, Section 4.

For visual purposes, let's just say that A and B are $2\times 3$ matrices and have elements $a_{ij}$ and $b_{ij}$

  $A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\end{array}\right],B=\left[\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\end{array}\right]$

Give the matrices arbitrary values.

  $A=\left[\begin{array}{ccc}0& 1& 2\\ 3& 4& 5\end{array}\right],B=\left[\begin{array}{ccc}6& 7& 8\\ 9& 10& 11\end{array}\right]$

To find $A+B$ , simply add the corresponding elements of each matrix (elements in the same position are added). That is, add the numbers with the same indices. The resultant matrix has the same dimensions as A and B.

  $\begin{array}{l}A+B=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\end{array}\right]+\left[\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\end{array}\right]\\ A+B=\left[\begin{array}{ccc}{a}_{11}+b_{11}& a_{12}+b_{12}& a_{13}+b_{13}\\ a_{21}+b_{21}& a_{22}+b_{22}& a_{23}+b_{23}\end{array}\right]\end{array}$

With actual values, this would be

  $\begin{array}{l}A+B=\left[\begin{array}{ccc}0& 1& 2\\ 3& 4& 5\end{array}\right]+\left[\begin{array}{ccc}6& 7& 8\\ 9& 10& 11\end{array}\right]\\ A+B=\left[\begin{array}{ccc}0+6& 1+7& 2+8\\ 3+9& 4+10& 5+11\end{array}\right]\\ A+B=\left[\begin{array}{ccc}6& 8& 10\\ 12& 14& 16\end{array}\right]\end{array}$

To find $A-B$ , subtract each element $b_{ij}$ from the corresponding element in A (which is $a_{ij}$ ). That is, subtract the numbers with the same indices. The resultant matrix has the same dimensions as A and B.

  $\begin{array}{l}A-B=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\end{array}\right]+\left[\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\end{array}\right]\\ A-B=\left[\begin{array}{ccc}{a}_{11}-b_{11}& a_{12}-b_{12}& a_{13}-b_{13}\\ a_{21}-b_{21}& a_{22}-b_{22}& a_{23}-b_{23}\end{array}\right]\end{array}$

With actual values, this would be

  $\begin{array}{l}A-B=\left[\begin{array}{ccc}0& 1& 2\\ 3& 4& 5\end{array}\right]-\left[\begin{array}{ccc}6& 7& 8\\ 9& 10& 11\end{array}\right]\\ A-B=\left[\begin{array}{ccc}0-6& 1-7& 2-8\\ 3-9& 4-10& 5-11\end{array}\right]\\ A-B=\left[\begin{array}{ccc}-6& -6& -6\\ -6& -6& -6\end{array}\right]\end{array}$

To find the product $kA$ , simply multiply each element of A, represented by $a_{ij}$ , by the constant k. The resultant matrix has the same dimensions as A and B.

  $\begin{array}{l}kA=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\end{array}\right]\\ kA=\left[\begin{array}{ccc}k{a}_{11}& k{a}_{12}& k{a}_{13}\\ k{a}_{21}& k{a}_{22}& k{a}_{23}\end{array}\right]\end{array}$

For our purposes, let k = 3. Putting in actual values,

  $\begin{array}{l}kA=\left[\begin{array}{ccc}0& 1& 2\\ 3& 4& 5\end{array}\right]\\ kA=3\cdot \left[\begin{array}{ccc}3\cdot 0& 3\cdot 1& 3\cdot 2\\ 3\cdot 3& 3\cdot 4& 3\cdot 5\end{array}\right]\\ kA=\left[\begin{array}{ccc}0& 3& 6\\ 9& 12& 15\end{array}\right]\end{array}$

Conclusion:

Therefore, the answer is

  $A+B=\left[\begin{array}{ccc}6& 8& 10\\ 12& 14& 16\end{array}\right],A-B=\left[\begin{array}{ccc}-6& -6& -6\\ -6& -6& -6\end{array}\right],kA=\left[\begin{array}{ccc}0& 3& 6\\ 9& 12& 15\end{array}\right]$