Inyalan0

2022-01-17

Compute the product AB by the definition of the product of matrices, where are computed separately, and by the row-comumn rule for computing AB.
$A=\left[\begin{array}{cc}-2& 3\\ 3& 5\\ 6& -2\end{array}\right]$,
$B=\left[\begin{array}{cc}5& -1\\ -1& 4\end{array}\right]$
Set up the product $A{b}_{1}$, where ${b}_{1}$ is the first column of B.
$A{b}_{1}=??$ where b1 is the first column of B.
Calculate is the first column of of B.
$A{b}_{1}=?$
Set up the product is the second column of B
$A{b}_{2}=?$
Calculate is the second column of B.
$A{b}_{2}=?$
Determine the numerical expression for the first entry in the first column of AB using the​ row-column rule.

porschomcl

Expert

We will use the product method of matrix to find solution to the problem.
Set up the product $A{b}_{1}$, where b, is the first column of B.
$A{b}_{1}=-2,5+3,-1$
$=-10+\left(-3\right)$
$=-10-3$
$=-13$
Calculate $A{b}_{1}$ where ${b}_{1}$ is the first column of B
$A{b}_{1}=-13$
Set up the product $A{b}_{2}$, where ${b}_{1}$ is the second column of B.
$A{b}_{2}=-2,-1+3,4$
$⇒2+12$
$⇒14$
Calculate $A{b}_{2}$, where ${b}_{2}$ is the second column of B.
$A{b}_{2}=14$
For the first entry,
$\left(\left(-2\right)\left(5\right)+\left(3\right)\left(-1\right)\right)$ is correct
Product $AB=$
$\left[\begin{array}{cc}-13& 14\\ 10& 17\\ 32& -14\end{array}\right]$

Archie Jones

Expert

The calculation of $A{b}_{1}$ is as follows:
$A{b}_{1}=$
$\left[\begin{array}{cc}-2& 2\\ 3& 4\\ 4& -3\end{array}\right]\left[\begin{array}{c}4\\ -1\end{array}\right]$
$=\left[\begin{array}{c}-2\left(4\right)+2\left(-1\right)\\ 3\left(4\right)+4\left(-1\right)\\ 4\left(4\right)+\left(-3\right)\left(-1\right)\end{array}\right]$
$=\left[\begin{array}{c}-8-2\\ 12-4\\ 16+3\end{array}\right]$
$\left[\begin{array}{c}-10\\ 8\\ 19\end{array}\right]$
Therefore, $A{b}_{1}=$ $\left[\begin{array}{c}-10\\ 8\\ 19\end{array}\right]$
The product $A{b}_{2}=$ $\left[\begin{array}{cc}-2& 2\\ 3& 4\\ 4& -3\end{array}\right]\left[\begin{array}{c}-1\\ 2\end{array}\right]$
The calculation of $A{b}_{1}$ is as follows:
$A{b}_{2}=ZSK\left[\begin{array}{cc}-2& 2\\ 3& 4\\ 4& -3\end{array}\right]\left[\begin{array}{c}-1\\ 2\end{array}\right]$
$=\left[\begin{array}{c}-2\left(-1\right)+2\left(2\right)\\ 3\left(-1\right)+4\left(2\right)\\ 4\left(-1\right)+\left(-3\right)\left(2\right)\end{array}\right]$
$=\left[\begin{array}{c}2+4\\ -3+8\\ -4-6\end{array}\right]$
$=\left[\begin{array}{c}6\\ 5\\ -10\end{array}\right]$
Therefore, $A{b}_{2}=ZSK\left[\begin{array}{c}6\\ 5\\ -10\end{array}\right]$
Write the matrix AB as
$$AB=$ $\left[\begin{array}{cc}-2\left(4\right)+2\left(-1\right)& -2\left(-1\right)+2\left(2\right)\\ 3\left(4\right)+4\left(-1\right)& 3\left(-1\right)+4\left(2\right)\\ 4\left(4\right)+\left(-3\right)\left(-1\right)& 4\left(-1\right)+\left(-3\right)\left(2\right)\end{array}\right]$$
Thus, the first entry in the first column of the matrix AB is $-2\left(4\right)+2\left(-1\right)$.
Hence, the correct option is (C).
The final product of the matrices A and B is $AB=$ $\left[\begin{array}{cc}-10& 6\\ 8& 5\\ 19& -10\end{array}\right]$

Do you have a similar question?