Question

$A=\left[\begin{array}{ccc}5& 7& 2\\ 4& -1& 3\\ 6& 8& -5\end{array}\right]$ and $B=\left[\begin{array}{ccc}6& 11& -4\\ 2& 1& -5\\ 3& -9& 6\end{array}\right]$ Match each value to the correct entry in matrix $AB$.

$AB=\left[\begin{array}{ccc}{x}_{11}& x_{12}& x_{13}\\ x_{21}& x_{22}& x_{23}\\ x_{31}& x_{32}& x_{33}\end{array}\right]$

Column 1	Column 2
$x_{11}$	$37$
$x_{12}$	$31$
$x_{13}$	$-94$
$x_{21}$	$44$
$x_{22}$	$119$
$x_{23}$	$7$
$x_{31}$	$50$
$x_{32}$	$16$
$x_{33}$	$-43$

Answer 1

Explanation for $\left(A\right)$:

$x_{11}$ can be found by multiplying each element of the first row of matrix $A$ to the corresponding each element of the first column of $B$

$\therefore$ $x_{11}=5\times 6+7\times 2+2\times 3$

$=30+14+6$

$=50$

Therefore $\left(A\right)\to \left(7\right)$

Explanation for $\left(B\right)$:

$x_{12}$ can be found by multiplying each element of the first row of matrix $A$ to the corresponding element of the second column of $B$.

$\therefore$ $x_{12}=5\times 11+7\times 1+2\times (-9)$

$=55+7-18$

$=44$

Therefore $\left(B\right)\to \left(4\right)$

Explanation for $\left(C\right)$:

$x_{13}$ can be found by multiplying each element of the first row of matrix $A$ to the corresponding element of the third column of $B$.

$\therefore$ $x_{13}=5\times (-4)+7\times (-5)+2\times 6$

$=-20-35+12$

$=-43$

Therefore $\left(C\right)\to \left(9\right)$

Explanation for $\left(D\right)$:

$x_{21}$ can be found by multiplying each element of the second row of matrix $A$ to the corresponding element of the first column of $B$.

$\therefore$ $x_{21}=4\times 6+(-1)\times 2+3\times 3$

$=24-2+9$

$=31$

Therefore $\left(D\right)\to \left(2\right)$

Explanation for $\left(E\right)$:

$x_{22}$ can be found by multiplying each element of the second row of matrix $A$ to the corresponding element of the second column of $B$.

$\therefore$ $x_{22}=4\times 11+(-1)\times 1+3\times (-9)$

$=44-1-27$

$=16$

Therefore $\left(E\right)\to \left(8\right)$

Explanation for $\left(F\right)$:

$x_{23}$ can be found by multiplying each element of the second row of matrix $A$ to the corresponding element of the third column of $B$.

$\therefore$ $x_{23}=4\times (-4)+(-1)\times (-5)+3\times 6$

$=-16+5+18$

$=7$

Therefore $\left(F\right)\to \left(6\right)$

Explanation for $\left(G\right)$:

$x_{31}$ can be found by multiplying each element of the third row of matrix $A$ to the corresponding element of the first column of $B$.

$\therefore$ $x_{31}=6\times 6+8\times 2+(-5)\times 3$

$=36+16+(-15)$

$=37$

Therefore $\left(G\right)\to \left(1\right)$

Explanation for $\left(H\right)$:

$x_{32}$ can be found by multiplying each element of the third row of matrix $A$ to the corresponding element of the second column of $B$.

$\therefore$ $x_{32}=6\times 11+8\times 1+(-5)\times (-9)$

$=66+8+45$

$=119$

Therefore $\left(H\right)\to \left(5\right)$

Explanation for $\left(I\right)$:

$x_{33}$ can be found by multiplying each element of the third row of matrix $A$ to the corresponding element of the third column of $B$.

$\therefore$ $x_{33}=6\times (-4)+8\times (-5)+(-5)\times 6$

$=-24-40-30$

$=-94$

Therefore $\left(I\right)\to \left(3\right)$

Hence, $\left(A\right)\to \left(7\right)$

$\left(B\right)\to \left(4\right)$

$\left(C\right)\to \left(9\right)$

$\left(D\right)\to \left(2\right)$

$\left(E\right)\to \left(8\right)$

$\left(F\right)\to \left(6\right)$

$\left(G\right)\to \left(1\right)$

$\left(H\right)\to \left(5\right)$

$\left(I\right)\to \left(3\right)$