a coach purchases 47 hats for his players and their families at a total cost of $302. the cost of a small…

a coach purchases 47 hats for his players and their families at a total cost of $302. the cost of a small hat is $5.50. a medium hat costs $6.00. a large hat costs $7.00. he purchases three times as many medium hats as small hats. using matrices, how many large hats did the coach purchase?\no 6\no 11\no 18\no 23

a coach purchases 47 hats for his players and their families at a total cost of $302. the cost of a small hat is $5.50. a medium hat costs $6.00. a large hat costs $7.00. he purchases three times as many medium hats as small hats. using matrices, how many large hats did the coach purchase?\no 6\no 11\no 18\no 23

Answer

Explanation:

Step1: Define variables

Let $x$ be the number of small hats, $y$ be the number of medium hats, and $z$ be the number of large hats. We have the following system of equations: Equation 1 (total number of hats): $x + y+z=47$ Equation 2 (total cost of hats): $5.5x + 6y+7z = 302$ Equation 3 (relationship between medium and small hats): $y = 3x$ or $- 3x+y=0$ In matrix - form $AX = B$, where $A=\begin{bmatrix}1&1&1\5.5&6&7\ - 3&1&0\end{bmatrix}$, $X=\begin{bmatrix}x\y\z\end{bmatrix}$ and $B=\begin{bmatrix}47\302\0\end{bmatrix}$

Step2: Calculate the determinant of $A$

$\text{det}(A)=1\times\begin{vmatrix}6&7\1&0\end{vmatrix}-1\times\begin{vmatrix}5.5&7\ - 3&0\end{vmatrix}+1\times\begin{vmatrix}5.5&6\ - 3&1\end{vmatrix}$ $=1\times(0 - 7)-1\times(0 + 21)+1\times(5.5+18)$ $=-7-21 + 23.5=-4.5$

Step3: Replace the third - column of $A$ with $B$ to get $A_z$

$A_z=\begin{bmatrix}1&1&47\5.5&6&302\ - 3&1&0\end{bmatrix}$ $\text{det}(A_z)=1\times\begin{vmatrix}6&302\1&0\end{vmatrix}-1\times\begin{vmatrix}5.5&302\ - 3&0\end{vmatrix}+47\times\begin{vmatrix}5.5&6\ - 3&1\end{vmatrix}$ $=1\times(0 - 302)-1\times(0 + 906)+47\times(5.5 + 18)$ $=-302-906+47\times23.5$ $=-302-906 + 1104.5=96.5$

Step4: Calculate $z$ using Cramer's rule

$z=\frac{\text{det}(A_z)}{\text{det}(A)}=\frac{96.5}{-4.5}\text{(wrong approach)}$

Let's use substitution method instead. Substitute $y = 3x$ into $x + y+z=47$ and $5.5x + 6y+7z = 302$ We get $x+3x + z=47$ or $4x+z=47$ (Equation 4) and $5.5x+6\times3x+7z = 302$ or $5.5x + 18x+7z = 302$ or $23.5x+7z = 302$ (Equation 5) From Equation 4, $z = 47 - 4x$ Substitute $z = 47 - 4x$ into Equation 5: $23.5x+7(47 - 4x)=302$ $23.5x+329-28x = 302$ $23.5x-28x=302 - 329$ $-4.5x=-27$ $x = 6$ Since $z = 47-(x + y)$ and $y = 3x$, then $y = 3\times6 = 18$ $z=47-(6 + 18)=23$

Answer:

23