Answer
(a) $\begin{bmatrix} 0 &1&-1\\-2 &3/2&0 \\1 &-3/2&1 \end{bmatrix}$
(b) $1$ oz of type-A, $1$ oz of type-B, $2$ oz of type-C,
(c) $2$ oz of type-A, $0$ oz of type-B, $1$ oz of type-C.
(d) no solution.
Work Step by Step
(a) To find the inverse of the matrix:
1. Write the matrix together with an identify as:
$\begin{array}( \\A|I= \\ \\ \end{array}
\begin{bmatrix} 3 & 1 &3 &| &1 &0&0\\4 & 2 &4 &| &0 &1&0 \\3 & 2 &4 &| &0 &0&1\end{bmatrix} \begin{array}( R_1\times4\\R_2\times3 \\R_3\times4 \\ \end{array}$
2. Use row operations to transform the left half into a reduced row-echelon:
$\begin{bmatrix} 12 & 4 &12 &| &4 &0&0\\12 & 6 &12 &| &0 &3&0 \\12 & 8 &16 &| &0 &0&4 \end{bmatrix} \begin{array}( \\R_2-R_1\to R_2 \\(R_3-R_1)/2\to R_3 \\ \end{array}$
$\begin{bmatrix} 12 & 4 &12 &| &4 &0&0\\0 & 2 &0 &| &-4 &3&0 \\0 & 2 &2 &| &-2 &0&2 \end{bmatrix} \begin{array}( \\R_2\times2 \\(R_3-R_2)\times6 \\ \end{array}$
$\begin{bmatrix} 12 & 4 &12 &| &4 &0&0\\0 & 4 &0 &| &-8 &6&0 \\0 & 0 &12 &| &12 &-18&12 \end{bmatrix} \begin{array}( (R_1-R_2-R_3)/12\\R_2/4 \\R_3/12 \\ \end{array}$
$\begin{bmatrix} 1 & 0 &0 &| &0 &1&-1\\0 & 1 &0 &| &-2 &3/2&0 \\0 & 0 &1 &| &1 &-3/2&1 \end{bmatrix}$
$\begin{array}( \\A^{-1}= \\ \\ \end{array}
\begin{bmatrix} 0 &1&-1\\-2 &3/2&0 \\1 &-3/2&1 \end{bmatrix}$
(b) Assume we need $x$ oz of type-A, $y$ oz of type-B, and $z$ oz of type-C, we can set up the following system to equations $AX=B$:
$\begin{bmatrix} 3 & 1 &3 \\4 & 2 &4 \\3 & 2 &4 \end{bmatrix} \begin{bmatrix}x \\y \\z\end{bmatrix}
\begin{array}( \\= \\ \\ \end{array}
\begin{bmatrix}10 \\14 \\13\end{bmatrix} $
Since we know $A^{-1}$, we have $X=A^{-1}B$ and
$\begin{bmatrix}x \\y \\z\end{bmatrix}
\begin{array}( \\= \\ \\ \end{array}
\begin{bmatrix} 0 &1&-1\\-2 &3/2&0 \\1 &-3/2&1 \end{bmatrix} \begin{bmatrix}10 \\14 \\13\end{bmatrix} \begin{array}( \\= \\ \\ \end{array}
\begin{bmatrix}1 \\1 \\2\end{bmatrix}$
Thus , we need $1$ oz of type-A, $1$ oz of type-B, and $2$ oz of type-C,
(c) Similar to case (b), we have:
$\begin{bmatrix}x \\y \\z\end{bmatrix}
\begin{array}( \\= \\ \\ \end{array}
\begin{bmatrix} 0 &1&-1\\-2 &3/2&0 \\1 &-3/2&1 \end{bmatrix} \begin{bmatrix}9 \\12 \\10\end{bmatrix} \begin{array}( \\= \\ \\ \end{array}
\begin{bmatrix}2 \\0 \\1\end{bmatrix}$
Thus , we need $2$ oz of type-A, $0$ oz of type-B, and $1$ oz of type-C for this case.
(d) Similarly:
$\begin{bmatrix}x \\y \\z\end{bmatrix}
\begin{array}( \\= \\ \\ \end{array}
\begin{bmatrix} 0 &1&-1\\-2 &3/2&0 \\1 &-3/2&1 \end{bmatrix} \begin{bmatrix}2 \\4 \\11\end{bmatrix} \begin{array}( \\= \\ \\ \end{array}
\begin{bmatrix}-7 \\2 \\7\end{bmatrix}$
Since a negative number does not make sense, we do not have solutions for this case.
