A = \(\begin{bmatrix}-1 & -3 & 2\\ -5 & -9 & 1\end{bmatrix}\), and v = \(\begin{bmatrix}5\\ -3\\ -2\end{bmatrix}\)
Then show that v is null of A.
\(\begin{bmatrix}3 & 6 & -4\\ 0 & 0 & -6\\ 0 & 0 & -2 \end{bmatrix}\)
T(e1) = \(\begin{bmatrix}5\\ 1\\ -2\end{bmatrix}\) and T(e2) = \(\begin{bmatrix}0\\ -1\\ 8\end{bmatrix}\)
find a formula for the image of an arbitrary x in R2. That is, find T(x) for x in R2.
a = \(\begin{bmatrix}1\\ -3\\ 2\end{bmatrix}\), b = \(\begin{bmatrix}0\\ -1\\ 3\end{bmatrix}\)
A = \(\begin{bmatrix}-2 & 8 & -1\\ 0 & 0 & 0\\ 0 & -5 & 3\end{bmatrix}\)
C = \(\begin{bmatrix}0.5 & 0.4 & 0.2\\ 0.2 & 0.3 & 0.1\\ 0.1 & 0.1 & 0.3\end{bmatrix}\)
and the final demand is 50 units for manufacturing, 30 units for agriculture and 20 units for service. Find the production level x that will satisfy this demand.
x1 – 2x2 – x3 + 3x4 = 0
-2x1 + 4x2 + 5x3 – 5x4 = 3
3x1 – 6x2 – 6x3 + 8x4 = 2
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