If X \(=\begin{bmatrix} 8 &-1&-3 \\-5 &1&2\\10&-1&-4 \end{bmatrix}\)  and Y = \(\begin{bmatrix} 2 & 1 & -1\\0 & 2 & 1\\ 5& p & q \end{bmatrix}\) then, find p, q if Y =  X^-1

If \(A=\left[ \begin{matrix} 2 & 4 & 4 \\ 2 & 5 & 4 \\ 2 & 5 & 3 \end{matrix} \right] \) then find A ^-1.

If \(A=\left[ \begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} 0 & -1 \\ 1 & 2 \end{matrix} \right] \)then, show that (AB)^-1 = B^-1A^-1

Solve by using matrix inversion method:
\(3 x-2 y+3 z=8 ; 2 x+y-z=1\)
\(4 x-3 y+2 z=4\)

An economy produces only coal and steel. These two commodities serve as intermediate inputs in each other’s production. 0.4 tonne of steel and 0.7 tonne of coal are needed to produce a tonne of steel. Similarly 0.1 tonne of steel and 0.6 tonne of coal are required to produce a tonne of coal. No capital inputs are needed. Do you think that the system is viable? 2 and 5 labour days are required to produce a tonnes of coal and steel respectively. If economy needs 100 tonnes of coal and 50 tonnes of steel, calculate the gross output of the two commodities and the total labour days required.