Let P=100410841 and I be the identity matrix of order 3. If Q=[qij] is a matrix such that p500−Q=I, thenq31+q32q21 equals

A
995
B
1001
C
1000
D
1005

Solution:

$P = I + B, B = [\begin{array}{lll} 0 & 0 & 0 \\ 4 & 0 & 0 \\ 8 & 4 & 0 \end{array}]$

$B^{2} = [\begin{array}{lll} 0 & 0 & 0 \\ 4 & 0 & 0 \\ 8 & 4 & 0 \end{array}] [\begin{array}{lll} 0 & 0 & 0 \\ 4 & 0 & 0 \\ 8 & 4 & 0 \end{array}] = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 16 & 0 & 0 \end{array}]$

$B^{3} = 0$

$P^{500} = (I + B)^{500} = I + 500 B + 250 \times 499 B^{2}$

$\Rightarrow P^{500} I - I = 500 B + 250 \times 499 B^{2}$

$\frac{500 \times 8 + 250 \times 499 \times 16 + 500 \times 4}{500 \times 4 + 0} = 1001$

Let $P = [\begin{matrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 8 & 4 & 1 \end{matrix}]$ and I be the identity matrix of order 3. If $Q = [q_{ij}]$ is a matrix such that $p^{500} - Q = I, then \frac{q_{31} + q_{32}}{q_{21}} equals$

Solution:

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