If A=1000110â24,6Aâ1=A2+cA+dI then (c,d) is

A
$(â 6, 11)$
B
$(â 11, 6)$
C
$(11, 6)$
D
$(6, 11)$

Solution:

$6 A^{â 1} = A^{2} + c A + d I$

$â 6 I = A^{3} + c A^{2} + d A$

We have

$A^{2} = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & â 2 & 4 \end{array}] [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & â 2 & 4 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & â 1 & 5 \\ 0 & â 10 & 14 \end{array}]$

and $\begin{aligned} A^{3} = A^{2} A \end{aligned}$

$= [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & â 1 & 5 \\ 0 & â 10 & 14 \end{array}] [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & â 2 & 4 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & â 11 & 19 \\ 0 & â 38 & 46 \end{array}]$

Now,

$\begin{array}{l} 6 I = A^{3} + c A^{2} + d A \\ 6 = 1 + c + d, 0 = 19 + 5 c + d \\ 6 = â 11 â c + d \\ 6 = 46 + 14 c + 4 d, 0 = â 38 â 10 c â 2 d \\ â d = 11, c = â 6 \end{array}$

If $A = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & â 2 & 4 \end{array}], 6 A^{â 1} = A^{2} + c A + d I$ then (c,d) is

Solution:

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