- Given matrix
A=[−43−11]
We need
det(A2016−2A2015−A2014).
- Factor out the common power of A
Since matrix powers of the same matrix commute,
A2016−2A2015−A2014=A2014(A2−2A−I).
So,
det(A2016−2A2015−A2014)=det(A2014)det(A2−2A−I).
Thus we need det(A) and A2−2A−I.
- Compute det(A)
det(A)=(−4)(1)−(−1)(3)=−4+3=−1.
Hence,
det(A2014)=(detA)2014=(−1)2014=1.
So the required determinant becomes
det(A2−2A−I).
- Compute A2
\begin{bmatrix}-4 & -1\\ 3 & 1\end{bmatrix}$$
Calculate entries:
- First row, first column:
$$(-4)(-4)+(-1)(3)=16-3=13$$
- First row, second column:
$$(-4)(-1)+(-1)(1)=4-1=3$$
- Second row, first column:
$$(3)(-4)+(1)(3)=-12+3=-9$$
- Second row, second column:
$$(3)(-1)+(1)(1)=-3+1=-2$$
Therefore,
$$A^2=\begin{bmatrix}13 & 3\\ -9 & -2\end{bmatrix}.$$
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5. **Compute $A^2-2A-I$**
First,
$$2A=\begin{bmatrix}-8 & -2\\ 6 & 2\end{bmatrix}.$$
So,
$$A^2-2A=
\begin{bmatrix}13 & 3\\ -9 & -2\end{bmatrix}
-
\begin{bmatrix}-8 & -2\\ 6 & 2\end{bmatrix}
=
\begin{bmatrix}21 & 5\\ -15 & -4\end{bmatrix}.$$
Now subtract identity matrix $I$:
$$A^2-2A-I=
\begin{bmatrix}21 & 5\\ -15 & -4\end{bmatrix}
-
\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}
=
\begin{bmatrix}20 & 5\\ -15 & -5\end{bmatrix}.$$
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6. **Find its determinant**
$$\det\begin{bmatrix}20 & 5\\ -15 & -5\end{bmatrix}
=20(-5)-5(-15)=-100+75=-25.$$
Therefore,
$$\det\left(A^{2016}-2A^{2015}-A^{2014}\right)=-25.$$
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7. **Match with options**
The correct option is:
**D: $-25$**
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8. **Comparison with stored answer**
Stored correct answer: **D**
Our derived answer: **D**
So they agree.