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Question: If the matrix $A = \begin{bmatrix} -1 & 0 & 5 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ satisfies the...

If the matrix A=[−105030001]A = \begin{bmatrix} -1 & 0 & 5 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix} satisfies the equation A27+aA26+bA=[−40500000−2]A^{27} + aA^{26} + bA = \begin{bmatrix} -4 & 0 & 5 \\ 0 & 0 & 0 \\ 0 & 0 & -2 \end{bmatrix} for some real numbers aa and bb, then

Answer

a=-3, b=0

Explanation

Solution

The given matrix is A=[−105030001]A = \begin{bmatrix} -1 & 0 & 5 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix}.
We need to find real numbers aa and bb such that A27+aA26+bA=[−40500000−2]A^{27} + aA^{26} + bA = \begin{bmatrix} -4 & 0 & 5 \\ 0 & 0 & 0 \\ 0 & 0 & -2 \end{bmatrix}.

First, let's compute the powers of AA. A1=[−105030001]A^1 = \begin{bmatrix} -1 & 0 & 5 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix}

A2=A⋅A=[−105030001][−105030001]=[100090001]A^2 = A \cdot A = \begin{bmatrix} -1 & 0 & 5 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 5 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 1 \end{bmatrix}

A3=A2⋅A=[100090001][−105030001]=[−1050270001]A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 5 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 5 \\ 0 & 27 & 0 \\ 0 & 0 & 1 \end{bmatrix}

A4=A3⋅A=[−1050270001][−105030001]=[1000810001]A^4 = A^3 \cdot A = \begin{bmatrix} -1 & 0 & 5 \\ 0 & 27 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 5 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 81 & 0 \\ 0 & 0 & 1 \end{bmatrix}

We observe a pattern for AnA^n:

For odd n≥1n \ge 1: An=[(−1)n0503n0001n]=[−10503n0001]A^n = \begin{bmatrix} (-1)^n & 0 & 5 \\ 0 & 3^n & 0 \\ 0 & 0 & 1^n \end{bmatrix} = \begin{bmatrix} -1 & 0 & 5 \\ 0 & 3^n & 0 \\ 0 & 0 & 1 \end{bmatrix}

For even n≥2n \ge 2: An=[(−1)n0003n0001n]=[10003n0001]A^n = \begin{bmatrix} (-1)^n & 0 & 0 \\ 0 & 3^n & 0 \\ 0 & 0 & 1^n \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3^n & 0 \\ 0 & 0 & 1 \end{bmatrix}

Using this pattern for n=27n=27 (odd) and n=26n=26 (even):

A27=[−10503270001]A^{27} = \begin{bmatrix} -1 & 0 & 5 \\ 0 & 3^{27} & 0 \\ 0 & 0 & 1 \end{bmatrix}

A26=[10003260001]A^{26} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3^{26} & 0 \\ 0 & 0 & 1 \end{bmatrix}

Substitute these into the given matrix equation:

[−10503270001]+a[10003260001]+b[−105030001]=[−40500000−2]\begin{bmatrix} -1 & 0 & 5 \\ 0 & 3^{27} & 0 \\ 0 & 0 & 1 \end{bmatrix} + a \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3^{26} & 0 \\ 0 & 0 & 1 \end{bmatrix} + b \begin{bmatrix} -1 & 0 & 5 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -4 & 0 & 5 \\ 0 & 0 & 0 \\ 0 & 0 & -2 \end{bmatrix}

Perform the matrix addition and scalar multiplication on the left side:

[(−1)+a(1)+b(−1)0+0+05+a(0)+b(5)0+0+0327+a(326)+b(3)0+0+00+a(0)+b(0)0+0+01+a(1)+b(1)]=[−40500000−2]\begin{bmatrix} (-1) + a(1) + b(-1) & 0+0+0 & 5+a(0)+b(5) \\ 0+0+0 & 3^{27}+a(3^{26})+b(3) & 0+0+0 \\ 0+a(0)+b(0) & 0+0+0 & 1+a(1)+b(1) \end{bmatrix} = \begin{bmatrix} -4 & 0 & 5 \\ 0 & 0 & 0 \\ 0 & 0 & -2 \end{bmatrix}

[−1+a−b05+5b0327+a326+3b0001+a+b]=[−40500000−2]\begin{bmatrix} -1 + a - b & 0 & 5 + 5b \\ 0 & 3^{27} + a3^{26} + 3b & 0 \\ 0 & 0 & 1 + a + b \end{bmatrix} = \begin{bmatrix} -4 & 0 & 5 \\ 0 & 0 & 0 \\ 0 & 0 & -2 \end{bmatrix}

Equating the corresponding entries of the matrices:

From entry (1,1): −1+a−b=−4  ⟹  a−b=−3-1 + a - b = -4 \implies a - b = -3 (Equation 1)

From entry (1,3): 5+5b=5  ⟹  5b=0  ⟹  b=05 + 5b = 5 \implies 5b = 0 \implies b = 0 (Equation 2)

From entry (2,2): 327+a326+3b=03^{27} + a3^{26} + 3b = 0 (Equation 3)

From entry (3,3): 1+a+b=−2  ⟹  a+b=−31 + a + b = -2 \implies a + b = -3 (Equation 4)

We have a system of linear equations for aa and bb.

From Equation 2, we have b=0b=0.

Substitute b=0b=0 into Equation 1: a−0=−3  ⟹  a=−3a - 0 = -3 \implies a = -3.

Substitute b=0b=0 into Equation 4: a+0=−3  ⟹  a=−3a + 0 = -3 \implies a = -3.

Both Equation 1 and Equation 4 give a=−3a=-3 and b=0b=0.

Finally, check if these values satisfy Equation 3:

327+a326+3b=327+(−3)326+3(0)=327−31⋅326+0=327−327=03^{27} + a3^{26} + 3b = 3^{27} + (-3)3^{26} + 3(0) = 3^{27} - 3^1 \cdot 3^{26} + 0 = 3^{27} - 3^{27} = 0.

Equation 3 is satisfied.

Thus, the values of aa and bb are a=−3a = -3 and b=0b = 0.