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Question: If $A+B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ and $A-2B = \begin{bmatrix} -1 & 1 \\ 0 & -1...

If A+B=[1011]A+B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} and A−2B=[−110−1]A-2B = \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}, then A

A

[1121]\begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix}

B

[2/31/31/32/3]\begin{bmatrix} 2/3 & 1/3 \\ 1/3 & 2/3 \end{bmatrix}

C

[1/32/32/31/3]\begin{bmatrix} 1/3 & 2/3 \\ 2/3 & 1/3 \end{bmatrix}

D

[1−111]\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}

Answer

[1/31/32/31/3]\begin{bmatrix} 1/3 & 1/3 \\ 2/3 & 1/3 \end{bmatrix}

Explanation

Solution

We are given the following matrix equations:

  1. A+B=[1011]A+B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}

  2. A−2B=[−110−1]A-2B = \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}

We can solve this system of equations for the matrix A. Multiply the first equation by 2:

2(A+B)=2[1011]2(A+B) = 2 \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}

2A+2B=[2022]2A + 2B = \begin{bmatrix} 2 & 0 \\ 2 & 2 \end{bmatrix} (Equation 3)

Now, add Equation 2 and Equation 3:

(A−2B)+(2A+2B)=[−110−1]+[2022](A-2B) + (2A+2B) = \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 2 & 2 \end{bmatrix}

(A+2A)+(−2B+2B)=[−1+21+00+2−1+2](A+2A) + (-2B+2B) = \begin{bmatrix} -1+2 & 1+0 \\ 0+2 & -1+2 \end{bmatrix}

3A+0=[1121]3A + \mathbf{0} = \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix}

3A=[1121]3A = \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix}

To find A, multiply the matrix by the scalar 13\frac{1}{3}:

A=13[1121]A = \frac{1}{3} \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix}

A=[13×113×113×213×1]A = \begin{bmatrix} \frac{1}{3} \times 1 & \frac{1}{3} \times 1 \\ \frac{1}{3} \times 2 & \frac{1}{3} \times 1 \end{bmatrix}

A=[1/31/32/31/3]A = \begin{bmatrix} 1/3 & 1/3 \\ 2/3 & 1/3 \end{bmatrix}

Comparing this result with the given options, we find that none of the options exactly match the calculated matrix A.