Question

If A+B = $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ and A-2B = $\begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}$, then A

• A. $\begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix}$
• B. $\begin{bmatrix} 2/3 & 1/3 \\ 1/3 & 2/3 \end{bmatrix}$
• C. $\begin{bmatrix} 1/3 & 2/3 \\ 2/3 & 1/3 \end{bmatrix}$
• D. $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$

Answer 1

$\begin{bmatrix} 1/3 & 1 \\ 2/3 & 1/3 \end{bmatrix}$

Answer 2

Let the given equations be:

1. $A + B = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$

2. $A - 2B = \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}$

We have a system of two linear equations with two matrix variables, A and B. We can solve this system using methods similar to solving algebraic equations.

Multiply equation (1) by 2:

$2(A + B) = 2 \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$

$2A + 2B = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$ (Equation 3)

Now, add equation (3) and equation (2):

$(2A + 2B) + (A - 2B) = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}$

Combine the terms involving A and B on the left side:

$(2A + A) + (2B - 2B) = 3A + \mathbf{0} = 3A$

Add the matrices on the right side:

$\begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 2 + (-1) & 2 + 1 \\ 2 + 0 & 2 + (-1) \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}$

So, we have:

$3A = \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}$

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

$A = \frac{1}{3} \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}$

$A = \begin{bmatrix} \frac{1}{3} \times 1 & \frac{1}{3} \times 3 \\ \frac{1}{3} \times 2 & \frac{1}{3} \times 1 \end{bmatrix}$

$A = \begin{bmatrix} 1/3 & 1 \\ 2/3 & 1/3 \end{bmatrix}$

The calculated matrix A is $\begin{bmatrix} 1/3 & 1 \\ 2/3 & 1/3 \end{bmatrix}$. This does not match any of the provided options. There appears to be an error in the provided options.