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Problem 16

# Can the matrix $E = \begin{array}{ll}\mid 3 & 1 \mid \\ \mid 1 & -1 \mid\end{array}$ be written as a linear combination of the matrices $\begin{array}{rl}\mathrm{m}_{1}=\mid 1 & 1\left|, \quad \mathrm{~m}_{2}=\right| 0 & 0 \mid & \mathrm{m}_{3}=\mid 0 & 2 \\ \mid 1 & 0 \mid & \mid 1 & 1 \mid & \mid 0 & -1 \mid\end{array}$ and $$\begin{array}{rc}\mathrm{m}_{4}=\mid 0 & 1 \mid \\ \mid 1 & 0\end{array}$$.

Expert verified
Yes, the matrix $$E = \begin{bmatrix} 3 & 1 \\ 1 & -1 \end{bmatrix}$$ can be written as a linear combination of the matrices $$m_1, m_2, m_3,$$ and $$m_4$$. The linear combination is: $$E = 3 \cdot m_1 - 1 \cdot m_2 + 0 \cdot m_3 - 1 \cdot m_4 = \begin{bmatrix} 3 & 1 \\ 1 & -1 \end{bmatrix}$$
See the step by step solution

## Step 1: Write down the linear combination equation

We need to find coefficients (a, b, c, and d) such that the matrix E can be written as: E = a * m_1 + b * m_2 + c * m_3 + d * m_4 Which is equivalent to: $$\begin{bmatrix} 3 & 1 \\ 1 & -1 \end{bmatrix} = a \cdot \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} + b \cdot \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} + c \cdot \begin{bmatrix} 0 & 2 \\ 0 & -1 \end{bmatrix} + d \cdot \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$ Now let's find these coefficients.

## Step 2: Create a system of equations from the linear combination equation

We can create a system of equations from the linear combination equation as follows: For element (1,1) of E: 3 = a * 1 + b * 0 + c * 0 + d * 0 For element (1,2) of E: 1 = a * 1 + b * 0 + c * 2 + d * 1 For element (2,1) of E: 1 = a * 1 + b * 1 + c * 0 + d * 1 For element (2,2) of E: -1 = a * 0 + b * 1 + c * (-1) + d * 0 Now we have the following system of equations: 1) a = 3 2) a + 2c + d = 1 3) a + b + d = 1 4) b - c = -1

## Step 3: Solve the system of equations

Let's solve the system of equations from step 2: 1) a = 3 2) 3 + 2c + d = 1 3) 3 + b + d = 1 4) b - c = -1 Now we have: 2) 2c + d = -2 3) b + d = -2 4) b - c = -1 From equation 3, we get: b = -2 - d Now substitute b in equation 4: (-2 - d) - c = -1 Now we have: c + d = -1 Now we have two equations with two variables (c and d): 5) 2c + d = -2 6) c + d = -1 Solving equations 5 and 6, we get: c = 0 and d = -1 Now substitute c and d back into the equation for b: b = -2 - (-1) = -1

## Step 4: Write down the linear combination

Now that we have found the coefficients (a = 3, b = -1, c = 0, and d = -1), we can write down the linear combination of matrices E: $$\begin{bmatrix} 3 & 1 \\ 1 & -1 \end{bmatrix} = 3 \cdot \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} - 1 \cdot \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} + 0 \cdot \begin{bmatrix} 0 & 2 \\ 0 & -1 \end{bmatrix} - 1 \cdot \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$ So, yes, the matrix E can be written as a linear combination of the matrices m_1, m_2, m_3, and m_4 with the coefficients a = 3, b = -1, c = 0, and d = -1.

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