Linear Algebra: Linear Independence and writing Matrices as linear combinations

1. The problem statement, all variables and given/known data

If linearly dependent, write one matrix as a linear combination of the rest.

$$\left[\begin{array}{cc} 1&1 \\ 2&1 \end{array}\right]$$ $$\left[\begin{array}{cc} 1&0 \\ 0&2 \end{array}\right]$$ $$\left[\begin{array}{cc} 0&3 \\ 2&1 \end{array}\right]$$ $$\left[\begin{array}{cc} 4&6 \\ 8&6 \end{array}\right]$$

2. Relevant equations

3. The attempt at a solution

Choosing coefficients a, b, c, d for the matrices, respectively, I set up 4 equations and 4 unknowns:

row 1 column 1: $$a + b + 4d = 0$$
row 1 column 2: $$a+ 3c + 6d = 0$$
row 2 column 1: $$2a + 2c + 8d = 0$$
row 2 column 2: $$a + 2b + c + 6d = 0$$

From this, I create a 4x5 matrix and using Gauss-Jordan elimination arrive at:

Edit: this is wrong. See below

$$\left[\begin{array}{ccccc} 1&0&0&2&0 \\ 0&1&0&2&0 \\ 0&0&1&\frac{4}{3}&0 \\ 0&0&0&0&0 \end{array}\right]$$

From this, it is clear that the 4 matrices are linearly dependent.

What I do not understand: how do I, given this last matrix, write one of the matrices as a linear combination of the others?

Thanks

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 Mentor Your last matrix means this: a = -2d b = -2d c = -4/3 d From this we can deduce that d is arbitrary, so let d = 1. Then a = -2, b = -2, and c = -4/3. Put these values into the equation you set up to determine linear dependence and you'll see that one of them is a particular linear combination of the other three matrices. BTW, when you solved the 4 equations in 4 unknowns, you could have used a 4 x 4 matrix instead of a 4 x 5 augmented matrix. Your 5th column started as all zeroes and never changed.

 Quote by Mark44 Your last matrix means this: a = -2d b = -2d c = -4/3 d From this we can deduce that d is arbitrary, so let d = 1. Then a = -2, b = -2, and c = -4/3. Put these values into the equation you set up to determine linear dependence and you'll see that one of them is a particular linear combination of the other three matrices.
I attempted to check this, but it appears to not work, as the equation generated is not true.

$$-2a - 2b + \frac{4}{3}c = d$$

Have I reduced the 4x5 matrix incorrectly or is the problem more fundamental?

Linear Algebra: Linear Independence and writing Matrices as linear combinations

 Going from the 4 equations/unknowns to the matrix I made a silly mistake causing the row reduced matrix to be incorrect. The (I think) correct row reduced matrix is $$\left[\begin{array}{ccccc} 1&0&0&3 \\ 0&1&0&1 \\ 0&0&1&1 \\ 0&0&0&0 \end{array}\right]$$ From here, I thought I would read out of the matrix the following equations: $$a = -3d$$ $$b = -d$$ $$c = -d$$ where d is any arbitrary constant. Assigning d = 1, and checking to see if $$a * M1 + b * M2 + c * M3$$ was in fact $$d * M4$$, I end up with exactly the wrong signs: $$-3\left[\begin{array}{cc}1&1\\2&1\end{array}\right] + -1\left[\begin{array}{cc}1&0\\0&2\end{array}\right] + -1\left[\begin{array}{cc}0&3\\2&1\end{array}\right] = \left[\begin{array}{cc}-4&-6\\-8&-6\end{array}\right]$$ This must mean that a should equal 3d, b should equal d, c should equal d, without the negatives, but I do not understand why.