# Elementary Matrix Multiplication and Gauss Elimination

I have the following matrix: $$A = \begin{bmatrix} 2 & 1 & 1 & 0 \\ 4 & 3 & 2 & 1 \\ 8 & 7 & 8 & 5 \\ 6 & 7 & 9 & 8 \\ \end{bmatrix}$$

and I want to apply Gauss elemination through a series of elementary matrix multiplications.

To cancel out the numbers under the first diagonal element of $$A$$, $$E_1 = \begin{bmatrix} 1 & 0 & 0 & 0 \\ -2 & 1 & 0 & 0 \\ -4 & 0 & 1 & 0 \\ -3 & 0 & 0 & 1 \\ \end{bmatrix}$$

Applying $$E_1$$, it works as expected

$$A^1 = E_1A = \begin{bmatrix} 2 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 3 & 4 & 5 \\ 0 & 4 & 6 & 8 \\ \end{bmatrix}$$

For the second column, however, starting with $$E_2 = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$$, how do I eliminate numbers under the second diagonal element of $$A$$?

If I want to apply on $$E_2$$ the row operations that would make the numbers under the second diagonal element of $$A$$ zero, then I would get for the third row of $$E_2$$, assuming it starts as an identity matrix, $$I_n$$ wouldn't I get:

$$E_2 = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -7 & 0 & 1 & 0 \\ -7 & 0 & 0 & 1 \\ \end{bmatrix}$$

The row operations in the context of A are apply $$-7R_1 + R_3$$ and $$-7R_1 + R_4$$ to cancel out the numbers under the second diagonal element of $$A$$ - so, the above shows I applied those to $$I_n$$

Doing the multiplication though, I get

$$A^2 = E_2E_1A = \begin{bmatrix} 2 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ -14 & -4 & -3 & 5 \\ -14 & -3 & -2 & 8 \\ \end{bmatrix}$$

What am I doing wrong?

Edit: Calculation error in $$A^1$$ at position $$(4,3)$$. It should be $$9 + (-3)\times 1 = 6 \neq 5$$

First of all, you are applying $$E_2$$ to $$A^1$$. Hence, while designing $$E_2$$, you should be looking at $$A^1$$ (and not at $$A$$). While looking at $$A_1$$, you know that you want to cancel $$3$$ and $$4$$ at position $$(3,2)$$ and $$(4,2)$$ respectively (and not $$-7$$ of $$A$$).

You can try $$E_2$$ as follows: $$E_2 = \left[ \begin{array}{cccc} 1& 0 & 0&0\\ 0&1&0&0\\ 0&-3&1&0\\ 0&-4&0&1 \end{array} \right]$$

The reason to select this $$E_2$$ is that you want to operate $$-3R_2 + R_3$$ at $$R_3$$ and $$-4R_2 + R_4$$ at $$R_4$$. Hence, the third row of $$E_2$$ should have $$-3$$ and $$+1$$ in the second and third column respectively. Similarly, the fourth row of $$E_2$$ should have $$-4$$ and $$+1$$ in the second and fourth column.

In addition to the reason mentioned in the beginning, your $$E_2$$ also doesn't work because $$-7$$ in your $$E_2$$ operates on first column. If you want to limit the operation on the second column, the first column of $$E_2$$ should remain $$[1\ 0\ 0\ 0]$$.

At the end, you should get $$A^2 = E_2 E_1 A = \left[ \begin{array}{cccc} 2& 1 & 1&0\\ 0&1&0&1\\ 0&0&4&2\\ 0&0&6&4 \end{array} \right]$$

Now, you can design $$E_3$$ by using the operation $$-\frac{3}{2}R_3 + R_4$$ at $$R_4$$ of $$A^2$$.

• If I was doing this without elementary matrices and was applying direct row operations on $A$ , wouldn't $-7R_1 + R_3$ and $-7R_1 + R_4$ work? That is, if I cleared out the sub diagonal elements in column 1?
– Art
Feb 9, 2019 at 3:04
• So, it is important to check if you are operating on $A$, or $A^1$. Also, once you have cleared the first column of $A$, you should be avoiding to use $R_1$ operations. All $R_1$ operations will add some numbers in the first column (like $-7$) and if you try to clear them, you will get the original matrix. You can try for your understanding. Feb 9, 2019 at 3:08