# Matrices – Transposition, Determinants, and Augmentation

Hello and welcome to this video about matrix manipulation! In this video, we will cover:

• Transposing matrices
• Finding determinants of matrices
• Augmenting matrices

A quick preface before we get started. Generally speaking, the larger the matrices you are working with are, the more tedious the work becomes. On one hand, it can be great practice to get a better conceptual understanding about what matrices do, since they are part of an interesting branch of mathematics. But on the other hand, once the size exceeds 3 x 3 or so, the amount of time and paper required for some of the operations by hand is quite large. Technology of some sort is definitely recommended.

One way matrices can be manipulated is to transpose them.
The transpose of a matrix is denoted by a T. So the transpose of [A] is [A]T.
To transpose a matrix, reflect all the elements over the main diagonal. In other words, row 1 of the original becomes column 1 of the transposed matrix, row 2 of the original becomes column 2 of the transposed matrix and so on.

You will transpose most often with square matrices. Let’s look at a couple of examples:

Non-square matrices can also be transposed:

Sometimes, transposing a matrix yields an equal matrix. Consider [M] and [N] here:

[M] and [N] are symmetric. [M]=[M]T and [N]=[N]T.

Another thing we can do with matrices is calculate their determinants. Determinants (which are always a single number) will only be found when dealing with square matrices.
The determinant of [A] can be denoted det(A) or |A|.

To find the determinant of a 3 x 3 matrix, we will start by multiplying the elements on the top row by the determinants of the 2×2 matrices created by using the four elements not in the original element’s row or column. So if we start with a, we will then multiply it by the determinant of [e, f, h, i] because those are the four elements not in a’s row or column. We will then alternate our signs, so since this first part is positive, we will subtract the second part of our equation. This time we will start with b. We will multiply b by the determinant of [d, f, h, i] because those are the four elements not included in b’s row or column. Finally, we will add the last part of our equation. Remember, we do this because we are alternating signs: positive, negative, positive. So we have c times the determinant of [d, e, g, h]. Now we are able to simplify this even further since we already know how to find the determinants of 2 x 2 matrices, so our final equation will look like this.

Determinants can be found for any size square matrix, but this is an operation where access to technology can save much time. Usually, we are not as concerned with the calculation of the determinant as we are with what the determinant tells us about the matrix.

Sometimes it will be necessary to augment matrices. Augmenting a matrix means to combine the columns of two separate matrices into a single matrix that can be manipulated as one. Note that the number of rows in both matrices must match in order to augment them.

The notation for augmentation is a bar. For example, so if we want to find the augment of matrix A and matrix C we would do this: Remember the bar means to augment. So we start with matrix A and we just copy it down: 1, 2, 3, 4. Then since we’re augmenting it with matrix C, we go up here and we see that we have 7, 8. All we’re going to do is add that to the end of our other matrix: 7, 8. That gives us our augmented matrix.

Now let’s augment matrix A and matrix D. So A augment D looks like this: remember, start with A and just copy it down. Then we simply add D to the end of A. So negative 3, negative 4, 6, 2
And that’s it! That’s how you augment matrices.

Thanks for watching, and happy studying!

## Practice Questions

Question #1:

Find the transpose of the matrix below.
$$\begin{bmatrix} 4&5&6\\ -1&2&9\\ 4&-7&1 \end{bmatrix}$$

$$\begin{bmatrix} 4&-1&4\\ 5&2&-7\\ 6&9&1 \end{bmatrix}$$
$$\begin{bmatrix} 3&-1&3\\ 5&1&-7\\ 2&9&1 \end{bmatrix}$$
$$\begin{bmatrix} 4&-1&4\\ 1&2&-7\\ 6&0&1 \end{bmatrix}$$
$$\begin{bmatrix} 4&-1&4\\ 0&2&-7\\ 0&0&1 \end{bmatrix}$$

To transpose the matrix, elements are reflected over the main diagonal. This means that the $$5$$ and the $$-1$$ will change places, the $$6$$ and the $$4$$ will change places, and the $$9$$ and the $$-7$$ will change places.
$$\begin{bmatrix} \bf4&5&6\\ -1&\bf2&9\\ 4&-7&\bf1 \end{bmatrix}$$ becomes
$$\begin{bmatrix} 4&-1&4\\ 5&2&-7\\ 6&9&1 \end{bmatrix}$$

Question #2:

Which matrix will yield an equal matrix when transposed?

$$\begin{bmatrix} 4&6&2\\ 6&5&3\\ 2&3&7 \end{bmatrix}$$
$$\begin{bmatrix} 3&5&7\\ 2&3&3\\ 5&4&3 \end{bmatrix}$$
$$\begin{bmatrix} 0&0&1\\ 1&3&0\\ 5&4&3 \end{bmatrix}$$
$$\begin{bmatrix} 3&0&0\\ 1&7&8\\ 3&2&6 \end{bmatrix}$$
$$\begin{bmatrix} \bf4&6&2\\ 6&\bf5&3\\ 2&3&\bf7 \end{bmatrix}$$

The original matrix is symmetric. This means that when the elements are reflected over the main diagonal, the new matrix is identical to the original matrix.

Question #3:

Calculate the determinant of the following matrix:
$$\begin{bmatrix} 1&5\\ 2&7 \end{bmatrix}$$

$$-5$$
$$-4$$
$$-3$$
$$-2$$

The formula $$ad-bc$$ can be used to calculate the determinant of the matrix. In the original matrix, $$a=1,b=5,c=2,$$ and $$d=7$$. When these values are plugged into the formula, $$ad-bc$$ becomes $$(1)(7)-(5)(2)$$, which simplifies to $$7-10$$, or $$-3$$.

Question #4:

$$\begin{bmatrix} R|Y \end{bmatrix}$$
The bar between R and Y indicates the notaiton for _________________.

row reduction

determinants

transposition

augmentation

The bar symbol indicates augmentation. Augmenting a matrix means combining the columns into a single matrix that can be manipulated as one.

Question #5:

Augment matrix E and matrix K.
[E]
$$\begin{bmatrix} 3&5\\ 2&1 \end{bmatrix}$$

[K]
$$\begin{bmatrix} 7\\ 0\end{bmatrix}$$

[E|K]
$$\left[\begin{matrix} 4&5\\ 2&1\end{matrix} \left|\,\begin{matrix}7\\1\end{matrix}\right.\right]$$

[E|K]
$$\left[\begin{matrix} 3&5\\ 2&1\end{matrix} \left|\,\begin{matrix}7\\0\end{matrix}\right.\right]$$

[E|K]
$$\left[\begin{matrix} 7&3\\ 2&1\end{matrix} \left|\,\begin{matrix}5\\0\end{matrix}\right.\right]$$

[E|K]
$$\left[\begin{matrix} 2&1\\ 3&7\end{matrix} \left|\,\begin{matrix}0\\5\end{matrix}\right.\right]$$