Matrices: Transposition, Determinants, and Augmentation

Matrices: Transposition, Determinants, and Augmentation Video

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, 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 3×3 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.

Transposing Matrices

One way matrices can be manipulated is to transpose them. The transpose of a matrix is denoted by aT. 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:

Example #1

Here’s [A]:

\[
\left[ A \right] =
\begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix}
\]

If we want to find the transpose of matrix A here’s how we’d do that. The transpose of matrix A is also going to be a 2×2 matrix. So we start with the first element—that one’s going to stay the same. Then, from there all we’re going to do is flip our elements over the diagonal.

With 2 and 3, we’re going to flip them (2 becomes 3, and 3 becomes 2). Then, 4 is going to stay the same.

\[
\left[ A \right]^T =
\begin{bmatrix}
1 & 3 \\
2 & 4
\end{bmatrix}
\]

That’s our transposed [A]!

Example #2

Now if we want to transpose [B], we’ll do something similar.

\[
\left[ B \right] =
\begin{bmatrix}
-1 & 3 & -5 \\
-4 & 7 & 1 \\
0 & 2 & 9
\end{bmatrix}
\]

So the transpose of matrix B is going to be another 3×3 matrix. Remember the first element stays the same, -1. Then the next diagonal flips, so 3 and -4 are going to flip: 3 becomes -4, and -4 becomes 3.

Now for our main diagonal we’re going to do the same thing; our three elements are going to flip. So -5 is going to go down to where the 0 is—so -5 becomes 0, 7 stays the same, and 0 becomes -5.

One more time let’s now flip these two elements; 1 becomes 2, 2 becomes 1, and 9 stays the same.

\[
\left[ B \right]^T =
\begin{bmatrix}
-1 & -4 & 0 \\
3 & 7 & 2 \\
-5 & 1 & 9
\end{bmatrix}
\]

That’s our transposed [B]!

Example #3

Here’s [C]:

\[
\left[ C \right] =
\begin{bmatrix}
9 & 8 & 7 \\
4 & 5 & 6
\end{bmatrix}
\]

To find the transpose of [C], it’s going to look just a teeny bit different from what we did last time.

Remember we want our first row to be our first column now. So instead of 9, 8, and 7 being a row, it’s going to be our column. Then, our second row (4, 5, 6) is going to become our second column.

\[
\left[ C \right]^T =
\begin{bmatrix}
9 & 4 \\
8 & 5 \\
7 & 6
\end{bmatrix}
\]

That’s all there is to it!

Example #4: Equal Matrix

Sometimes, transposing a matrix yields an equal matrix.

Consider [M] and [N] here:

\[
\left[ M \right] =
\begin{bmatrix}
3 & 7 \\
7 & 2
\end{bmatrix}
\quad \text{and} \quad
\left[ M \right]^T =
\begin{bmatrix}
3 & 7 \\
7 & 2
\end{bmatrix}
\]
\[
\left[ N \right] =
\begin{bmatrix}
1 & 3 & 2 \\
3 & 5 & 6 \\
2 & 6 & 8
\end{bmatrix}
\quad \text{and} \quad
\left[ N \right]^T =
\begin{bmatrix}
1 & 3 & 2 \\
3 & 5 & 6 \\
2 & 6 & 8
\end{bmatrix}
\]

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

Calculating Determinants

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 as det(A) or |A|.

Example #1

For a 2×2 matrix, the calculation of the determinant is \(ad-bc\).

For example, let’s look at [A].

If \( \left[ A \right]^T = \begin{bmatrix}1 & 3 \\ 2 & 4\end{bmatrix} \), then \(det(A)\)\( = (1)(4) – (3)(2)\)\( = 4 – 6 = -2\).

So the determinant of [A] is -2.

Example #2

For a 3×3 matrix, the calculation of the determinant becomes significantly larger and involves determinants of some of the 2×2 matrices within the larger matrix.

\[
\begin{bmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{bmatrix}
\]

To find the determinant of a 3×3 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 |e, f, h, i| because those are the four elements not in a’s row or column.

\[
\left|
\begin{array}{cc}
a & f \\
h & i
\end{array}
\right|
\]

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 |d, f, g, i| because those are the four elements not included in b’s row or column.

\[
\left|
\begin{array}{ccc}
a & b & c \\
d & e & f \\
g & h & i
\end{array}
\right|
=
a \left|
\begin{array}{cc}
e & f \\
h & i
\end{array}
\right|

b \left|
\begin{array}{cc}
d & f \\
g & i
\end{array}
\right|
\]

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 |d, e, g, h|.

\[
a \left|
\begin{array}{cc}
e & f \\
h & i
\end{array}
\right|

b \left|
\begin{array}{cc}
d & f \\
g & i
\end{array}
\right|
+
c \left|
\begin{array}{cc}
d & e \\
g & h
\end{array}
\right|
\]

Now we are able to simplify this even further since we already know how to find the determinants of 2×2 matrices, so our final equation will look like this:

\(a(ei-fh)-b(di-fg)\)\(+c(dh-eg)\)

 
Plugging your numbers into this equation would then give you the determinant of this 3×3 matrix.

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.

Augmenting Matrices

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.

\[
\left[ A \right] =
\begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix}
\quad
\left[ C \right] =
\begin{bmatrix}
7 \\
8
\end{bmatrix}
\quad
\left[ D \right] =
\begin{bmatrix}
-3 & -4 \\
6 & 2
\end{bmatrix}
\]

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

\[
\left[ A \mid C \right] =
\begin{bmatrix}
1 & 2 & 7 \\
3 & 4 & 8
\end{bmatrix}
\]

Now let’s augment matrix A and matrix D. Remember, start with A and just copy it down. Then we simply add D to the end of A.

\[
\left[ A \mid D \right] =
\begin{bmatrix}
1 & 2 & -3 & -4 \\
3 & 4 & 6 & 2
\end{bmatrix}
\]

And that’s it! That’s how you augment matrices.

Thanks for watching, and happy studying!

Matrix Practice Questions

Question #1:

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

\(\begin{bmatrix} 4&-1&4\\ 5&2&-7\\ 6&9&1 \end{bmatrix}\)
\(\begin{bmatrix} 3&-1&3\\ 5&1&-7\\ 2&9&1 \end{bmatrix}\)
\(\begin{bmatrix} 4&-1&4\\ 1&2&-7\\ 6&0&1 \end{bmatrix}\)
\(\begin{bmatrix} 4&-1&4\\ 0&2&-7\\ 0&0&1 \end{bmatrix}\)
Answer:

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?

\(\begin{bmatrix} 4&6&2\\ 6&5&3\\ 2&3&7 \end{bmatrix}\)
\(\begin{bmatrix} 3&5&7\\ 2&3&3\\ 5&4&3 \end{bmatrix}\)
\(\begin{bmatrix} 0&0&1\\ 1&3&0\\ 5&4&3 \end{bmatrix}\)
\(\begin{bmatrix} 3&0&0\\ 1&7&8\\ 3&2&6 \end{bmatrix}\)
Answer:

\(\begin{bmatrix}
\bf4&6&2\\
6&\bf5&3\\
2&3&\bf7
\end{bmatrix}\)

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:
\(\begin{bmatrix}
1&5\\
2&7
\end{bmatrix}\)

\(-5\)
\(-4\)
\(-3\)
\(-2\)
Answer:

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 notation for _________________.

row reduction
determinants
transposition
augmentation
Answer:

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] \(\left[\begin{matrix} 2&1\\ 3&7\end{matrix} \left|\,\begin{matrix}0\\5\end{matrix}\right.\right]\)
Answer:

Augmenting matrices occurs when the columns of two separate matrices are combined into a single matrix that can be manipulated. The number of rows must be the same in order for the matrices to be augmented.

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by Mometrix Test Preparation | Last Updated: April 3, 2025