Rank Of Matrix A = [[2, 3], [4, 6]]

Hey guys, let's dive into the fascinating world of linear algebra and figure out the rank of a matrix! Today, we're tackling a specific one: $A = \begin{bmatrix} 2 & 3 \\ 4 & 6 \end{bmatrix}$. Understanding the rank of a matrix is super important because it tells us a lot about the matrix's properties, like its invertibility and the dimension of the vector space it spans. So, buckle up, and let's break down how to find the rank of this matrix $A$. We'll explore different methods, from row reduction to looking at determinants of submatrices, and figure out which of the options (A) 3, (B) 2, (C) 1, or (D) 0 is the correct answer. By the end of this, you'll not only know the rank of $A$ but also feel more confident in tackling similar problems yourself. Let's get started!

Understanding Matrix Rank

Alright, first things first, what is the rank of a matrix, anyway? Think of it as the maximum number of linearly independent rows or columns in that matrix. It's a fundamental concept that gives us insights into the 'size' or 'dimensionality' of the linear transformation represented by the matrix. For a matrix $A$, its rank, often denoted as $rank(A)$, is the same whether you count linearly independent rows or columns – pretty neat, right? So, if a matrix has $n$ rows and $m$ columns, its rank can be at most $\min(n, m)$. In our case, matrix $A$ is a 2x2 matrix, meaning it has 2 rows and 2 columns. So, the rank of $A$ can be at most $\min(2, 2) = 2$. This immediately tells us that option (A) 3 is impossible. We're now narrowing down our possibilities to 0, 1, or 2.

There are several ways to determine the rank. One common method is Gaussian elimination, where we transform the matrix into its row echelon form (or reduced row echelon form). The rank is then simply the number of non-zero rows in the echelon form. Another approach involves looking at the determinants of submatrices (minors). The rank is the size of the largest square submatrix that has a non-zero determinant. If the determinant of the matrix itself is non-zero, its rank is equal to its dimension. If it's zero, we look at smaller submatrices. Let's apply these methods to our matrix $A = \begin{bmatrix} 2 & 3 \\ 4 & 6 \end{bmatrix}$. We're going to crunch some numbers and see what we get!

Method 1: Gaussian Elimination (Row Reduction)

This is arguably the most straightforward method for finding the rank, especially for larger matrices. The goal is to use elementary row operations to get the matrix into row echelon form. These operations are: swapping two rows, multiplying a row by a non-zero scalar, or adding a multiple of one row to another row. These operations don't change the rank of the matrix. Let's take our matrix $A = \begin{bmatrix} 2 & 3 \\ 4 & 6 \end{bmatrix}$.

Our first step is to get a '1' in the top-left corner, but we already have a 2 there. That's fine. We want to make the entry below it a zero. We can achieve this by performing the operation: $R_2 \leftarrow R_2 - 2R_1$. This means we take the second row and subtract twice the first row from it.

So, the new second row will be:

$(4 - 2*2, 6 - 2*3) = (4 - 4, 6 - 6) = (0, 0)$.

Our matrix now looks like this:

$A \sim \begin{bmatrix} 2 & 3 \\ 0 & 0 \end{bmatrix}$

This matrix is now in row echelon form. Why? Because all the zero rows (if any) are at the bottom, and the leading non-zero entry (the pivot) of each non-zero row is to the right of the leading entry of the row above it. In this case, we have one non-zero row: $[2 3]$. The second row is entirely zeros.

The rank of a matrix is the number of non-zero rows in its row echelon form. In our transformed matrix $\begin{bmatrix} 2 & 3 \\ 0 & 0 \end{bmatrix}$, we have only one non-zero row. Therefore, the rank of matrix $A$ is 1.

This method is super reliable, and once you get the hang of row operations, it's a breeze. We've found our answer using this technique, but let's explore another method to confirm our results and solidify our understanding. This is why multiple approaches are so valuable in math, guys!

Method 2: Determinants of Submatrices (Minors)

This method involves calculating the determinants of all possible square submatrices. The rank of the matrix is the order (size) of the largest square submatrix that has a non-zero determinant. If the determinant of the entire matrix is non-zero, then the rank is equal to the dimension of the matrix. If it's zero, we check the determinants of all possible submatrices of the next smaller size.

Our matrix is $A = \begin{bmatrix} 2 & 3 \\ 4 & 6 \end{bmatrix}$. It's a 2x2 matrix. So, first, let's calculate the determinant of $A$ itself. The determinant of a 2x2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $ad - bc$.

For our matrix $A$, the determinant is:

$det(A) = (2 \times 6) - (3 \times 4)$

$det(A) = 12 - 12$

$det(A) = 0$

Since the determinant of the entire 2x2 matrix $A$ is 0, its rank is not 2. This means we need to look at smaller submatrices. The next smaller size is 1x1.

A 1x1 submatrix is just a single element of the original matrix. The determinant of a 1x1 matrix $[x]$ is simply $x$. We need to check if any of these 1x1 submatrices have a non-zero determinant (i.e., are non-zero elements).

Let's list the 1x1 submatrices and their determinants:

1. Submatrix $[2]$: Determinant is 2.
2. Submatrix $[3]$: Determinant is 3.
3. Submatrix $[4]$: Determinant is 4.
4. Submatrix $[6]$: Determinant is 6.

We found several 1x1 submatrices with non-zero determinants (2, 3, 4, and 6). The largest size of a square submatrix with a non-zero determinant is 1x1.

Therefore, the rank of matrix $A$ is 1.

This second method confirms our result from Gaussian elimination. It's great to see how different mathematical approaches can lead to the same correct answer. This reinforces our understanding and builds confidence. Whether you prefer row operations or determinant calculations, the rank is consistently 1 for this matrix $A$. Pretty cool, huh?

Why Is the Rank 1?

So, why did we end up with a rank of 1? Let's dig a little deeper into what that means for our specific matrix $A = \begin{bmatrix} 2 & 3 \\ 4 & 6 \end{bmatrix}$. A rank of 1 means that all the rows (and columns) are essentially scalar multiples of a single non-zero vector. In simpler terms, one row (or column) can be obtained by multiplying the other row (or column) by a constant.

Let's look at the rows of $A$: the first row is $[2 3]$ and the second row is $[4 6]$. Do you see the relationship? If you multiply the first row $[2 3]$ by 2, you get $[2 imes 2 2 imes 3] = [4 6]$, which is exactly the second row!

This means the two rows are linearly dependent. They don't provide independent information. In the world of vectors, they point in the same or opposite direction. For our matrix $A$, the vector $[4 6]$ is just $2$ times the vector $[2 3]$. Because one row is just a multiple of the other, they only span a one-dimensional space. Hence, the rank is 1.

Similarly, let's look at the columns. The first column is $\begin{bmatrix} 2 \\ 4 \end{bmatrix}$ and the second column is $\begin{bmatrix} 3 \\ 6 \end{bmatrix}$. Is there a relationship here? Yes! The second column $\begin{bmatrix} 3 \\ 6 \end{bmatrix}$ is $\frac{3}{2}$ times the first column $\begin{bmatrix} 2 \\ 4 \end{bmatrix}$ (since $\frac{3}{2} imes 2 = 3$ and $\frac{3}{2} imes 4 = 6$). So, the columns are also linearly dependent.

The rank of a matrix is the dimension of the column space (the span of its column vectors) and also the dimension of the row space (the span of its row vectors). Since our rows and columns are linearly dependent and are multiples of each other, they only span a space of dimension 1. This is precisely why the rank is 1.

Understanding linear dependence is key here. If the rows (or columns) were completely unrelated, like $[1 0]$ and $[0 1]$, they would be linearly independent, and the rank would be higher. But in our case, the strong relationship between the rows (and columns) limits the dimension of the space they can span, resulting in a rank of 1.

Conclusion: What is the Rank(A)?

We've explored two solid methods – Gaussian elimination and checking determinants of submatrices – and they both consistently point to the same answer. For the matrix $A = \begin{bmatrix} 2 & 3 \\ 4 & 6 \end{bmatrix}$, we found that:

1. Using Gaussian elimination, we reduced $A$ to row echelon form $\begin{bmatrix} 2 & 3 \\ 0 & 0 \end{bmatrix}$, which has one non-zero row. Thus, $rank(A) = 1$.
2. Calculating the determinant of $A$ gave us 0, indicating the rank is less than 2. Then, checking the 1x1 submatrices revealed non-zero elements, meaning the largest submatrix with a non-zero determinant is of size 1x1. Thus, $rank(A) = 1$.

Furthermore, we observed that the rows (and columns) of $A$ are linearly dependent, with the second row being twice the first row, and the second column being 3/2 times the first column. This linear dependence is the fundamental reason why the rank is 1 – it signifies that the matrix spans a space of dimension 1.

So, to answer the question: What is rank(A)?

Looking at our options:

A) 3 B) 2 C) 1 D) 0

The correct answer is (C) 1.

I hope this breakdown was super clear and helpful, guys! Understanding matrix rank is a cornerstone in linear algebra, and applying these methods should make you feel much more comfortable tackling similar problems. Keep practicing, and you'll master it in no time! Let me know if you have any other math puzzles you'd like to solve together.