Matrices With Inverses How To Identify Them

Hey there, matrix enthusiasts! Ever wondered if you can 'undo' a matrix? That's where the concept of an inverse matrix comes into play. Not all matrices have inverses, and figuring out which ones do is a fundamental skill in linear algebra. So, let's dive into the question: Which of the following matrices has an inverse? We'll explore the options and break down the criteria for a matrix to be invertible.

Understanding Invertible Matrices

Before we jump into the specific matrices, let's establish what it means for a matrix to have an inverse. Simply put, an invertible matrix, also known as a non-singular matrix, is a square matrix that, when multiplied by another matrix (its inverse), results in the identity matrix. Think of it like multiplying a number by its reciprocal – you get 1. The identity matrix is the matrix equivalent of '1' in the world of matrices.

Key Criteria for Invertibility

There are a couple of key things to remember when determining if a matrix has an inverse:

1. Square Matrices Only: The most crucial requirement is that the matrix must be square, meaning it has the same number of rows and columns. This is because the inverse, when multiplied, needs to result in the identity matrix, which is always square.
2. Non-Zero Determinant: A matrix is invertible if and only if its determinant is not equal to zero. The determinant is a special value that can be calculated from the elements of a square matrix. If the determinant is zero, the matrix is called singular and does not have an inverse. The determinant gives us critical information about the matrix's properties and whether it can be inverted.

Why is the Determinant Important?

The determinant is a scalar value that encapsulates crucial information about a matrix. For a 2x2 matrix, the determinant can be easily calculated, and for larger matrices, there are various methods like cofactor expansion. The determinant's non-zero value indicates that the matrix represents a transformation that doesn't collapse space into a lower dimension, ensuring the matrix can be 'undone.' In simpler terms, a non-zero determinant signifies the matrix's invertibility.

Now, with these criteria in mind, let's examine the matrices presented in the question.

Analyzing the Matrices

We have four matrices to consider:

1. $\left[\begin{array}{c}-6 \ 3\end{array}\right]$
2. $\left[\begin{array}{lll}4 & -2 & 1\end{array}\right]$
3. $\left[\begin{array}{cc}6 & -9 \ 2 & 1\end{array}\right]$
4. $\left[\begin{array}{ccc}-3 & 5 & 1 & 0 & -2 & 4 & 7\end{array}\right]$

Let's go through them one by one, checking for the square matrix requirement and then the determinant.

Matrix 1: $\left[\begin{array}{c}-6 \ 3\end{array}\right]$

This matrix is a 2x1 matrix, meaning it has 2 rows and 1 column. It's not a square matrix, so right away, we know it does not have an inverse. Remember, only square matrices can be invertible. This matrix represents a column vector in a two-dimensional space, and it cannot be 'undone' in the same way a square matrix transformation can.

Matrix 2: $\left[\begin{array}{lll}4 & -2 & 1\end{array}\right]$

This matrix is a 1x3 matrix (1 row, 3 columns). Again, this is not a square matrix. Therefore, it cannot have an inverse. This matrix represents a row vector in a three-dimensional space, and similar to the previous case, it lacks the necessary dimensions to be invertible.

Matrix 3: $\left[\begin{array}{cc}6 & -9 \ 2 & 1\end{array}\right]$

Ah, finally! This is a 2x2 matrix, meaning it is a square matrix. So, we're one step closer. Now, we need to calculate the determinant to see if it's non-zero. For a 2x2 matrix $\left[\begin{array}{cc}a & b \ c & d\end{array}\right]$, the determinant is calculated as ad - bc.

In this case, our determinant is (6 * 1) - (-9 * 2) = 6 + 18 = 24. Since the determinant is 24, which is not equal to zero, this matrix does have an inverse! This matrix represents a transformation in a two-dimensional space that can be reversed, making it invertible.

Matrix 4: `$\left[\begin{array}{ccc}-3 & 5 & 1

& 0 & -2 & 4 & 7\end{array}\right]