When Is AB Not Invertible? Finding The Value Of M

In linear algebra, the concept of matrix inverses plays a crucial role in solving systems of linear equations and understanding linear transformations. A matrix has an inverse if and only if its determinant is non-zero. When we consider the product of two matrices, $AB$, the invertibility of the resulting matrix is closely tied to the invertibility of the individual matrices $A$ and $B$. In this article, we delve into the condition under which the product of two matrices, specifically $AB$, does not have an inverse. We will explore this concept through a detailed example where $A=\begin{pmatrix} m & 3 \\ 2 & 1 \end{pmatrix}$ and $B=\begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}$. Our main objective is to find the value(s) of $m$ for which $AB$ is not invertible. This involves understanding how the determinant of $AB$ relates to the determinants of $A$ and $B$, and ultimately, how the value of $m$ affects the invertibility of the product matrix. This exploration will not only enhance our understanding of matrix algebra but also provide a practical approach to solving related problems.

Condition for Non-Invertibility

A matrix is invertible if and only if its determinant is non-zero. Conversely, a matrix is non-invertible (or singular) if its determinant is zero. For the product of two matrices $AB$, the determinant of the product is the product of the determinants, i.e., $\det(AB) = \det(A) \cdot \det(B)$. Therefore, $AB$ is non-invertible if and only if $\det(AB) = 0$, which occurs if either $\det(A) = 0$ or $\det(B) = 0$ (or both). Understanding this principle is fundamental to solving the problem at hand. To determine when $AB$ does not have an inverse, we need to calculate the determinants of matrices $A$ and $B$ and analyze the conditions under which their product becomes zero. This involves algebraic manipulation and a clear understanding of how determinants are computed for 2x2 matrices. We will apply this knowledge to the given matrices $A$ and $B$ to find the specific value(s) of $m$ that make $AB$ non-invertible.

Detailed Calculation

Given the matrices $A=\begin{pmatrix} m & 3 \\ 2 & 1 \end{pmatrix}$ and $B=\begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}$, we first calculate the determinant of $B$. The determinant of a 2x2 matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is given by $ad - bc$. Thus,

$\det(B) = (1)(4) - (3)(2) = 4 - 6 = -2$.

Since $\det(B) = -2 \neq 0$, matrix $B$ is invertible. Now, we compute the determinant of matrix $A$:

$\det(A) = (m)(1) - (3)(2) = m - 6$.

For $AB$ to be non-invertible, we require $\det(AB) = \det(A) \cdot \det(B) = 0$. Since $\det(B) = -2$, we need $\det(A) = 0$ for the product to be zero. Therefore,

$m - 6 = 0$,

which gives us $m = 6$. This result is crucial because it pinpoints the exact value of $m$ that makes matrix $A$ singular, and consequently, makes the product $AB$ non-invertible. The calculation clearly demonstrates the relationship between the determinant of a matrix and its invertibility, and how this relationship extends to the product of matrices.

Step-by-Step Solution

To find the value of $m$ for which $AB$ does not have an inverse, we follow these steps:

1. Calculate the determinant of matrix B: $\det(B) = (1)(4) - (3)(2) = 4 - 6 = -2$.
2. Calculate the determinant of matrix A: $\det(A) = (m)(1) - (3)(2) = m - 6$.
3. Set the determinant of AB to zero: Since $\det(AB) = \det(A) \cdot \det(B)$, we have $\det(A) \cdot \det(B) = 0$.
4. Solve for m: Given $\det(B) = -2$, we need $\det(A) = 0$. Thus, $m - 6 = 0$, which implies $m = 6$.

This step-by-step solution provides a clear and concise method for determining the value of $m$. By breaking down the problem into smaller, manageable steps, we can easily see how each calculation contributes to the final answer. This approach is particularly helpful for students learning linear algebra, as it reinforces the fundamental concepts and techniques involved in matrix operations and determinant calculations.

Verification and Implications

To verify our result, we can substitute $m = 6$ back into matrix $A$ and compute $AB$. If $m = 6$, then $A = \begin{pmatrix} 6 & 3 \\ 2 & 1 \end{pmatrix}$. Now, let's compute the product $AB$:

$AB = \begin{pmatrix} 6 & 3 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} (6)(1) + (3)(2) & (6)(3) + (3)(4) \\ (2)(1) + (1)(2) & (2)(3) + (1)(4) \end{pmatrix} = \begin{pmatrix} 12 & 30 \\ 4 & 10 \end{pmatrix}$.

Now, let's calculate the determinant of the resulting matrix $AB$:

$\det(AB) = (12)(10) - (30)(4) = 120 - 120 = 0$.

Since $\det(AB) = 0$, the matrix $AB$ is indeed non-invertible when $m = 6$. This verification step is crucial as it confirms the correctness of our solution. It also highlights the practical implications of understanding matrix invertibility, particularly in fields such as computer graphics, engineering, and economics, where matrices are used to model and solve real-world problems.

Conclusion

In conclusion, we have successfully determined the value of $m$ for which the matrix product $AB$ does not have an inverse. By understanding the relationship between the determinant of a matrix and its invertibility, we found that $AB$ is non-invertible when $m = 6$. This problem illustrates an important concept in linear algebra, emphasizing the significance of determinants in determining the invertibility of matrices and their products. The step-by-step approach used in this solution provides a clear methodology for tackling similar problems, reinforcing the fundamental principles of matrix algebra. Furthermore, the verification step ensures the accuracy of our solution, highlighting the importance of thoroughness in mathematical problem-solving. Understanding these concepts is crucial for anyone working with matrices in various fields, as it allows for the effective manipulation and application of matrices in solving complex problems.