Matrix Equation: Solve For X And Y

In the fascinating world of mathematics, solving systems of linear equations is a fundamental skill, and matrices offer an elegant and powerful way to tackle these problems. Today, we're diving into a specific matrix equation: $\left[\begin{array}{ll} 2 & 1 \ 3 & 2 \end{array}\right]\left[\begin{array}{l} x \ y \end{array}\right]=\left[\begin{array}{l} 3 \ 3 \end{array}\right]$. This equation, at its heart, represents a system of two linear equations with two variables, $x$ and $y$. The beauty of using matrices is that it transforms a sometimes-tedious algebraic process into a more structured and often quicker computational one. We'll unravel this equation step-by-step, showing you how to find the values of $x$ and $y$ that satisfy this matrix relationship. This method is not just a mathematical curiosity; it's a cornerstone of many fields, from computer graphics and engineering to economics and data science. Understanding how to solve such equations is like unlocking a secret code that allows us to interpret and manipulate relationships between different quantities. So, get ready to explore the world of matrix algebra and discover how this particular equation yields its unique solution.

Understanding the Matrix Equation

Before we jump into solving, let's break down what our equation, $\left[\begin{array}{ll} 2 & 1 \ 3 & 2 \end{array}\right]\left[\begin{array}{l} x \ y \end{array}\right]=\left[\begin{array}{l} 3 \ 3 \end{array}\right]$, actually means. On the left side, we have the multiplication of two matrices. The first matrix, $\left[\begin{array}{ll} 2 & 1 \ 3 & 2 \end{array}\right]$, is a 2x2 matrix (2 rows, 2 columns). The second matrix, $\left[\begin{array}{l} x \ y \end{array}\right]$, is a 2x1 matrix (2 rows, 1 column), often called a column vector, representing our unknown variables. The result of this multiplication is another 2x1 matrix, $\left[\begin{array}{l} 3 \ 3 \end{array}\right]$, which is a column vector containing the constants. This matrix multiplication is performed by taking the dot product of each row of the first matrix with the column vector. Specifically, the first row of the first matrix (2, 1) multiplied by the column vector $\left[\begin{array}{l} x \ y \end{array}\right]$ gives us $2x + 1y$. Similarly, the second row of the first matrix (3, 2) multiplied by the same column vector yields $3x + 2y$. Therefore, the matrix equation can be translated back into a system of linear equations:

$2x + y = 3$ $3x + 2y = 3$

This is the familiar form of a system of linear equations that many of us learned in algebra. The power of the matrix form is its conciseness and its suitability for computational methods. It allows us to represent complex systems compactly and to apply powerful mathematical tools like inverse matrices or Gaussian elimination to find the solution.

Method 1: Using the Inverse Matrix

One of the most direct ways to solve a matrix equation of the form $AX = B$, where $A$ is a square matrix, $X$ is the vector of unknowns, and $B$ is the constant vector, is by using the inverse of matrix $A$. If the inverse of $A$, denoted as $A^{-1}$, exists, then we can multiply both sides of the equation by $A^{-1}$ on the left: $A^{-1}(AX) = A^{-1}B$. Since matrix multiplication is associative, this simplifies to $(A^{-1}A)X = A^{-1}B$. And because $A^{-1}A$ is the identity matrix ($I$), we get $IX = A^{-1}B$, which further simplifies to $X = A^{-1}B$. Thus, if we can find the inverse of matrix $A$, we can directly compute the solution vector $X$.

For our specific equation, $A = \left[\begin{array}{ll} 2 & 1 \ 3 & 2 \end{array}\right]$, $X = \left[\begin{array}{l} x \ y \end{array}\right]$, and $B = \left[\begin{array}{l} 3 \ 3 \end{array}\right]$. First, let's find the determinant of matrix $A$. For a 2x2 matrix $\left[\begin{array}{ll} a & b \ c & d \end{array}\right]$, the determinant is $ad - bc$. In our case, the determinant is $(2)(2) - (1)(3) = 4 - 3 = 1$. Since the determinant is non-zero (it's 1!), the inverse matrix $A^{-1}$ exists.

To find the inverse of a 2x2 matrix $\left[\begin{array}{ll} a & b \ c & d \end{array}\right]$, we use the formula: $A^{-1} = \frac{1}{ad-bc}\left[\begin{array}{cc} d & -b \ -c & a \end{array}\right]$. Applying this to our matrix $A$:

$A^{-1} = \frac{1}{1}\left[\begin{array}{cc} 2 & -1 \ -3 & 2 \end{array}\right] = \left[\begin{array}{cc} 2 & -1 \ -3 & 2 \end{array}\right]$.

Now, we can find the solution vector $X$ by multiplying $A^{-1}$ by $B$:

$X = A^{-1}B = \left[\begin{array}{cc} 2 & -1 \ -3 & 2 \end{array}\right]\left[\begin{array}{l} 3 \ 3 \end{array}\right]$.

Performing this matrix multiplication:

$x = (2)(3) + (-1)(3) = 6 - 3 = 3$ $y = (-3)(3) + (2)(3) = -9 + 6 = -3$

So, the solution is $x=3$ and $y=-3$. This method is incredibly efficient, especially for larger systems where direct algebraic manipulation becomes cumbersome.

Method 2: Using Cramer's Rule

Another systematic approach to solve systems of linear equations, particularly when using matrices, is Cramer's Rule. This rule provides a direct formula for each variable's solution using determinants. It's especially handy when you need to find a specific variable without necessarily solving for all of them, although for our case, we need both $x$ and $y$. Cramer's Rule is applicable to systems where the number of equations equals the number of variables, and the determinant of the coefficient matrix is non-zero (which we've already established for our problem).

Recall our system of equations derived from the matrix:

$2x + y = 3$ $3x + 2y = 3$

Let $A$ be the coefficient matrix: $A = \left[\begin{array}{ll} 2 & 1 \ 3 & 2 \end{array}\right]$. The determinant of $A$, denoted as $\det(A)$ or $|A|$, is $1$, as calculated previously. Now, to find $x$, we create a new matrix, let's call it $A_x$, by replacing the first column of $A$ (the coefficients of $x$) with the constant vector $B = \left[\begin{array}{l} 3 \ 3 \end{array}\right]$. So, $A_x = \left[\begin{array}{ll} 3 & 1 \ 3 & 2 \end{array}\right]$. The determinant of $A_x$ is $\det(A_x) = (3)(2) - (1)(3) = 6 - 3 = 3$.

According to Cramer's Rule, the solution for $x$ is given by the ratio of the determinant of $A_x$ to the determinant of $A$:

$x = \frac{\det(A_x)}{\det(A)} = \frac{3}{1} = 3$.

Similarly, to find $y$, we create another matrix, $A_y$, by replacing the second column of $A$ (the coefficients of $y$) with the constant vector $B$. Thus, $A_y = \left[\begin{array}{ll} 2 & 3 \ 3 & 3 \end{array}\right]$. The determinant of $A_y$ is $\det(A_y) = (2)(3) - (3)(3) = 6 - 9 = -3$.

Using Cramer's Rule for $y$:

$y = \frac{\det(A_y)}{\det(A)} = \frac{-3}{1} = -3$.

This yields the same solution we found using the inverse matrix method: $x=3$ and $y=-3$. Cramer's Rule is a powerful tool, especially in theoretical contexts or when dealing with systems where symbolic solutions are required. It beautifully illustrates the relationship between the coefficients, constants, and the solutions of a linear system through the elegant concept of determinants.

Method 3: Substitution or Elimination (Algebraic Approach)

While matrices provide a sophisticated way to solve systems of equations, it's always good to remember the more traditional algebraic methods, as they are often the foundation upon which matrix methods are built. For our specific problem, solving the system $2x + y = 3$ and $3x + 2y = 3$ using substitution or elimination is quite straightforward and can serve as a great way to verify our matrix-derived solutions. Let's use the elimination method.

We have the two equations:

1. $2x + y = 3$
2. $3x + 2y = 3$

Our goal is to eliminate one of the variables, either $x$ or $y$, by making their coefficients opposites in the two equations. Let's choose to eliminate $y$. We can multiply the first equation by $-2$ so that the coefficient of $y$ becomes $-2$, which is the opposite of the coefficient of $y$ in the second equation (which is $2$).

Multiplying equation (1) by $-2$: $-2(2x + y) = -2(3)$ $-4x - 2y = -6$ (Equation 3)

Now, we add Equation 3 to Equation 2: $(-4x - 2y) + (3x + 2y) = -6 + 3$ $-4x + 3x - 2y + 2y = -3$ $-x = -3$

Multiplying by $-1$ to solve for $x$: $x = 3$

Now that we have the value of $x$, we can substitute it back into either of the original equations to find $y$. Let's use the first equation ($2x + y = 3$): $2(3) + y = 3$ $6 + y = 3$

Subtract 6 from both sides to solve for $y$: $y = 3 - 6$ $y = -3$

This algebraic method also confirms our solution: $x=3$ and $y=-3$. It's reassuring to see that all three methods—inverse matrix, Cramer's Rule, and elimination—lead to the exact same result. This consistency underscores the robustness of mathematical principles. While matrix methods might seem more complex at first glance, they offer a standardized framework for solving systems, which is invaluable in more advanced mathematical and computational contexts.

Conclusion

We have successfully solved the matrix equation $\left[\begin{array}{ll} 2 & 1 \ 3 & 2 \end{array}\right]\left[\begin{array}{l} x \ y \end{array}\right]=\left[\begin{array}{l} 3 \ 3 \end{array}\right]$ using three distinct methods: the inverse matrix approach, Cramer's Rule, and the traditional algebraic elimination method. Each method, grounded in fundamental mathematical principles, yielded the same unambiguous solution: $x=3$ and $y=-3$. This exploration highlights the versatility and power of linear algebra, demonstrating how matrix operations can elegantly represent and solve systems of equations that are commonplace in various scientific and technological disciplines. Whether you're a student grappling with introductory linear algebra or a professional applying these concepts in real-world problems, mastering these techniques is crucial. The ability to translate problems into matrix form and solve them efficiently can unlock deeper insights and streamline complex analyses.

For further exploration into the world of linear algebra and matrix operations, I highly recommend visiting Khan Academy's Linear Algebra section. It offers a comprehensive and accessible resource for learning and reinforcing these essential mathematical concepts.