Problem 164 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Given \(A=\left[\begin{array}{ccc}1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1\end{array}\right]\) and \(B=\left[\begin{array}{ccc}3 & -1 & 2 \\ 4 & 2 & 5 \\\ 2 & 0 & 3\end{array}\right]\). Find the matrix \(C\) such that \(A+2 C=B\).

Step-by-Step Solution

Answer

The solution is the matrix \( C \) obtained after performing the series of operations specified in the steps above. This matrix is obtained by subtracting each element of matrix \( A \) from the corresponding element of matrix \( B \), and then halving each resulting element.

Step 1: Understand the given equation

The given equation is \( A+2C=B \). We are to solve for matrix \( C \). This is similar to a basic algebraic equation, but with matrices instead of plain numbers.

Step 2: Rearrange the equation to solve for \( C \)

In order to isolate \( C \), subtract \( A \) from both sides of the equation to get \( 2C=B-A \). Then, divide both sides by 2 to isolate \( C \), giving us \( C=\frac{1}{2}(B-A) \).

Step 3: Subtract the matrices \( A \) and \( B \)

Perform the subtraction operation on the matrices \( B \) and \( A \). Each element of matrix \( B \) is subtracted by the corresponding element of matrix \( A \). The result is another matrix, say \( D \). The operation will look like the following: \( D_{ij}=B_{ij}-A_{ij} \).

Step 4: Scale the resulting matrix \( D \) by \( \frac{1}{2} \)

Multiply each element of the resulting matrix \( D \) by \( \frac{1}{2} \) to get the matrix \( C \). The operation will look like the following: \( C_{ij}=\frac{1}{2}*D_{ij} \).

Step 5: Finalize the result

The resulting matrix \( C \) is the answer to the problem. It should have the same dimensions as the original matrices \( A \) and \( B \)

Key Concepts

Matrix Addition

Matrix addition is a fundamental operation in matrix algebra where two matrices of the same dimensions are added together to create a new matrix. For instance, given two matrices, \( A \) and \( B \), with the same number of rows and columns, their sum, \( A + B \), results in a matrix where each element, \( (A + B)_{ij} \), is the sum of the corresponding elements \( A_{ij} \) and \( B_{ij} \).

For example, if we have \( A=\begin{bmatrix}1 & 3\ 2 & 5\end{bmatrix} \) and \( B=\begin{bmatrix}4 & 2\ 6 & -1\end{bmatrix} \), their sum will be \( A + B = \begin{bmatrix}(1+4) & (3+2)\ (2+6) & (5-1)\end{bmatrix} = \begin{bmatrix}5 & 5\ 8 & 4\end{bmatrix} \).

It is important to emphasize that matrix addition is commutative, meaning that \( A + B \) is always equal to \( B + A \) for any two matrices \( A \) and \( B \) with the same size.

Matrix Subtraction

The concept of matrix subtraction parallels that of matrix addition, where you subtract corresponding elements of the matrices. To subtract matrix \( B \) from matrix \( A \), again, both matrices must have the same dimensions. The entry in the resultant matrix \( A - B \), at position \( (i, j) \), will be the result of \( A_{ij} - B_{ij} \).

Continuing our example from matrix addition, the subtraction of \( B \) from \( A \) would be: \( A - B = \begin{bmatrix}(1-4) & (3-2)\ (2-6) & (5+1)\end{bmatrix} = \begin{bmatrix}-3 & 1\ -4 & 6\end{bmatrix} \).

Unlike addition, matrix subtraction is not commutative, meaning that \( A - B \) does not equal \( B - A \). In the exercise provided, the subtraction \( B-A \) is a crucial step to isolate and subsequently solve for matrix \( C \). Understanding the non-commutative nature of subtraction helps identify potential errors when manipulating matrix equations.

Matrix Equation

Dealing with a matrix equation involves finding a matrix variable that satisfies the given equation. Just like solving for a variable in arithmetic, solving for a matrix entails performing operations that simplify the equation until the unknown matrix is isolated.

To illustrate this, consider the matrix equation from the exercise, \( A+2C=B \). The solution approach begins by reorganizing the equation to isolate the unknown matrix, leading to \( 2C=B-A \), followed by \( C=\frac{1}{2}(B-A) \). Clearing the fraction by multiplying both sides by \( \frac{1}{2} \) results in matrix \( C \).

It's valuable to appreciate the similarity between matrix equations and standard algebraic equations. In both cases, gradual isolation of the unknown and systematically performing equivalent operations on both sides of the equation lead to a solution. This foundational knowledge eases the understanding of more complex matrix operations and their applications in fields such as systems of linear equations, vector spaces, and transforming geometric objects.