Question

For I =\(\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}\), if X and Y are square matrices of order 2 such that XY = X and Y X = Y , then (Y² + 2Y ) equals to:

• A. 2Y
• B. I+3X
• C. I+3Y
• D. 3Y

Hint:

When dealing with matrix equations involving products like \( XY \) or \( YX \), factorization can be a helpful tool. In this case, we used the identity \( Y(X - I) = 0 \), which gave us two possible solutions. Once you find the values of the variables, be sure to substitute them into the original expressions to simplify further and verify the results. Matrix algebra can be tricky, but breaking down the problem step by step makes it more manageable.

Answer 1

The Correct Option is D

To solve for \((Y^2 + 2Y)\) given \(XY = X\) and \(YX = Y\), let's explore the properties of matrices \(X\) and \(Y\) alongside the identity matrix \(I\). The conditions state:

• \(XY = X\)
• \(YX = Y\)

First, let's find \(Y^2\):

If we left multiply both sides of \(YX = Y\) by \(Y\), we have:

• \(Y(YX) = Y^2 = Y^2X\)
• Since \(YX = Y\), it simplifies to \(Y^2 = Y\).

Next, consider the expression \(Y^2 + 2Y\):

• Substituting \(Y^2 = Y\) into the expression, we get:
• \((Y^2 + 2Y) = (Y + 2Y) = 3Y\)

Thus, the expression \(Y^2 + 2Y\) simplifies to: 3Y

So, the correct option is 3Y.

Answer 2

The given conditions are:

\[ XY = X \quad \text{and} \quad YX = Y. \]

Step 1: Analyze the second equation \( YX = Y \):

From \( YX = Y \), we can factorize as: \[ Y(X - I) = 0, \] where \( I \) is the identity matrix.

Step 2: Solve the factorized equation:

Thus, we have two possibilities: \[ Y = 0 \quad \text{or} \quad X = I. \] Since \( X \neq 0 \) by the given conditions, we take \( X = I \) (the identity matrix).

Step 3: Substitute \( X = I \) into the expression \( Y^2 + 2Y \):

Substituting \( X = I \) into \( Y^2 + 2Y \) gives: \[ Y^2 + 2Y = Y(Y + 2). \]

Step 4: Use the equation \( YX = Y \) to further simplify:

From \( YX = Y \) and \( X = I \), we have: \[ YI = Y, \] which simplifies to: \[ Y^2 = Y. \]

Step 5: Substitute \( Y^2 = Y \) into \( Y^2 + 2Y \):

Substituting \( Y^2 = Y \) into the expression \( Y^2 + 2Y \), we get: \[ Y^2 + 2Y = Y + 2Y = 3Y. \]

Conclusion: Thus, the equation simplifies to: \[ Y^2 + 2Y = 3Y. \]