Question

Statement- \(1:\) If \(a, b, c, d\) are real number and \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) and \(A^{3}=0\), then \(A^{2}=O\). Statement- 2: : For matrix \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) we have \(\mathrm{A}^{2}-(\mathrm{a}+\mathrm{d}) \mathrm{A}+(\mathrm{ad}-\mathrm{bc}) \mathrm{I}=\mathrm{O}\) (1) Statement \(-1\) is True, Statement - 2 is True Statement \(-2\) is a correct explanation for 1 Statement \(-1\) (2) Statement- 1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement- 1 (3) Statement \(-1\) is True, Statement \(-2\) is False (4) Statement \(-1\) is False, Statement \(-2\) is True

Short answer

1: True, 2: True; Statement 2 explains Statement 1.

Step-by-step solution

Understand the Problem

Given two statements about a matrix \(A = \left[ \begin{array}{ll} a & b \ c & d \end{array} \right]\). You need to determine whether each statement is true or false and whether the second statement explains the first.

Analyze Statement 1

Statement 1 claims that if \(A^3 = 0\), then \(A^2 = O\). We need to prove or disprove this. Remember that if \(A^3 = 0\), then \(A\) is a nilpotent matrix, and we need to check whether \(A^2 = 0\).

Analyze Statement 2

Statement 2 is the Cayley-Hamilton Theorem for a 2x2 matrix: \( A^2 - (\text{tr}A)A + (\text{det}A)I = O\). We need to verify this for the given matrix.

Verify Cayley-Hamilton Theorem

Apply the Cayley-Hamilton theorem to the given matrix. Let \( A = \left[ \begin{array}{ll} a & b \ c & d \end{array} \right] \). Then, \( \text{tr}A = a + d \) and \( \text{det}A = ad - bc \). Verify that \(A^2 - (a+d)A + (ad-bc)I = O\).

Validate Statement 2

Check the validity of Statement 2. This can be done by substituting \(A = \left[ \begin{array}{ll} a & b \ c & d \end{array} \right] \) into the equation to see if it holds for a general 2x2 matrix.

Relate Statement 2 to Statement 1

Verify if the Cayley-Hamilton theorem implies that \(A^3 = 0\) leads to \(A^2 = 0\). If \(A^3 = 0\), then \(A\) is nilpotent, and we need to confirm if this implies \(A^2 = 0\).

Reach Conclusion

Analyze findings to determine the truth values of Statement 1 and Statement 2. Check if the true Statement 2 explains true Statement 1 based on Cayley-Hamilton theorem and nilpotent property.

Key concepts

Cayley-Hamilton Theorem

The Cayley-Hamilton Theorem is a fundamental result in matrix theory. It states that every square matrix satisfies its own characteristic equation. For a 2x2 matrix, say \( A = \left[ \begin{array}{ll} a & b \ c & d \end{array} \right] \), the characteristic equation is given by the polynomial:\[ p_A(\lambda) = \lambda^2 - (\text{tr}A)\lambda + (\text{det}A) = 0 \] where \( \text{tr}A \) is the trace of matrix \( A \) (the sum of its diagonal elements, \( a + d \)) and \( \text{det}A \) is the determinant of \( A \) (calculated as \( ad - bc \)).
Plugging the matrix \( A \) into this equation, we get the Cayley-Hamilton expression:
\[ A^2 - (\text{a} + \text{d})A + (ad - bc)I = 0 \] This equation essentially means that if you substitute the actual matrix \( A \) into this polynomial equation, the result will be the zero matrix. It validates the statement \( A^2 - (a+d)A + (ad - bc)I = O \), thus proving the second part correct.

Nilpotent Matrices

A matrix \( A \) is termed nilpotent if there exists some positive integer \( k \) such that \( A^k = 0 \), where 0 is the zero matrix. In the given problem, we have an important claim that if \( A^3 = 0 \), then \( A^2 \) must also be zero.
Let's unpack this a little more:

• If \( A \) is nilpotent, specifically if \( A^3 = 0 \), it inherently means that multiplying the matrix \( A \) with itself three times results in a zero matrix.
• This progresses from the property that if a lower power of \( k \) can produce zero, then all higher powers must result in zero as well.
• Therefore, for a 2x2 matrix that is nilpotent and \( A^3 = 0 \), it must be true that \( A^2 = 0 \), reinforcing the statement.

This validates statement 1 in our exercise.

Matrix Multiplication

Matrix multiplication forms the backbone of understanding relationships between matrices, such as in our given problem.
Here's a quick rundown on how matrix multiplication works:

• To multiply two matrices, say \( A = \left[ \begin{array}{ll} a & b \ c & d \end{array} \right] \) and \( B = \left[ \begin{array}{ll} e & f \ g & h \end{array} \right] \), we take the dot product of rows from the first matrix with columns from the second matrix.
  
• For example, the element in the first row and first column of the product matrix would be calculated as: \( ae + bg \).
  
• Continuing this for all elements, the resulting matrix \( C = AB \) would look like: \[ C = \left[ \begin{array}{ll} ae + bg & af + bh \ ce + dg & cf + dh \end{array} \right] \].

Understanding matrix multiplication helps in verifying nilpotency and applying the Cayley-Hamilton theorem effectively. When we multiply matrices in scenarios like nilpotent matrices, each step incrementally validates results.