Solving a Matrix Equation: Finding Constants a and b

This problem asks us to find the values of constants 'a' and 'b' that satisfy a given matrix equation involving matrix A and the identity matrix I₂. The equation is \((aI_2 + bA)^2 = A\), where A is a specific 2x2 matrix.

Let's first write down the matrices involved:

The identity matrix of order 2 is \(I_2 = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]\).

The given matrix A is \(A = \left[\begin{array}{cc}0 & 1 \\ −1 & 0\end{array}\right]\).

Now, let's calculate the expression \((aI_2 + bA)\):

\(aI_2 = a \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}a & 0 \\ 0 & a\end{array}\right]\)

\(bA = b \left[\begin{array}{cc}0 & 1 \\ −1 & 0\end{array}\right] = \left[\begin{array}{cc}0 & b \\ −b & 0\end{array}\right]\)

Adding these two matrices:

\(aI_2 + bA = \left[\begin{array}{cc}a & 0 \\ 0 & a\end{array}\right] + \left[\begin{array}{cc}0 & b \\ −b & 0\end{array}\right] = \left[\begin{array}{cc}a+0 & 0+b \\ 0+(-b) & a+0\end{array}\right] = \left[\begin{array}{cc}a & b \\ −b & a\end{array}\right]\)

Next, we need to square the matrix \((aI_2 + bA)\):

\((aI_2 + bA)^2 = \left[\begin{array}{cc}a & b \\ −b & a\end{array}\right] \left[\begin{array}{cc}a & b \\ −b & a\end{array}\right]\)

Performing matrix multiplication:

\((aI_2 + bA)^2 = \left[\begin{array}{cc} (a)(a) + (b)(-b) & (a)(b) + (b)(a) \\ (−b)(a) + (a)(−b) & (−b)(b) + (a)(a) \end{array}\right]\)

\((aI_2 + bA)^2 = \left[\begin{array}{cc} a^2 − b^2 & ab + ba \\ −ba − ab & −b^2 + a^2 \end{array}\right]\)

\((aI_2 + bA)^2 = \left[\begin{array}{cc} a^2 − b^2 & 2ab \\ −2ab & a^2 − b^2 \end{array}\right]\)

According to the question, \((aI_2 + bA)^2 = A\). So, we set the calculated matrix equal to matrix A:

\(\left[\begin{array}{cc} a^2 − b^2 & 2ab \\ −2ab & a^2 − b^2 \end{array}\right] = \left[\begin{array}{cc} 0 & 1 \\ −1 & 0\end{array}\right]\)

For two matrices to be equal, their corresponding elements must be equal. This gives us a system of equations:

1. \(a^2 - b^2 = 0\)
2. \(2ab = 1\)
3. \(-2ab = -1\)
4. \(a^2 - b^2 = 0\)

Equations 1 and 4 are identical. Equations 2 and 3 are also equivalent (\(2ab = 1\) is the same as \(-2ab = -1\)). So we have two independent equations:

1. \(a^2 - b^2 = 0\)
2. \(2ab = 1\)

From Equation 1, \(a^2 = b^2\). This implies that either \(a = b\) or \(a = -b\).

Let's consider these two cases using Equation 2 (\(2ab = 1\)).

Case 1: \(a = b\)

Substitute \(b=a\) into Equation 2:

\(2a(a) = 1\)

\(2a^2 = 1\)

\(a^2 = \frac{1}{2}\)

\(a = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}}\)

If \(a = \frac{1}{\sqrt{2}}\), then since \(a=b\), we have \(b = \frac{1}{\sqrt{2}}\). Let's check this solution in Equation 2: \(2(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) = 2(\frac{1}{2}) = 1\). This works.

If \(a = -\frac{1}{\sqrt{2}}\), then since \(a=b\), we have \(b = -\frac{1}{\sqrt{2}}\). Let's check this solution in Equation 2: \(2(-\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}}) = 2(\frac{1}{2}) = 1\). This also works.

Case 2: \(a = -b\)

Substitute \(b=-a\) into Equation 2:

\(2a(-a) = 1\)

\(-2a^2 = 1\)

\(a^2 = -\frac{1}{2}\)

For real values of a, \(a^2\) cannot be negative. Thus, there are no real solutions for a and b in this case.

Considering only real solutions, we have two possibilities from Case 1: \(a=b=\frac{1}{\sqrt{2}}\) or \(a=b=-\frac{1}{\sqrt{2}}\).

We check the given options to see which one matches our results. The options are:

• a = b = \( \sqrt{2} \)
• a = b = \( \frac{1}{\sqrt{2}} \)
• a = b = \( \frac{1}{\sqrt{3}} \)
• a = b = \( \sqrt{3} \)

The option \(a = b = \frac{1}{\sqrt{2}}\) matches one of our valid solutions.

Therefore, the values for a and b that satisfy the matrix equation are \(a = b = \frac{1}{\sqrt{2}}\).

Revision Table: Key Steps for Solving Matrix Equations

Additional Information: Properties of Matrices and Matrix Operations

Solving matrix equations like \((aI_2 + bA)^2 = A\) requires a good understanding of matrix algebra, including scalar multiplication, matrix addition, and matrix multiplication. Here are some related concepts:

• Identity Matrix (\(I\)): The identity matrix is a square matrix with ones on the main diagonal and zeros elsewhere. When multiplied by any matrix of the same size, it leaves the original matrix unchanged (i.e., \(AI = IA = A\)). In this problem, \(I_2\) is the 2x2 identity matrix.
• Scalar Multiplication: Multiplying a matrix by a scalar (a single number like 'a' or 'b') involves multiplying every element of the matrix by that scalar.
• Matrix Addition/Subtraction: Matrices of the same dimensions can be added or subtracted by adding or subtracting their corresponding elements.
• Matrix Multiplication: The product of two matrices \(X\) and \(Y\) is defined only if the number of columns in \(X\) equals the number of rows in \(Y\). The element in the \(i\)-th row and \(j\)-th column of the product \(XY\) is found by multiplying the elements of the \(i\)-th row of \(X\) by the corresponding elements of the \(j\)-th column of \(Y\) and summing the results. Matrix multiplication is generally not commutative (i.e., \(XY \neq YX\)).
• Squaring a Matrix: Squaring a matrix \(M\) means multiplying it by itself, i.e., \(M^2 = M \times M\). This is only possible if the matrix is square.
• Equating Matrices: Two matrices are equal if and only if they have the same dimensions and every element in one matrix is equal to the corresponding element in the other matrix. This property is crucial for turning a matrix equation into a system of algebraic equations.

In this specific problem, the matrix \(A = \left[\begin{array}{cc}0 & 1 \\ −1 & 0\end{array}\right]\) has a special property: \(A^2 = \left[\begin{array}{cc}0 & 1 \\ −1 & 0\end{array}\right] \left[\begin{array}{cc}0 & 1 \\ −1 & 0\end{array}\right] = \left[\begin{array}{cc} (0)(0)+(1)(-1) & (0)(1)+(1)(0) \\ (-1)(0)+(0)(-1) & (-1)(1)+(0)(0) \end{array}\right] = \left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right] = -I_2\). This means \(A\) is related to the complex number 'i' representation in matrix form, where \(A^2 = -I_2\) is analogous to \(i^2 = -1\). While not directly used in the solution derived here, such properties can sometimes simplify matrix equations.