Let \(A = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}\) and (mI + nA) 2= A where m, n are positive real numbers and I is the identify matrix. What is (m + n) equal to?
Solving the Matrix Equation (mI + nA)<sup>2</sup> = A
The problem asks us to find the value of \(m + n\), given a matrix \(A\), the identity matrix \(I\), and the equation \((mI + nA)^2 = A\), where \(m\) and \(n\) are positive real numbers.
The given matrix \(A\) is:
\(A = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}\)
The identity matrix \(I\) of the same order (2x2) is:
\(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
Step 1: Calculate (mI + nA)
First, we find the expression for \(mI + nA\):
\(mI + nA = m\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + n\begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}\)
\(mI + nA = \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix} + \begin{bmatrix} 0 & 2n \\ -2n & 0 \end{bmatrix}\)
\(mI + nA = \begin{bmatrix} m + 0 & 0 + 2n \\ 0 + (-2n) & m + 0 \end{bmatrix}\)
\(mI + nA = \begin{bmatrix} m & 2n \\ -2n & m \end{bmatrix}\)
Step 2: Calculate (mI + nA)<sup>2</sup>
Next, we square the matrix \((mI + nA)\):
\((mI + nA)^2 = \begin{bmatrix} m & 2n \\ -2n & m \end{bmatrix} \begin{bmatrix} m & 2n \\ -2n & m \end{bmatrix}\)
To multiply these matrices, we follow the rule of matrix multiplication:
- (Row 1 of first matrix) \(\times\) (Column 1 of second matrix) = \(m \cdot m + 2n \cdot (-2n) = m^2 - 4n^2\)
- (Row 1 of first matrix) \(\times\) (Column 2 of second matrix) = \(m \cdot 2n + 2n \cdot m = 2mn + 2mn = 4mn\)
- (Row 2 of first matrix) \(\times\) (Column 1 of second matrix) = \((-2n) \cdot m + m \cdot (-2n) = -2mn - 2mn = -4mn\)
- (Row 2 of first matrix) \(\times\) (Column 2 of second matrix) = \((-2n) \cdot 2n + m \cdot m = -4n^2 + m^2\)
So, \((mI + nA)^2\) is:
\((mI + nA)^2 = \begin{bmatrix} m^2 - 4n^2 & 4mn \\ -4mn & m^2 - 4n^2 \end{bmatrix}\)
Step 3: Set (mI + nA)<sup>2</sup> Equal to A
According to the problem, \((mI + nA)^2 = A\). So, we equate the resulting matrix to matrix \(A\):
\(\begin{bmatrix} m^2 - 4n^2 & 4mn \\ -4mn & m^2 - 4n^2 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}\)
Step 4: Form and Solve Equations for m and n
By equating the corresponding elements of the matrices, we get a system of equations:
- \(m^2 - 4n^2 = 0\)
- \(4mn = 2\)
- \(-4mn = -2\)
- \(m^2 - 4n^2 = 0\)
Equations 1 and 4 are identical. Equations 2 and 3 are also equivalent (\(4mn=2\) and \(-4mn=-2\) both simplify to \(mn = \frac{1}{2}\)).
We have two independent equations:
- \(m^2 - 4n^2 = 0\)
- \(mn = \frac{1}{2}\)
From Equation 2, we can express \(m\) in terms of \(n\). Since \(m\) and \(n\) are positive real numbers, \(m = \frac{1}{2n}\).
Substitute this expression for \(m\) into Equation 1:
\(\left(\frac{1}{2n}\right)^2 - 4n^2 = 0\)
\(\frac{1}{4n^2} - 4n^2 = 0\)
To eliminate the denominator, multiply the entire equation by \(4n^2\) (note that \(n \neq 0\) since \(mn = 1/2\)):
\(4n^2 \left(\frac{1}{4n^2}\right) - 4n^2 (4n^2) = 0 \cdot 4n^2\)
\(1 - 16n^4 = 0\)
\(16n^4 = 1\)
\(n^4 = \frac{1}{16}\)
Since \(n\) is a positive real number, we take the positive fourth root:
\(n = \sqrt[4]{\frac{1}{16}}\)
\(n = \frac{1}{2}\)
Now substitute the value of \(n\) back into the equation \(m = \frac{1}{2n}\) to find \(m\):
\(m = \frac{1}{2 \cdot \frac{1}{2}}\)
\(m = \frac{1}{1}\)
\(m = 1\)
We found \(m = 1\) and \(n = \frac{1}{2}\). Both are positive real numbers, which satisfies the condition given in the problem.
Step 5: Calculate (m + n)
Finally, we need to find the value of \(m + n\):
\(m + n = 1 + \frac{1}{2}\)
\(m + n = \frac{2}{2} + \frac{1}{2}\)
\(m + n = \frac{3}{2}\)
The value of \(m + n\) is \(\frac{3}{2}\).
Revision Table: Key Matrix Concepts
| Concept | Description | Example/Property |
|---|---|---|
| Identity Matrix (I) | A square matrix with ones on the main diagonal and zeros elsewhere. Acts like '1' in scalar multiplication. | For 2x2: \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) For any matrix A, AI = IA = A. |
| Scalar Multiplication | Multiplying a matrix by a scalar (a number). Each element of the matrix is multiplied by the scalar. | If \(k\) is a scalar and \(B = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), then \(kB = \begin{bmatrix} ka & kb \\ kc & kd \end{bmatrix}\). |
| Matrix Addition/Subtraction | Adding or subtracting matrices of the same dimensions. Corresponding elements are added or subtracted. | If \(B = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\) and \(C = \begin{bmatrix} e & f \\ g & h \end{bmatrix}\), then \(B+C = \begin{bmatrix} a+e & b+f \\ c+g & d+h \end{bmatrix}\). |
| Matrix Multiplication | A more complex operation where the element in the i-th row and j-th column of the product is the dot product of the i-th row of the first matrix and the j-th column of the second matrix. Requires specific dimensions. | For 2x2 matrices: \(\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} e & f \\ g & h \end{bmatrix} = \begin{bmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{bmatrix}\). |
| Matrix Equality | Two matrices are equal if and only if they have the same dimensions and their corresponding elements are equal. | \(\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} e & f \\ g & h \end{bmatrix}\) implies \(a=e, b=f, c=g, d=h\). |
Additional Information: Solving Matrix Equations
Solving matrix equations often involves using properties of matrix operations (addition, subtraction, multiplication) and equating corresponding elements to form systems of linear or non-linear algebraic equations. In this specific problem:
- We used scalar multiplication (\(mI\) and \(nA\)).
- We used matrix addition (\(mI + nA\)).
- We performed matrix multiplication (\((mI + nA)^2\)).
- We used matrix equality to set up algebraic equations.
- We solved a system of non-linear equations (\(m^2 - 4n^2 = 0\) and \(mn = 1/2\)) for the unknown variables \(m\) and \(n\).
It's important to remember that matrix multiplication is generally not commutative (AB \(\neq\) BA). However, in this case, we are squaring a single matrix \((mI + nA)\), so the order of multiplication relative to the square operation is fixed.
The condition that \(m\) and \(n\) are positive real numbers is crucial, as it helps select the correct values when solving equations that might yield multiple roots (like \(n^4 = 1/16\), which has four roots, but only one positive real root).
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