If \(A = \left( {\begin{array}{} 1&2\\ 2&3 \end{array}} \right)\) and A 2– kA – I 2= 0, where I 2is the 2 × 2 identity matrix, then what is the value of k?
4
Solving Matrix Equations to Find Scalar Values
We are given a matrix \(A\) and a matrix equation involving \(A\), its square \(A^2\), a scalar \(k\), and the 2x2 identity matrix \(I_2\). The goal is to find the value of \(k\) that satisfies the given equation.
The given matrix is:
\(A = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}\)
The identity matrix \(I_2\) is:
\(I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
The given matrix equation is:
\(A^2 - kA - I_2 = 0\)
Here, \(0\) represents the 2x2 zero matrix, \(\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\).
Step 1: Calculate A²
First, we need to find the square of the matrix \(A\), which is \(A \times A\).
\(A^2 = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} \times \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}\)
To multiply two matrices, we multiply the elements of each row of the first matrix by the corresponding elements of each column of the second matrix and sum the products.
| Element | Calculation |
|---|---|
| (1,1) | \((1 \times 1) + (2 \times 2) = 1 + 4 = 5\) |
| (1,2) | \((1 \times 2) + (2 \times 3) = 2 + 6 = 8\) |
| (2,1) | \((2 \times 1) + (3 \times 2) = 2 + 6 = 8\) |
| (2,2) | \((2 \times 2) + (3 \times 3) = 4 + 9 = 13\) |
So, \(A^2\) is:
\(A^2 = \begin{pmatrix} 5 & 8 \\ 8 & 13 \end{pmatrix}\)
Step 2: Calculate kA
Next, we calculate \(kA\), which involves scalar multiplication. We multiply each element of the matrix \(A\) by the scalar \(k\).
\(kA = k \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} k \times 1 & k \times 2 \\ k \times 2 & k \times 3 \end{pmatrix} = \begin{pmatrix} k & 2k \\ 2k & 3k \end{pmatrix}\)
Step 3: Substitute into the Matrix Equation
Now, substitute the calculated values of \(A^2\), \(kA\), and \(I_2\) into the given equation \(A^2 - kA - I_2 = 0\).
\(\begin{pmatrix} 5 & 8 \\ 8 & 13 \end{pmatrix} - \begin{pmatrix} k & 2k \\ 2k & 3k \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)
Step 4: Perform Matrix Subtraction
To subtract matrices, we subtract the corresponding elements.
\(\begin{pmatrix} 5 - k - 1 & 8 - 2k - 0 \\ 8 - 2k - 0 & 13 - 3k - 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)
Simplify the elements on the left side:
\(\begin{pmatrix} 4 - k & 8 - 2k \\ 8 - 2k & 12 - 3k \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)
Step 5: Equate Corresponding Elements to Solve for k
For two matrices to be equal, their corresponding elements must be equal. We can set any element of the resulting matrix on the left side equal to the corresponding element of the zero matrix on the right side and solve for \(k\).
- From the element in the first row, first column: \(4 - k = 0 \implies k = 4\)
- From the element in the first row, second column: \(8 - 2k = 0 \implies 2k = 8 \implies k = 4\)
- From the element in the second row, first column: \(8 - 2k = 0 \implies 2k = 8 \implies k = 4\)
- From the element in the second row, second column: \(12 - 3k = 0 \implies 3k = 12 \implies k = 4\)
All equations yield the same value for \(k\), which is 4.
Thus, the value of \(k\) that satisfies the given matrix equation is 4.
Revision Table: Matrix Operations
| Concept | Description |
|---|---|
| Matrix Multiplication | Product of rows of the first matrix with columns of the second. Requires compatible dimensions. |
| Scalar Multiplication | Multiplying every element of a matrix by a single number (scalar). |
| Matrix Addition/Subtraction | Adding or subtracting corresponding elements of matrices. Requires matrices of the same dimensions. |
| Matrix Equation | An equation where the variables are matrices or scalars within matrix expressions. Solved by performing matrix operations and equating corresponding elements. |
Additional Information: Cayley-Hamilton Theorem Connection
The matrix equation \(A^2 - kA - I_2 = 0\) has a special connection to the Cayley-Hamilton Theorem. This theorem states that every square matrix satisfies its own characteristic equation. The characteristic equation of a 2x2 matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is given by \(det(A - \lambda I) = 0\), which expands to \(\lambda^2 - (a+d)\lambda + (ad-bc) = 0\). Here, \((a+d)\) is the trace of the matrix (\(Tr(A)\)) and \((ad-bc)\) is the determinant of the matrix (\(det(A)\)). So the equation is \(\lambda^2 - Tr(A)\lambda + det(A) = 0\).
According to the theorem, if we replace \(\lambda\) with \(A\), the matrix equation \(A^2 - Tr(A)A + det(A)I = 0\) holds true.
For the given matrix \(A = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}\):
- Trace: \(Tr(A) = 1 + 3 = 4\)
- Determinant: \(det(A) = (1 \times 3) - (2 \times 2) = 3 - 4 = -1\)
According to the Cayley-Hamilton Theorem, the characteristic equation is \(\lambda^2 - 4\lambda - 1 = 0\), and the matrix \(A\) should satisfy \(A^2 - 4A - 1I_2 = 0\).
Comparing this to the given equation \(A^2 - kA - I_2 = 0\), we can see that the value of \(k\) must be equal to the trace of the matrix \(A\).
In this case, \(k = Tr(A) = 4\).
This confirms our step-by-step calculation result and shows a more advanced concept related to this type of problem.
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