If A is the identity matrix of order 3 and B is its transpose, then what is the value of the determinant of the matrix C = A + B?
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Calculating the Determinant of Matrix Sum (Identity + Transpose)
The question asks us to find the determinant of a matrix C, which is formed by adding the identity matrix of order 3 (A) and its transpose (B). Let's break this down step by step.
Step 1: Identify Matrix A
Matrix A is the identity matrix of order 3. An identity matrix, denoted by I, is a square matrix with ones on the main diagonal and zeros elsewhere. For order 3, it looks like this:
$$ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$Step 2: Identify Matrix B
Matrix B is the transpose of matrix A. The transpose of a matrix is obtained by flipping the matrix over its diagonal; that is, swapping row and column indices \(A^T_{ij} = A_{ji}\). For the identity matrix, transposing it results in the same matrix.
$$ B = A^T = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}^T = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$So, matrix B is also the identity matrix of order 3.
Step 3: Calculate Matrix C = A + B
Matrix C is the sum of matrix A and matrix B. To add matrices, we add their corresponding elements.
$$ C = A + B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} + \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ $$ C = \begin{pmatrix} 1+1 & 0+0 & 0+0 \\ 0+0 & 1+1 & 0+0 \\ 0+0 & 0+0 & 1+1 \end{pmatrix} $$ $$ C = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} $$Matrix C is a diagonal matrix with 2s along the main diagonal.
Step 4: Calculate the Determinant of Matrix C
We need to find the determinant of matrix C, denoted as \(\det(C)\) or \(|C|\). For a 3x3 matrix \(\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}\), the determinant can be calculated as \(a(ei - fh) - b(di - fg) + c(dh - eg)\).
For our matrix \(C = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}\), we can use this formula:
$$ \det(C) = 2((2)(2) - (0)(0)) - 0((0)(2) - (0)(0)) + 0((0)(0) - (2)(0)) $$ $$ \det(C) = 2(4 - 0) - 0 + 0 $$ $$ \det(C) = 2 \times 4 $$ $$ \det(C) = 8 $$Alternatively, for a diagonal matrix, the determinant is simply the product of the elements on the main diagonal.
$$ \det(C) = 2 \times 2 \times 2 = 8 $$Thus, the value of the determinant of matrix C is 8.
| Matrix Operation | Result |
|---|---|
| Matrix A (Identity, order 3) | \(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) |
| Matrix B (Transpose of A) | \(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) |
| Matrix C = A + B | \(\begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}\) |
| Determinant of C | 8 |
The final answer is 8.
Revision Table: Matrix Concepts
| Concept | Definition/Property | Example (3x3) |
|---|---|---|
| Identity Matrix (I) | Square matrix with 1s on diagonal, 0s elsewhere. \(AI = IA = A\). | \(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) |
| Transpose (\(A^T\)) | Matrix obtained by swapping rows and columns. \((A^T)_{ij} = A_{ji}\). \((A^T)^T = A\). \((A+B)^T = A^T + B^T\). | If \(A=\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}\), then \(A^T=\begin{pmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{pmatrix}\). Note: For Identity Matrix, \(I^T=I\). |
| Matrix Addition (A+B) | Add corresponding elements. Matrices must be same order. \(A+B = B+A\). | \(\begin{pmatrix} a & b \\ c & d \end{pmatrix} + \begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} a+e & b+f \\ c+g & d+h \end{pmatrix}\) |
| Determinant (\(\det(A)\) or \(|A|\)) | A scalar value calculated from a square matrix. Represents scaling factor of linear transformation. Non-zero determinant means matrix is invertible. | For \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\), det is \(ad-bc\). For \(\begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix}\), det is \(abc\). |
Additional Information: Properties of Determinants
- The determinant of a matrix and its transpose are equal: \(\det(A^T) = \det(A)\).
- If a matrix has a row or column of zeros, its determinant is 0.
- If two rows or two columns of a matrix are identical, its determinant is 0.
- If a matrix is multiplied by a scalar \(k\), the determinant of the new matrix is \(k^n \times \det(A)\), where \(n\) is the order of the matrix. In our case, \(C = 2A\), so \(\det(C) = \det(2A) = 2^3 \det(A) = 8 \det(A)\). Since \(A\) is the identity matrix, \(\det(A) = 1\). Thus, \(\det(C) = 8 \times 1 = 8\). This property is another way to arrive at the same answer and confirms our result.
- The determinant of a product of matrices is the product of their determinants: \(\det(AB) = \det(A)\det(B)\). Note that \(\det(A+B)\) is generally NOT equal to \(\det(A) + \det(B)\).
Similar Questions
If \[ A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \] then what is \( AA^{T} \) equal to (where \( A^{T} \) is the transpose of \( A \))?
Let \(A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\). What is the least value of \(\theta\) for which \(A + A^{T} = I\), where \(I\) is the identity matrix of order 2?
Important Questions from Transpose of a Matrix
If $A$ and $B$ are two symmetric matrices of the same order, then $AB - BA$ is a:
If \[ A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \] then what is \( AA^{T} \) equal to (where \( A^{T} \) is the transpose of \( A \))?
Let \(A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\). What is the least value of \(\theta\) for which \(A + A^{T} = I\), where \(I\) is the identity matrix of order 2?
