If \( P = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \) and \( Q = \begin{bmatrix} 2 & -4 & 1 \end{bmatrix} \) are two matrices, then \( (PQ)' \) will be:
\(\begin{bmatrix} -2 & 4 & 2 \\ 4 & -8 & -4 \\ -1 & 2 & 1 \end{bmatrix}\)
Understanding the Matrix Problem: Calculating (PQ)'
The problem asks us to find the transpose of the product of two given matrices, \( P \) and \( Q \). First, we need to perform the matrix multiplication \( PQ \), and then find the transpose of the resulting matrix.
Given Matrices
We are given the matrices:
- Matrix \( P \): \( P = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \)
- Matrix \( Q \): \( Q = \begin{bmatrix} 2 & -4 & 1 \end{bmatrix} \)
Matrix \( P \) is a column matrix of size 3x1 (3 rows, 1 column). Matrix \( Q \) is a row matrix of size 1x3 (1 row, 3 columns).
Step-by-Step Matrix Multiplication (PQ)
To multiply two matrices, the number of columns in the first matrix must equal the number of rows in the second matrix. Here, P is 3x1 and Q is 1x3. The number of columns in P (1) equals the number of rows in Q (1). So, the multiplication \( PQ \) is possible. The resulting matrix \( PQ \) will have the dimensions of the number of rows of P and the number of columns of Q, which is 3x3.
Let \( PQ = R \). The element in the \( i \)-th row and \( j \)-th column of \( R \), denoted as \( R_{ij} \), is found by multiplying the elements of the \( i \)-th row of \( P \) by the corresponding elements of the \( j \)-th column of \( Q \) and summing the products. In this case, each "row" of P is just a single element, and each "column" of Q is also a single element.
Let's calculate each element of the 3x3 matrix \( PQ \):
- \( (PQ)_{11} = P_{11} \times Q_{11} = (-1) \times (2) = -2 \)
- \( (PQ)_{12} = P_{11} \times Q_{12} = (-1) \times (-4) = 4 \)
- \( (PQ)_{13} = P_{11} \times Q_{13} = (-1) \times (1) = -1 \)
- \( (PQ)_{21} = P_{21} \times Q_{11} = (2) \times (2) = 4 \)
- \( (PQ)_{22} = P_{21} \times Q_{12} = (2) \times (-4) = -8 \)
- \( (PQ)_{23} = P_{21} \times Q_{13} = (2) \times (1) = 2 \)
- \( (PQ)_{31} = P_{31} \times Q_{11} = (1) \times (2) = 2 \)
- \( (PQ)_{32} = P_{31} \times Q_{12} = (1) \times (-4) = -4 \)
- \( (PQ)_{33} = P_{31} \times Q_{13} = (1) \times (1) = 1 \)
So, the matrix \( PQ \) is:
| Column 1 | Column 2 | Column 3 | |
|---|---|---|---|
| Row 1 | -2 | 4 | -1 |
| Row 2 | 4 | -8 | 2 |
| Row 3 | 2 | -4 | 1 |
In matrix form:
\( PQ = \begin{bmatrix} -2 & 4 & -1 \\ 4 & -8 & 2 \\ 2 & -4 & 1 \end{bmatrix} \)
Calculating the Transpose (PQ)'
The transpose of a matrix is obtained by swapping its rows and columns. If a matrix \( A \) has elements \( A_{ij} \), its transpose \( A' \) (or \( A^T \)) will have elements \( A'_{ij} = A_{ji} \).
Let's find the transpose of the matrix \( PQ \):
\( (PQ)' = \begin{bmatrix} (-2)' & (4)' & (2)' \\ (4)' & (-8)' & (-4)' \\ (-1)' & (2)' & (1)' \end{bmatrix}_{swapped} \)
Swapping the elements based on the rule \( A'_{ij} = A_{ji} \):
- The element in row 1, column 1 of \( (PQ)' \) is the element from row 1, column 1 of \( PQ \), which is -2.
- The element in row 1, column 2 of \( (PQ)' \) is the element from row 2, column 1 of \( PQ \), which is 4.
- The element in row 1, column 3 of \( (PQ)' \) is the element from row 3, column 1 of \( PQ \), which is 2.
- The element in row 2, column 1 of \( (PQ)' \) is the element from row 1, column 2 of \( PQ \), which is 4.
- The element in row 2, column 2 of \( (PQ)' \) is the element from row 2, column 2 of \( PQ \), which is -8.
- The element in row 2, column 3 of \( (PQ)' \) is the element from row 3, column 2 of \( PQ \), which is -4.
- The element in row 3, column 1 of \( (PQ)' \) is the element from row 1, column 3 of \( PQ \), which is -1.
- The element in row 3, column 2 of \( (PQ)' \) is the element from row 2, column 3 of \( PQ \), which is 2.
- The element in row 3, column 3 of \( (PQ)' \) is the element from row 3, column 3 of \( PQ \), which is 1.
So, the transpose matrix \( (PQ)' \) is:
\( (PQ)' = \begin{bmatrix} -2 & 4 & 2 \\ 4 & -8 & -4 \\ -1 & 2 & 1 \end{bmatrix} \)
Final Answer
Comparing this result with the given options, we find that it matches one of them.
The matrix \( (PQ)' \) is \( \begin{bmatrix} -2 & 4 & 2 \\ 4 & -8 & -4 \\ -1 & 2 & 1 \end{bmatrix} \).
Revision Table: Matrix Operations
| Operation | Description | Example |
|---|---|---|
| Matrix Multiplication (AB) | Possible if number of columns in A = number of rows in B. Resulting matrix has dimensions (rows of A) x (columns of B). Element (AB)\(_{ij}\) is sum of products of elements from row i of A and column j of B. | If A is 2x3 and B is 3x4, AB is 2x4. |
| Matrix Transpose (A') | Obtained by swapping rows and columns. If A is m x n, A' is n x m. Element (A')\(_{ij}\) = A\(_{ji}\). | If A = \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \), then A' = \( \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \). |
Additional Information: Properties of Matrix Transpose
The transpose operation has several useful properties in matrix algebra:
- Double Transpose: The transpose of the transpose of a matrix is the original matrix itself. \( (A')' = A \).
- Transpose of a Sum: The transpose of a sum of matrices is the sum of their transposes. \( (A + B)' = A' + B' \). This applies if A and B have the same dimensions.
- Transpose of a Scalar Product: The transpose of a matrix multiplied by a scalar is the scalar multiplied by the transpose of the matrix. \( (kA)' = kA' \), where k is a scalar.
- Transpose of a Product: The transpose of a product of two matrices is the product of their transposes in reverse order. \( (AB)' = B'A' \). This is a very important property, and it was implicitly used in checking our calculation order, although direct calculation was easier here.
These properties are fundamental for solving various problems involving matrices and are frequently used in linear algebra and related fields.
Important Questions from Matrices
Three defective bulbs are mixed with 8 good ones. If three bulbs are drawn one by one with replacement, the probabilities of getting exactly 1 defective, more than 2 defective, no defective and more than 1 defective respectively are :
If A, B, and C are three singular matrices given by:
\[ A = \begin{bmatrix} 1 & 4 \\ 3 & 2a \end{bmatrix} \]
\[ B = \begin{bmatrix} 3b & 5 \\ a & 2 \end{bmatrix} \]
and
\[ C = \begin{bmatrix} a + b + c & c + 1 \\ a + c & c \end{bmatrix} \]
then the value of abc is:
The matrix
\[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
is a:
- \( \text{Scalar matrix} \)
- \( \text{Diagonal matrix} \)
- \( \text{Skew-symmetric matrix} \)
- \( \text{Symmetric matrix} \)
Choose the correct answer from the options given below:
If A is a square matrix and I is an identity matrix such that \(A^2 = A\), then \(A(I - 2A)^3 + 2A^3\) is equal to :
If \([A]_{3 \times 2} [B]_{x \times y} = [C]_{3 \times 1}\), then:
