If $A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ then the matrix $AB$ is equal to
2. $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$
Matrix Multiplication Calculation
This problem requires us to calculate the product of two given matrices, $A$ and $B$, denoted as $AB$. Matrix multiplication is a fundamental operation where the rows of the first matrix are multiplied by the columns of the second matrix.
Understanding Matrix Multiplication
To multiply two matrices, say $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$ and $B = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}$, we create a resultant matrix $C = \begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{bmatrix}$. Each element $c_{ij}$ in the resulting matrix $C$ is obtained by computing the dot product of the $i$-th row of matrix $A$ and the $j$-th column of matrix $B$. The formula is:
$ c_{ij} = \sum_{k=1}^{2} a_{ik} b_{kj} $
For a 2x2 matrix multiplication, the specific calculations are:
- $c_{11} = a_{11}b_{11} + a_{12}b_{21}$
- $c_{12} = a_{11}b_{12} + a_{12}b_{22}$
- $c_{21} = a_{21}b_{11} + a_{22}b_{21}$
- $c_{22} = a_{21}b_{12} + a_{22}b_{22}$
Calculating AB for the Given Matrices
The matrices provided in the question are:
$ A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} $
$ B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $
We compute the product $AB$ by applying the multiplication rules step-by-step:
| Calculation for Element | Result |
|---|---|
| Element (1,1): $ (1 \times 0) + (0 \times 1) $ | $ 0 + 0 = 0 $ |
| Element (1,2): $ (1 \times 1) + (0 \times 0) $ | $ 1 + 0 = 1 $ |
| Element (2,1): $ (0 \times 0) + (-1 \times 1) $ | $ 0 - 1 = -1 $ |
| Element (2,2): $ (0 \times 1) + (-1 \times 0) $ | $ 0 + 0 = 0 $ |
Based on these calculations, the resulting matrix $AB$ is:
$ AB = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} $
Final Answer Selection
Now, we compare our calculated matrix $AB$ with the given options:
- 1. $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
- 2. $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$
- 3. $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
- 4. $\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$
Our calculated result for $AB$ is $\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$. This result does not precisely match any of the provided options. However, Option 2, which is $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$, corresponds to the calculation of the matrix product $BA$. As per the provided correct answer, Option 2 is indicated as the solution.
Important Questions from Matrix Algebra
- If A = \( \left[\begin{array}{cc}0 & 1 \\ −1 & 0\end{array}\right]\) and (aI 2 + bA)2 = A, then
If A = \(\left[\begin{array}{cc}2 & −3 \\3 & 5\end{array}\right]\), then which of the following statements are correct?
A. A is a square matrix
B. A−1 exists
C. A is a symmetric matrix
D. |A| = 19
E. A is a null matrix
Choose the correct answer from the options given below.
If A is Square Matrix of order 3, then product of A and its transpose is
What is the transformation matrix M that transforms a square in the xy-plane defined by (1, 1) T, (-1, 1) T, (-1, -1) T and (1, -1) T to a parallelogram whose corresponding vertices are (2, 1) T, (0, 1) T, (-2, -1) T and (0, -1) T?
Let \(A = \left[ {\begin{array}{*{20}{c}} 1&1&0\\ 0&1&0\\ 1&1&0\\ 0&0&1 \end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}} 1&0&0&0\\ 0&1&1&0\\ 1&0&1&1\\ \end{array}} \right]\) Find the boolean product A ⊙ B of the two matrices.
