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Question

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

The correct answer is

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.

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Important Questions from Matrix Algebra

  1. Consider the system of equations: x + y = 2 and 2x + 2y = 5. This system has

  2. The standard ordered basis of R 3 is {e 1, e 2, e 3} Let T : R 3 → R 3 be the linear transformation such that T(e 1) = 7e 1- 5e 3, T (e 2) = -2e 2+ 9e 3, T(e 3) = e 1+ e 2+ e 3. The standard matrix of T is:

  3. The system of equations

    x + y + z = 6;

    x + 4y + 6z = 20;

    x + 4y + λz = μ

    has NO solution for values of λ and μ given by

  4. 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?

  5. If A = \( \left[\begin{array}{cc}0 & 1 \\ −1 & 0\end{array}\right]\)  and (aI 2  + bA)2  = A, then
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