Find the matrix A A $\begin {bmatrix} 0 & 1\\2 & -1 \end {bmatrix}$ = $\begin {bmatrix} 2 & -1\\1 & 0 \end {bmatrix}$

1. Analyze the Question

We are given the matrix equation $A \times \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}$. We need to find matrix $A$.

Let $X = \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}$. The equation is $AX = B$. To solve for $A$, we multiply both sides by the inverse of $X$ ($X^{-1}$) from the right: $A = BX^{-1}$

2. Explain the Concept

To find the inverse of a $2 \times 2$ matrix $X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, we use the formula: $X^{-1} = \frac{1}{\det(X)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ where $\det(X) = (ad - bc)$.

3. Show the Solution

Step 1: Find the determinant of $X$ $\det(X) = (0 \times -1) - (1 \times 2) = 0 - 2 = -2$

Step 2: Find the inverse of $X$ ($X^{-1}$) $X^{-1} = \frac{1}{-2} \begin{bmatrix} -1 & -1 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} 1/2 & 1/2 \\ 1 & 0 \end{bmatrix}$

Step 3: Calculate $A = BX^{-1}$ $A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \times \begin{bmatrix} 1/2 & 1/2 \\ 1 & 0 \end{bmatrix}$

Perform matrix multiplication (Row $\times$ Column):

• Top-left: $(2 \times 1/2) + (-1 \times 1) = 1 - 1 = 0$
• Top-right: $(2 \times 1/2) + (-1 \times 0) = 1 + 0 = 1$
• Bottom-left: $(1 \times 1/2) + (0 \times 1) = 1/2 + 0 = 1/2$
• Bottom-right: $(1 \times 1/2) + (0 \times 0) = 1/2 + 0 = 1/2$

So, $A = \begin{bmatrix} 0 & 1 \\ 1/2 & 1/2 \end{bmatrix}$

4. Identify the Answer

Comparing our result with the given options, it matches option B.

5. Tip

When solving $AX = B$, always remember that matrix multiplication is not commutative. You must multiply by the inverse on the side that "cancels out" the matrix you are trying to isolate. If $X$ is on the right of $A$, multiply by $X^{-1}$ on the right of $B$.