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Let ${A}$ be a $2 \times 2$ matrix such that ${A}$ * $\begin{pmatrix} 3 \\ -8 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \text{and} \quad {A} * \begin{pmatrix} 5 \\ 7 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.$ Find ${A}^{-1} *\begin{pmatrix} -2 \\ -1 \end{pmatrix}.$

What is the easiest way to start this problem?

Thanks

Rohini Sharma
  • 449

2 Answers2

3

Given

$$ \mathbf{A} \begin{pmatrix} 3 \\ -8 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \textrm{and} \quad \mathbf{A} \begin{pmatrix} 5 \\ 7 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$

IF $\mathbf{A}^{-1}$ exists, we can write

$$ \mathbf{A}^{-1} \mathbf{A} \begin{pmatrix} 3 \\ -8 \end{pmatrix} = \mathbf{A}^{-1} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \textrm{and} \quad \mathbf{A}^{-1} \mathbf{A} \begin{pmatrix} 5 \\ 7 \end{pmatrix} = \mathbf{A}^{-1} \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$

Note that

$$ \mathbf{A}^{-1} \begin{pmatrix} -2 \\ -1 \end{pmatrix} = -2 \mathbf{A}^{-1} \mathbf{A} \begin{pmatrix} 3 \\ -8 \end{pmatrix} - \mathbf{A}^{-1} \mathbf{A} \begin{pmatrix} 5 \\ 7 \end{pmatrix} $$

So

$$ \mathbf{A}^{-1} \begin{pmatrix} -2 \\ -1 \end{pmatrix} = -2 \begin{pmatrix} 3 \\ -8 \end{pmatrix} - \begin{pmatrix} 5 \\ 7 \end{pmatrix} = \begin{pmatrix} -11 \\ 9 \end{pmatrix} $$

Thus

$$ \bbox[16px,border:2px solid #800000] { \mathbf{A}^{-1} \begin{pmatrix} -2 \\ -1 \end{pmatrix} = \begin{pmatrix} -11 \\ 9 \end{pmatrix} } $$

johannesvalks
  • 6,407
1

First, notice that $A$ is invertible since it transforms a basis of $\Bbb R^2$ to a basis of $\Bbb R^2$. Now write

$$A^{-1 }\cdot(-2,-1)^T=-2A^{-1}\cdot(1,0)^T-A^{-1}\cdot(0,1)^T$$ and the result follows.