Matrix Range Yz Plane B17Bf9

1. **State the problem:** We need to find a $3\times 3$ matrix $A$ such that the linear transformation $T(x) = Ax$ has its range equal to the $yz$-plane. 2. **Recall the $yz$-plane:** The $yz$-plane consists of all vectors of the form $\begin{bmatrix}0 \\ y \\ z\end{bmatrix}$ where $y,z \in \mathbb{R}$. This means the first component of any vector in the range must be zero. 3. **Matrix form and range:** The range of $T$ is the column space of $A$. To have the range equal to the $yz$-plane, the columns of $A$ must lie in the $yz$-plane. 4. **Constructing $A$:** Each column of $A$ is a vector in $\mathbb{R}^3$. To ensure the range is the $yz$-plane, the first component of each column must be zero. 5. **General form:** $$ A = \begin{bmatrix} 0 & 0 & 0 \\ a & b & c \\ d & e & f \end{bmatrix} $$ where $a,b,c,d,e,f$ are real numbers. 6. **Ensuring the range is exactly the $yz$-plane:** The columns must span the $yz$-plane, so the vectors $\begin{bmatrix}a \\ d\end{bmatrix}$, $\begin{bmatrix}b \\ e\end{bmatrix}$, and $\begin{bmatrix}c \\ f\end{bmatrix}$ in $\mathbb{R}^2$ must span $\mathbb{R}^2$. This means at least two of these vectors must be linearly independent. 7. **Example:** Choose columns to be the standard basis vectors of the $yz$-plane: $$ A = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} $$ 8. **Verification:** The range of $T$ is the span of columns: $$ \text{Range}(T) = \text{span}\left\{ \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix} \right\} = \text{span}\left\{ \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix} \right\} $$ which is exactly the $yz$-plane. **Final answer:** $$ A = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} $$