Let C=[} 2&-1 0&1 ] Find DC if it is defined. Otherwise,click on "Undefined". DC=square

To determine if the product \( DC \) is defined, we need to check the dimensions of the matrices \( C \) and \( D \).<br /><br />Matrix \( C \) is a \( 2 \times 2 \) matrix:<br />\[ C = \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix} \]<br /><br />Matrix \( D \) is also a \( 2 \times 2 \) matrix:<br />\[ D = \begin{bmatrix} 7 & -4 \\ -3 & 4 \end{bmatrix} \]<br /><br />Since both matrices are \( 2 \times 2 \), the product \( DC \) is defined and will also be a \( 2 \times 2 \) matrix.<br /><br />Now, let's calculate the product \( DC \):<br /><br />\[ DC = \begin{bmatrix} 7 & -4 \\ -3 & 4 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix} \]<br /><br />To find each element of the resulting matrix, we use the formula for matrix multiplication:<br /><br />\[ (DC)_{ij} = \sum_{k=1}^{2} D_{ik} C_{kj} \]<br /><br />Let's calculate each element:<br /><br />1. \( (DC)_{11} = 7 \cdot 2 + (-4) \cdot 0 = 14 \)<br />2. \( (DC)_{12} = 7 \cdot (-1) + (-4) \cdot 1 = -7 - 4 = -11 \)<br />3. \( (DC)_{21} = -3 \cdot 2 + 4 \cdot 0 = -6 \)<br />4. \( (DC)_{22} = -3 \cdot (-1) + 4 \cdot 1 = 3 + 4 = 7 \)<br /><br />So, the resulting matrix \( DC \) is:<br />\[ DC = \begin{bmatrix} 14 & -11 \\ -6 & 7 \end{bmatrix} \]<br /><br />Thus, the final answer is:<br />\[ DC = \begin{bmatrix} 14 & -11 \\ -6 & 7 \end{bmatrix} \]