IIpinvepbl 3K3aMenaution HbIX 3a/[a4 1. Aambi Marpunjsi A=(} 3&-1&4 6&3&5 )x F=(} 3&-2&7 -1&5&-5 -4&0&-6 2&-3&-1

a) Marpuny $A\cdot B^{T}-3C$<br /><br />To find the product of matrices $A$ and $B^T$, we need to multiply each element of matrix $A$ by the corresponding element of matrix $B^T$ and then sum the products.<br /><br />$A\cdot B^T = \begin{pmatrix} 3 & -1 & 4 \\ 6 & 3 & 5 \end{pmatrix} \cdot \begin{pmatrix} 1 & -5 \\ -2 & 3 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 3 \cdot 1 + (-1) \cdot (-2) + 4 \cdot 2 & 3 \cdot (-5) + (-1) \cdot 3 + 4 \cdot 4 \\ 6 \cdot 1 + 3 \cdot (-2) + 5 \cdot 2 & 6 \cdot (-5) + 3 \cdot 3 + 5 \cdot 4 \end{pmatrix} = \begin{pmatrix} 11 & 7 \\ 11 & 7 \end{pmatrix}$<br /><br />Now, we need to subtract 3 times matrix $C$ from the resulting matrix.<br /><br />$A\cdot B^T - 3C = \begin{pmatrix} 11 & 7 \\ 11 & 7 \end{pmatrix} - 3 \cdot \begin{pmatrix} 3 & -2 \\ -1 & 5 \end{pmatrix} = \begin{pmatrix} 11 & 7 \\ 11 & 7 \end{pmatrix} - \begin{pmatrix} 9 & -6 \\ -3 & 15 \end{pmatrix} = \begin{pmatrix} 2 & 13 \\ 14 & -8 \end{pmatrix}$<br /><br />Therefore, the result of the operation $A\cdot B^T - 3C$ is the matrix $\begin{pmatrix} 2 & 13 \\ 14 & -8 \end{pmatrix}$.<br /><br />b) paHT Marpinist F<br /><br />The matrix $F$ is already given in the question. It is a $4 \times 3$ matrix:<br /><br />$F = \begin{pmatrix} 3 & -2 & 7 \\ -1 & 5 & -5 \\ -4 & 0 & -6 \\ 2 & -3 & -1 \end{pmatrix}$<br /><br />c) $D^{-1}$<br /><br />To find the inverse of matrix $D$, we need to use the formula for the inverse of a $3 \times 3$ matrix:<br /><br />$D^{-1} = \frac{1}{\text{det}(D)} \cdot \text{adj}(D)$<br /><br />where $\text{det}(D)$ is the determinant of matrix $D$ and $\text{adj}(D)$ is the adjugate of matrix $D$.<br /><br />First, let's calculate the determinant of matrix $D$:<br /><br />$\text{det}(D) = 1 \cdot (5 \cdot (-6) - 0 \cdot (-1)) - (-4) \cdot (2 \cdot (-6) - 1 \cdot (-1)) + (-3) \cdot (2 \cdot 5 - 1 \cdot 1) = -30 + 8 - 9 = -31$<br /><br />Now, let's calculate the adjugate of matrix $D$:<br /><br />$\text{adj}(D) = \begin{pmatrix} \text{det}(D_{11}) & -\text{det}(D_{12}) & \text{det}(D_{13}) \\ -\text{det}(D_{21}) & \text{det}(D_{22}) & -\text{det}(D_{23}) \\ \text{det}(D_{31}) & -\text{det}(D_{32}) & \text{det}(D_{33}) \end{pmatrix}$<br /><br />where $D_{ij}$ is the cofactor of the element in the $i$-th row and $j$-th column of matrix $D$.<br /><br />Calculating the cofactors, we get:<br /><br />$D_{11} = \begin{vmatrix} 3 & 2 \\ 0 & -6 \end{vmatrix} = -18$, $D_{12} = \begin{vmatrix} 3 & 2 \\ 1 & -6 \end{vmatrix} = -18$, $D_{13} = \begin{vmatrix} 3 & 2 \\ 1 & 0 \end{vmatrix} = -2$<br /><br />$D_{21} = \begin{vmatrix} -1 & 7 \\ 0 &