1. ## Matrix problem

Question : If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$ Find $C = A^T . B^T$ and the determinant

Solution :

$C = \begin{bmatrix} 23 & 31 \\ 34 & 46 \end{bmatrix}$.......Is this correct ???

Determinant

|C| = 4.............Is this correct ???

2. Originally Posted by zorro
Question : If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$ Find $C = A^T . B^T$ and the determinant

Solution :

$C = \begin{bmatrix} 23 & 31 \\ 34 & 46 \end{bmatrix}$.......Is this correct ???

Determinant

|C| = 4.............Is this correct ???
yes

3. Originally Posted by zorro
Question : If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$ Find $C = A^T . B^T$ and the determinant

Solution :

$C = \begin{bmatrix} 23 & 31 \\ 34 & 46 \end{bmatrix}$.......Is this correct ???

Determinant

|C| = 4.............Is this correct ???
$A^T = \left[\begin{matrix}1 & 3\\2 & 4\end{matrix}\right]$

$B^T = \left[\begin{matrix}5 & 7\\6 & 8\end{matrix}\right]$.

So $C = A^TB^T$

$= \left[\begin{matrix} 1 & 3\\2 & 4\end{matrix}\right]\left[\begin{matrix} 5 & 7\\6 & 8\end{matrix}\right]$

$= \left[\begin{matrix}23 & 31\\34 & 46\end{matrix}\right]$.

$|C| = \left|\begin{matrix}23 & 31\\34 & 46\end{matrix}\right|$

$= 23\cdot 46 - 31\cdot 34$

4. Thanks mite

5. Originally Posted by Prove It
$A^T = \left[\begin{matrix}1 & 3\\2 & 4\end{matrix}\right]$

$B^T = \left[\begin{matrix}5 & 7\\6 & 8\end{matrix}\right]$.

So $C = A^TB^T$

$= \left[\begin{matrix} 1 & 3\\2 & 4\end{matrix}\right]\left[\begin{matrix} 5 & 7\\6 & 8\end{matrix}\right]$

$= \left[\begin{matrix}23 & 31\\34 & 50\end{matrix}\right]$.

$|C| = \left|\begin{matrix}23 & 31\\34 & 50\end{matrix}\right|$

$= 23\cdot 50 - 31\cdot 34$
the last entri for matrix $C$ is $2 \times 7 + 4 \times 8 = 14 + 32 = 46$, not $50$