The problem asks us to multiply two matrices: $ \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 7 \\ 6 & 8 \end{bmatrix} $. We need to find the resulting matrix. The first entry (17) and second entry (23) are given. We have to find the third and fourth entries.

AlgebraMatrix MultiplicationLinear Algebra
2025/3/6

1. Problem Description

The problem asks us to multiply two matrices:
\begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix}
\begin{bmatrix}
5 & 7 \\
6 & 8
\end{bmatrix}
$.
We need to find the resulting matrix. The first entry (17) and second entry (23) are given. We have to find the third and fourth entries.

2. Solution Steps

Matrix multiplication is performed as follows:
For a 2x2 matrix multiplied by a 2x2 matrix:
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\begin{bmatrix}
e & f \\
g & h
\end{bmatrix}
=
\begin{bmatrix}
ae+bg & af+bh \\
ce+dg & cf+dh
\end{bmatrix}
We are given the matrices:
\begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix}
\begin{bmatrix}
5 & 7 \\
6 & 8
\end{bmatrix}
The resulting matrix will be:
\begin{bmatrix}
(1)(5) + (2)(6) & (1)(7) + (2)(8) \\
(3)(5) + (4)(6) & (3)(7) + (4)(8)
\end{bmatrix}
=
\begin{bmatrix}
5 + 12 & 7 + 16 \\
15 + 24 & 21 + 32
\end{bmatrix}
=
\begin{bmatrix}
17 & 23 \\
39 & 53
\end{bmatrix}
The third entry is 3(5)+4(6)=15+24=393(5) + 4(6) = 15 + 24 = 39.
The fourth entry is 3(7)+4(8)=21+32=533(7) + 4(8) = 21 + 32 = 53.

3. Final Answer

\begin{bmatrix}
17 & 23 \\
39 & 53
\end{bmatrix}
39
53

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