Solution: Total sum is $5 \times 22 = 110$. Sum of known values: $20 + 21 + 23 + 24 = 88$. The fifth value is $110 - 88 = 22$. \boxed{22}Question: Find the matrix $\mathbf{M}$ such that $\mathbf{M} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ and $\mathbf{M} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$.

["Understanding Matrix Transformation Through Column Summation", "When analyzing linear transformations, matrices encode how basis vectors are transformed. Given a matrix $\mathbf{M}$ such that:", "$$\n\mathbf{M} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 2 \ -1 \end{pmatrix} \quad \ ext{and} \quad \mathbf{M} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 4 \end{pmatrix},\n$$", "we can construct $\mathbf{M}$ using a foundational linearity principle: the columns of $\mathbf{M}$ are exactly the images of the standard basis vectors under the transformation.", "### Step 1: Build the matrix from transformed basis vectors", "By definition of matrix multiplication:", "- The first column of $\mathbf{M}$ is $\mathbf{M} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 2 \ -1 \end{pmatrix}$,\n- The second column is $\mathbf{M} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 4 \end{pmatrix}$.", "Thus, combining these columns:", "$$\n\mathbf{M} = \begin{pmatrix} 2 & 3 \ -1 & 4 \end{pmatrix}\n$$", "### Step 2: Verify using vector sum insight", "Interestingly, the total sum of matrix elements from column-wise sum idea relates to matrix trace and sum principles. The sum of all entries in $\mathbf{M}$ is:", "$$\n(2 + 3) + (-1 + 4) = 5 + 3 = 8\n$$", "But our known total sum $5 \ imes 22 = 110$ mentioned in prior context appears unrelated to this specific matrix — however, the core idea of constructing matrices from basis transformations remains vital.", "For our matrix $\mathbf{M} = \begin{pmatrix} 2 & 3 \ -1 & 4 \end{pmatrix}$, the calculation aligns perfectly with the transformation rules:", "$$\n\mathbf{M} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 2 \cdot 1 + 3 \cdot 0 \ -1 \cdot 1 + 4 \cdot 0 \end{pmatrix} = \begin{pmatrix} 2 \ -1 \end{pmatrix}, \quad\n\mathbf{M} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 2 \cdot 0 + 3 \cdot 1 \ -1 \cdot 0 + 4 \cdot 1 \end{pmatrix} = \begin{pmatrix} 3 \ 4 \end{pmatrix}\n$$", "Both conditions are satisfied.", "### Conclusion", "The matrix $\mathbf{M}$ that satisfies the given transformations is:", "$$\n\boxed{ \begin{pmatrix} 2 & 3 \ -1 & 4 \end{pmatrix} }\n$$", "This method exemplifies how linear algebra leverages basis vectors to define transformation matrices. While the sum $5 \ imes 22 = 110$ provided in the initial prompt serves as an illustrative numerical example, the key takeaway here is constructing matrices directly from transformation rules — a fundamental and powerful approach in linear algebra."]