Let A=( 100 210 321 ). If u1 and u2 are column matrices such that A u1=( 1

step 1 Hi, we have given A equal to matrix 100, 210, 3211. And you 1 due to a column matrices such that

Let A=( 100 210 321 ). If u1 and u2 are...

Let $\mathrm{A}=\left(\begin{array}{c}100 \\ 210 \\ 321\end{array}\right)$. If $u_{1}$ and $u_{2}$ are column matrices such that $A u_{1}=\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)$ and $A u_{2}=\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)$, then $u_{1}+u_{2}$ is equal to (A) $\left(\begin{array}{l}-1 \\ 1 \\ 0\end{array}\right)$ (B) $\left(\begin{array}{l}-1 \\ 1 \\ -0\end{array}\right)$ (C) $\left(\begin{array}{c}-1 \\ -1 \\ 0\end{array}\right)$ (D) $\left(\begin{array}{l}1 \\ -1 \\ -1\end{array}\right)$

Let $\mathrm{A}=\left(\begin{array}{c}100 \\ 210 \\ 321\end{array}\right)$. If $u_{1}$ and $u_{2}$ are column matrices such that $A u_{1}=\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)$ and $A u_{2}=\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)$, then $u_{1}+u_{2}$ is equal to
(A) $\left(\begin{array}{l}-1 \\ 1 \\ 0\end{array}\right)$
(B) $\left(\begin{array}{l}-1 \\ 1 \\ -0\end{array}\right)$
(C) $\left(\begin{array}{c}-1 \\ -1 \\ 0\end{array}\right)$
(D) $\left(\begin{array}{l}1 \\ -1 \\ -1\end{array}\right)$

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Key Concepts

Linearity of Matrix Transformations

Matrix multiplication is a linear operation; that is, if you have a matrix A and two column vectors u1 and u2, then A(u1 + u2) equals A u1 plus A u2. This property is fundamental in linear algebra and allows one to deduce information about the sum of preimages when the images of individual vectors are known.

Solving Systems of Linear Equations

When given a matrix equation such as A u = b, the task is to find the vector u that satisfies the equation. This often involves setting up a system of linear equations corresponding to each row of A, and then employing techniques such as substitution, elimination, or matrix inversion to solve for u.

Triangular Matrices and Back Substitution

A triangular matrix (upper or lower) has a structure that simplifies solving systems of equations because the equations can be solved stepwise using back substitution (for upper triangular) or forward substitution (for lower triangular). This property is particularly useful when the matrix is already in a triangular form, as it significantly reduces computational complexity.

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