Solve the problem.
-Let $A = \left[ \begin{array} { r r r } 1 & - 3 & 2 \ - 2 & 5 & - 1 \ 3 & - 4 & 5 \end{array} \right]$ and $\mathbf { b } = \left[ \begin{array} { l } b _ { 1 } \ b _ { 2 } \ b _ { 3 } \end{array} \right]$.
Determine if the equation $\mathrm { Ax } = \mathrm { b }$ is consistent for all possible $\mathrm { b } _ { 1 } , \mathrm {~b} _ { 2 } , \mathrm {~b} _ { 3 }$. If the equation is not consistent for all possible $b _ { 1 } , b _ { 2 } , b _ { 3 }$, give a description of the set of all $\mathbf { b }$ for which the equation is consistent (i.e., a condition which must be satisfied by $b _ { 1 } , b _ { 2 } , b _ { 3 }$ ).
A) Equation is consistent for all $b _ { 1 } , b _ { 2 } , b _ { 3 }$ satisfying $- 3 b _ { 1 } + b _ { 3 } = 0$.
B) Equation is consistent for all $b _ { 1 } , b _ { 2 } , b _ { 3 }$ satisfying $2 b _ { 1 } + b _ { 2 } = 0$.
C) Equation is consistent for all possible $b _ { 1 } , b _ { 2 } , b _ { 3 }$.
D) Equation is consistent for all $b _ { 1 } , b _ { 2 }$, $b _ { 3 }$ satisfying $7 b _ { 1 } + 5 b _ { 2 } + b _ { 3 } = 0$.

Question

Quizplus · Accepted Answer

To determine if the equation $Ax = b$ is consistent for all possible $b_1, b_2, b_3$, we need to check if the matrix $A$ is invertible (i.e., has a non-zero determinant). If $A$ is invertible, then for every $b$, there exists a unique solution $x$, making the system consistent for all $b$. Calculating the determinant of $A$ and finding it non-zero would confirm this, but since the specific calculation isn't shown here, we rely on the principle that an $n \times n$ matrix with a non-zero determinant is invertible, thus ensuring consistency for all $b$.

Solve the Problem A=[1−32−25−13−45]A = \left[ \begin{array} { r r r } 1 & - 3 & 2 \\ - 2 & 5 & - 1 \\ 3 & - 4 & 5 \end{array} \right]A=​1−23​−35−4​2−15​​

Solve the Problem $A = \left[ \begin{array} { r r r } 1 & - 3 & 2 \\ - 2 & 5 & - 1 \\ 3 & - 4 & 5 \end{array} \right]$