Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 3 (a)/Solution 1

The intersection point ${\mathbf {x}}$, satisfies both equations, i.e.,

${\mathbf {x}}=t{\mathbf {a}}={\mathbf {q}}+s_{1}{\mathbf {b}}_{1}+s_{2}{\mathbf {b}}_{2}$

for some $t$, $s_{1}$ and $s_{2}$.

Therefore, to find the intersection point ${\mathbf {x}}$, we first find $t$, $s_{1}$ and $s_{2}$ which solve

$t{\mathbf {a}}={\mathbf {q}}+s_{1}{\mathbf {b}}_{1}+s_{2}{\mathbf {b}}_{2}$.

We first plug the given vectors into the equation. Thus,

$t{\begin{bmatrix}0\\1\\2\end{bmatrix}}={\begin{bmatrix}1\\1\\1\end{bmatrix}}+s_{1}{\begin{bmatrix}0\\1\\0\end{bmatrix}}+s_{2}{\begin{bmatrix}1\\2\\-1\end{bmatrix}}$

$\implies t{\begin{bmatrix}0\\1\\2\end{bmatrix}}-s_{1}{\begin{bmatrix}0\\1\\0\end{bmatrix}}-s_{2}{\begin{bmatrix}1\\2\\-1\end{bmatrix}}={\begin{bmatrix}1\\1\\1\end{bmatrix}}$

$\implies t{\begin{bmatrix}0\\1\\2\end{bmatrix}}+s_{1}{\begin{bmatrix}0\\-1\\0\end{bmatrix}}+s_{2}{\begin{bmatrix}-1\\-2\\1\end{bmatrix}}={\begin{bmatrix}1\\1\\1\end{bmatrix}}.$

Now, we write the system of linear equations in the form $A{\begin{bmatrix}t\\s_{1}\\s_{2}\end{bmatrix}}={\mathbf {b}}$:

${\begin{bmatrix}0&0&-1\\1&-1&-2\\2&0&1\end{bmatrix}}{\begin{bmatrix}t\\s_{1}\\s_{2}\end{bmatrix}}={\begin{bmatrix}1\\1\\1\end{bmatrix}}.$

Solving this system by Gaussian elimination now gives

$\left[{\begin{array}{ccc|c}0&0&-1&1\\1&-1&-2&1\\2&0&1&1\end{array}}\right]\to \left[{\begin{array}{ccc|c}0&0&-1&1\\-2&2&4&-2\\2&0&1&1\end{array}}\right]\to \left[{\begin{array}{ccc|c}0&0&-1&1\\0&2&5&-1\\2&0&1&1\end{array}}\right]\to \left[{\begin{array}{ccc|c}0&0&-1&1\\0&2&0&4\\2&0&0&2\end{array}}\right]\to \left[{\begin{array}{ccc|c}1&0&0&1\\0&1&0&2\\0&0&1&-1\end{array}}\right]$

and it follows that $t=1$, $s_{1}=2$, and $s_{2}=-1.$

Hence,

${\mathbf {x}}=t{\mathbf {a}}={\mathbf {a}}=\color {blue}{\begin{bmatrix}0\\1\\2\end{bmatrix}}.$