Question

Question: Let \({\bf{u}} = \left( {\begin{array}{*{20}{c}}7\\2\\5\end{array}} \right)\), \({\bf{v}} = \left( {\begin{array}{*{20}{c}}3\\1\\3\end{array}} \right)\), and \({\bf{w}} = \left( {\begin{array}{*{20}{c}}6\\1\\0\end{array}} \right)\). It can be shown that \(3{\bf{u}} - 5{\bf{v}} - {\bf{w}} = 0\). Use this fact (and no row operations) to find \({x_1}\) and \({x_2}\) that satisfy the equation\(\left( {\begin{array}{*{20}{c}}7&3\\2&1\\5&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\1\\0\end{array}} \right)\).

Short answer

The required valuesare \({x_1} = 3\) and \({x_2} = - 5\).

Step-by-step solution

Describing the relation \(3{\bf{u}} - 5{\bf{v}} - {\bf{w}} = 0\)

Substitute the vectors \({\bf{u}} = \left( {\begin{aligned}{{}}7\\2\\5\end{aligned}} \right)\), \({\bf{v}} = \left( {\begin{aligned}{{}}3\\1\\3\end{aligned}} \right)\), and \({\bf{w}} = \left( {\begin{aligned}{{}}6\\1\\0\end{aligned}} \right)\) in the relation \(3{\bf{u}} - 5{\bf{v}} - {\bf{w}} = 0\).

\(\begin{aligned}{}3{\bf{u}} - 5{\bf{v}} - {\bf{w}} &= 3\left( {\begin{aligned}{{}}7\\2\\5\end{aligned}} \right) - 5\left( {\begin{aligned}{{}}3\\1\\3\end{aligned}} \right) - \left( {\begin{aligned}{{}}6\\1\\0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{21}\\6\\{15}\end{aligned}} \right) - \left( {\begin{aligned}{{}}{15}\\5\\{15}\end{aligned}} \right) - \left( {\begin{aligned}{{}}6\\1\\0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{21 - 15 - 6}\\{6 - 5 - 1}\\{15 - 15 - 0}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}0\\0\\0\end{aligned}} \right)\end{aligned}\)

Thus, \(3{\bf{u}} - 5{\bf{v}} - {\bf{w}} = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right)\).

Simplifying the above equation

Add vector w on both sides of the equation \(3{\bf{u}} - 5{\bf{v}} - {\bf{w}} = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right)\).

\(\begin{array}{c}3{\bf{u}} - 5{\bf{v}} - {\bf{w}} + {\bf{w}} = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right) + {\bf{w}}\\3{\bf{u}} - 5{\bf{v}} = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right) + {\bf{w}}\end{array}\)

Thus, \(3{\bf{u}} - 5{\bf{v}} = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right) + {\bf{w}}\).

Step 3:Writing the above equation in vector form

The left part of the equation \(3{\bf{u}} - 5{\bf{v}} = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right) + {\bf{w}}\) is \(3{\bf{u}} - 5{\bf{v}}\).

Using the matrix-vector product,\(3{\bf{u}} - 5{\bf{v}}\) can be written as shown below:

\(3{\bf{u}} - 5{\bf{v}} = \left( {\begin{array}{*{20}{c}}{\bf{u}}&{\bf{v}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}3\\{ - 5}\end{array}} \right)\)

The equation becomes

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}{\bf{u}}&{\bf{v}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}3\\{ - 5}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right) + {\bf{w}}\\\left( {\begin{array}{*{20}{c}}{\bf{u}}&{\bf{v}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}3\\{ - 5}\end{array}} \right) = {\bf{w}}.\end{array}\)

Thus, \(\left( {\begin{array}{*{20}{c}}{\bf{u}}&{\bf{v}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}3\\{ - 5}\end{array}} \right) = {\bf{w}}\).

Obtaining the values of unknowns

Compare the equation \(\left( {\begin{array}{*{20}{c}}{\bf{u}}&{\bf{v}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}3\\{ - 5}\end{array}} \right) = {\bf{w}}\) with the equation \(\left( {\begin{array}{*{20}{c}}7&3\\2&1\\5&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\1\\0\end{array}} \right)\)to get\({\bf{u}} = \left( {\begin{array}{*{20}{c}}7\\2\\5\end{array}} \right)\), \({\bf{v}} = \left( {\begin{array}{*{20}{c}}3\\1\\3\end{array}} \right)\),and \({\bf{w}} = \left( {\begin{array}{*{20}{c}}6\\1\\0\end{array}} \right)\).

Also, \(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3\\{ - 5}\end{array}} \right)\).

Thus, the required values are\({x_1} = 3\) and \({x_2} = - 5\).