Question

Let \({{\mathop{\rm a}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}1\\4\\{ - 2}\end{array}} \right],{{\mathop{\rm a}\nolimits} _2} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 3}\\7\end{array}} \right],\) and \({\rm{b = }}\left[ {\begin{array}{*{20}{c}}4\\1\\h\end{array}} \right]\). For what values(s) of \(h\) is \({\mathop{\rm b}\nolimits} \) in the plane spanned by \({{\mathop{\rm a}\nolimits} _1}\) and \({{\mathop{\rm a}\nolimits} _2}\)?

Short answer

\({\mathop{\rm b}\nolimits} \) in the plane is spanned by \({a_1}\) and \({a_2}\) if and only if \(h = - 17\).

Step-by-step solution

Rewrite the matrix into a vector equation 

In \({\mathbb{R}^2}\), the sum of two vectors \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) is thevector addition\({\mathop{\rm u}\nolimits} + v\), which is obtained by adding the corresponding entries of \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \).

The scalar multiple of a vector \({\mathop{\rm u}\nolimits} \) by real number \(c\) is the vector \(c{\mathop{\rm u}\nolimits} \) obtained by multiplying each entry in \({\mathop{\rm u}\nolimits} \) by \(c\).

Use scalar multiplication and vector addition to rewrite the matrix into a vector equation \(\begin{aligned}{c}{x_1}\left[ {\begin{array}{*{20}{c}}1\\4\\{ - 2}\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 3}\\7\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}4\\1\\h\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{{x_1} - 2{x_2}}\\{4{x_1} - 3{x_2}}\\{ - 2{x_1} + 7{x_2}}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}4\\1\\h\end{array}} \right]\end{aligned}\).

Write the matrix into a vector equation

The vectors on the left and right sides are equal if and only if their corresponding entries are both equal. Thus,\({x_1}\)and\({x_2}\)make the vector equation\({x_1}{a_1} + {x_2}{a_2} = b\)if and only if\({x_1}\)and\({x_2}\)satisfy the system

Write the matrix into a vector equation.

\(\begin{array}{c}{x_1} - 2{x_2} = 4\\4{x_1} - 3{x_2} = 1\\ - 2{x_2} + 7{x_2} = h\end{array}\)

Convert the vector equation into an augmented matrix

A vector equation \({{\mathop{\rm x}\nolimits} _1}{a_1} + {x_2}{a_2} + ... + {x_n}{a_n} = b\) has the same solution set as the linear system whoseaugmented matrix is \(\left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{...}&{{a_n}}&b\end{array}} \right]\).

The augmented matrix for the vector equations \({x_1} - 2{x_2} = 4,4{x_1} - 3{x_2} = 1\) and \( - 2{x_2} + 7{x_2} = h\) is represented as:

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&4\\4&{ - 3}&1\\{ - 2}&7&h\end{array}} \right]\)

Apply row operation

Perform an elementaryrow operation to produce the first augmented matrix.

Replace row 3 by adding 2 times row 1 to row 3.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&4\\4&{ - 3}&1\\0&3&{8 + h}\end{array}} \right]\)

Apply row operation

Perform an elementaryrow operation to produce a second augmented matrix.

Replace row 2 by adding - 4 times row 1 to row 2.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&4\\0&5&{ - 15}\\0&3&{8 + h}\end{array}} \right]\)

Apply row operation

Perform an elementary row operation to produce a third augmented matrix.

Multiply row 2 by \(\frac{1}{5}\).

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&4\\0&1&{ - 3}\\0&3&{8 + h}\end{array}} \right]\)

Apply row operation

Perform an elementary row operation to produce a third augmented matrix.

Replace row 3 by adding -3 times row 2 to row 3.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&4\\0&1&{ - 3}\\0&0&{17 + h}\end{array}} \right]\)

Convert the matrix into the equation

The vector\({\mathop{\rm y}\nolimits} \)defined by\(y = {c_1}{v_1} + .... + {c_p}{v_p}\)is called alinear combination of\({{\mathop{\rm v}\nolimits} _1},{v_2},...,{v_p}\)with weights\({c_1},{c_2},...,{c_p}\).

To obtain the solution of the system of equations, you have to convert the augmented matrix into the system of equations.

Write the obtained matrix \(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&4\\0&1&{ - 3}\\0&0&{17 + h}\end{array}} \right]\) into the equation notation.

\(\begin{array}{l}{x_1} - 2{x_2} = 4\\{x_2} = - 3\\0 = 17 + h\end{array}\)

If \(17 + h = 0\),i.e., \(h = - 17\), then the system is consistent; there exists a solution.

Thus, \({\mathop{\rm b}\nolimits} \)in the plane is spanned by \({a_1}\) and \({a_2}\) if and only if \(h = - 17\).