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Chapter 1: Q25E (page 1)

Question 25: Note that \(\left[ {\begin{array}{*{20}{c}}4&{ - 3}&1\\5&{ - 2}&5\\{ - 6}&2&{ - 3}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\\2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 7}\\{ - 3}\\{10}\end{array}} \right]\). Use this fact (and no row operations) to find scalars \({c_1},{c_2},{c_3}\) such that \(\left[ {\begin{array}{*{20}{c}}{ - 7}\\{ - 3}\\{10}\end{array}} \right] = {c_1}\left[ {\begin{array}{*{20}{c}}4\\5\\{ - 6}\end{array}} \right] + {c_2}\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 2}\\2\end{array}} \right] + {c_3}\left[ {\begin{array}{*{20}{c}}1\\5\\{ - 3}\end{array}} \right]\).

Short Answer

Expert verified

The values of \({c_1}\), \({c_2}\), and \({c_3}\)are given as \( - 3\), \( - 1\,\),and \(2\), respectively.

Step by step solution

01

Multiply the scalar into the matrix

When the scalar is multiplied by the matrix, it is multiplied by every element of the matrix.

\(\left[ {\begin{array}{*{20}{c}}{ - 7}\\{ - 3}\\{10}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{4{c_1}}\\{5{c_1}}\\{ - 6{c_1}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 3{c_2}}\\{ - 2{c_2}}\\{2{c_2}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{{c_3}}\\{5{c_3}}\\{ - 3{c_3}}\end{array}} \right]\)

02

Sum of the matrix

The sum of the matrix is calculated by adding all the corresponding elements of the matrix.

\(\left[ {\begin{array}{*{20}{c}}{ - 7}\\{ - 3}\\{10}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{4{c_1} - 3{c_2} + {c_3}}\\{5{c_1} - 2{c_2} + 5{c_3}}\\{ - 6{c_1} + 2{c_2} - 3{c_3}}\end{array}} \right]\)

03

Product of matrices

Write the linear combination of the matrix into the product of matrices.

\(\left[ {\begin{array}{*{20}{c}}{ - 7}\\{ - 3}\\{10}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4&{ - 3}&1\\5&{ - 2}&5\\{ - 6}&2&{ - 3}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\{{c_3}}\end{array}} \right]\)

04

Values of scalar

Now, compare the above matrix with the given matrix\(\left[ {\begin{array}{*{20}{c}}4&{ - 3}&1\\5&{ - 2}&5\\{ - 6}&2&{ - 3}\end{array}} \right]\left( {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\\2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 7}\\{ - 3}\\{10}\end{array}} \right]\).

On comparing, you get;

\(\begin{array}{l}{c_1} = - 3\\{c_2} = - 1\\{c_3} = 2\end{array}\)

Hence, the values of \({c_1}\), \({c_2}\), and \({c_3}\)are given as \( - 3\), \( - 1\,\),and \(2\), respectively.

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Most popular questions from this chapter

Suppose \({{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}\) are distinct points on one line in \({\mathbb{R}^3}\). The line need not pass through the origin. Show that \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is linearly dependent.

Write the reduced echelon form of a \(3 \times 3\) matrix A such that the first two columns of Aare pivot columns and

\(A = \left( {\begin{aligned}{*{20}{c}}3\\{ - 2}\\1\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}0\\0\\0\end{aligned}} \right)\).

Suppose the coefficient matrix of a linear system of three equations in three variables has a pivot position in each column. Explain why the system has a unique solution.

In Exercises 11 and 12, determine if \({\rm{b}}\) is a linear combination of \({{\mathop{\rm a}\nolimits} _1},{a_2}\) and \({a_3}\).

11.\({a_1} = \left[ {\begin{array}{*{20}{c}}1\\{ - 2}\\0\end{array}} \right],{a_2} = \left[ {\begin{array}{*{20}{c}}0\\1\\2\end{array}} \right],{a_3} = \left[ {\begin{array}{*{20}{c}}5\\{ - 6}\\8\end{array}} \right],{\mathop{\rm b}\nolimits} = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\6\end{array}} \right]\)

In Exercises 3 and 4, display the following vectors using arrows

on an \(xy\)-graph: u, v, \( - {\bf{v}}\), \( - 2{\bf{v}}\), u + v , u - v, and u - 2v. Notice that u - vis the vertex of a parallelogram whose other vertices are u, 0, and \( - {\bf{v}}\).

4. u and v as in Exercise 2
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