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

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

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

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

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

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

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

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.