Question

Solve the equation \left| {\begin{array}{*{20}{c}} {x - 2}&{2x - 3}&{3x - 4} \\\ {x - 4}&{2x - 9}&{3x - 16} \\\ {x - 8}&{2x - 27}&{3x - 64} \end{array}} \right| = 0

Answer 1

It is advisable that we open the determinant using properties else directly opening the determinant will be cumbersome and will be computationally tough for us. The use of properties before opening it will help us to reduce it into smaller terms.

Complete step-by-step solution:
Let us use the properties of determinants to reduce the determinant into smaller terms,
${C_2} \to {C_2} - 2{C_1}$

{x - 2}&{2x - 3 - (2x - 4)}&{3x - 4} \\\ {x - 4}&{2x - 9 - (2x - 8)}&{3x - 16} \\\ {x - 8}&{2x - 27 - (2x - 16)}&{3x - 64} \end{array}} \right| = 0 \\\ \Rightarrow \left| {\begin{array}{*{20}{c}} {x - 2}&1&{3x - 4} \\\ {x - 4}&{ - 1}&{3x - 16} \\\ {x - 8}&{ - 11}&{3x - 64} \end{array}} \right| = 0 $$ $${C_3} \to {C_3} - 3{C_1}$$ $$ \Rightarrow \left| {\begin{array}{*{20}{c}} {x - 2}&1&{3x - 4 - (3x - 6)} \\\ {x - 4}&{ - 1}&{3x - 16 - (3x - 12)} \\\ {x - 8}&{ - 11}&{3x - 64 - (3x - 24)} \end{array}} \right| = 0 \\\ \Rightarrow \left| {\begin{array}{*{20}{c}} {x - 2}&1&2 \\\ {x - 4}&{ - 1}&{ - 4} \\\ {x - 8}&{ - 11}&{ - 40} \end{array}} \right| = 0 $$ Let us do another transformation in the above obtained determinant to get a simpler one, $${R_2} \to {R_2} - {R_1}$$ $$ \left| {\begin{array}{*{20}{c}} {x - 2}&1&2 \\\ {x - 4 - x + 2}&{ - 1 - 1}&{ - 4 - 2} \\\ {x - 8}&{ - 11}&{ - 40} \end{array}} \right| = 0 \\\ \Rightarrow \left| {\begin{array}{*{20}{c}} {x - 2}&1&2 \\\ { - 2}&{ - 2}&{ - 6} \\\ {x - 8}&{ - 11}&{ - 40} \end{array}} \right| = 0 $$ $${R_3} \to {R_3} - {R_1}$$ $$ \Rightarrow \left| {\begin{array}{*{20}{c}} {x - 2}&1&2 \\\ { - 2}&{ - 2}&{ - 6} \\\ {x - 8 - x + 2}&{ - 11 - 1}&{ - 40 - 2} \end{array}} \right| = 0 \\\ \Rightarrow \left| {\begin{array}{*{20}{c}} {x - 2}&1&2 \\\ { - 2}&{ - 2}&{ - 6} \\\ { - 6}&{ - 12}&{ - 42} \end{array}} \right| = 0 $$ Now let us expand the determinant along the row $${R_1}$$and equate it to zero i.e. $$ \Rightarrow (x - 2)(84 - 72) - 1.(84 - 36) + 2.(24 - 12) = 0 \\\ \Rightarrow (x - 2).12 - 48 + 24 = 0 \\\ \Rightarrow 12x - 24 - 24 = 0 \\\ \Rightarrow 12x - 48 = 0 \\\ \Rightarrow 12x = 48 \\\ \Rightarrow x = 4 $$ **Hence the solution for the above equation is $$x = 4$$.** **Additional information:** A matrix is an array of many numbers. For a rectangular matrix, i.e. a matrix with the same wide variety of rows and columns, one can seize crucial facts approximately the matrix in a just single variety, known as the determinant. The determinant is useful for solving linear equations, taking pictures of how linear transformations change region or volume, and exchanging variables in integrals. The determinant can be viewed as a characteristic whose input is a square matrix and whose output is quite a number. If $$n$$ is the quantity of rows and columns in the matrix, remember, we're managing rectangular matrices, we are able to call our matrix an $$n \times n$$ matrix. The best rectangular matrix is a $$1 \times 1$$ matrix, which is not very thrilling because it consists of just a single number. The determinant of a $$1 \times 1$$ matrix is that range itself. **Note:** It is very important and advised that the determinant with higher numbered degree terms should be reduced to smaller terms. Thus it helps us to compute the determinant easier and evaluate it in an easier way with the help of properties.