Question

Find the determinant of the given matrix \left[ {\begin{array}{*{20}{c}} {18}&{40}&{89} \\\ {40}&{89}&{198} \\\ {89}&{198}&{440} \end{array}} \right] =
A. -8
B. -6
C. -1
D. 0

Answer 1

Here we use the row operations of a matrix to solve for the value of the given determinant. We use row operations and transpose (switch columns into rows and vice-versa) the matrix when required.

• In a row operation we change the elements of a row by adding, subtracting, multiplying or dividing the elements of the row with reference to other row. Example: ${R_n} \to {R_n} - 2{R_m}$where m, n are row numbers.
• Determinant of a matrix is same as determinant of transpose of a matrix.
• We use the formula to solve a determinant

a&b;&c; \\\ d&e;&f; \\\ g&h;&i; \end{array}} \right| = a\left| {\begin{array}{*{20}{c}} e&f; \\\ h&i; \end{array}} \right| - b\left| {\begin{array}{*{20}{c}} d&f; \\\ g&i; \end{array}} \right| + c\left| {\begin{array}{*{20}{c}} d&e; \\\ g&h; \end{array}} \right|.$$ Then use the following formula to calculate the determinant of order $2 \times 2.$ $$\left| {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right| = ad - bc.$$ **Complete step-by-step answer:** We have the matrix $$\left[ {\begin{array}{*{20}{c}} {18}&{40}&{89} \\\ {40}&{89}&{198} \\\ {89}&{198}&{440} \end{array}} \right]$$ To find the determinant of the matrix we make the first row as simple as possible using row and column operations because when the elements of first row will be simple or minimum, then calculations will become easy. $${R_1},{R_2},{R_3}$$are the three rows from top to bottom and $${C_1},{C_2},{C_3}$$are the three columns from left to right. Now we take the matrix $$\left[ {\begin{array}{*{20}{c}} {18}&{40}&{89} \\\ {40}&{89}&{198} \\\ {89}&{198}&{440} \end{array}} \right]$$ We apply the row operation $${R_2} \to {R_2} - 2{R_1}$$ $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} {18}&{40}&{89} \\\ {40}&{89}&{198} \\\ {89}&{198}&{440} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {18}&{40}&{89} \\\ 4&9&{20} \\\ {89}&{198}&{440} \end{array}} \right]$$ Again we apply row operation $${R_3} \to {R_3} - 5{R_1}$$ $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} {18}&{40}&{89} \\\ {40}&{89}&{198} \\\ {89}&{198}&{440} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {18}&{40}&{89} \\\ 4&9&{20} \\\ { - 1}&{ - 2}&{ - 5} \end{array}} \right]$$ Now we apply row operation $${R_1} \to {R_1} + 20{R_3}$$ $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} {18}&{40}&{89} \\\ {40}&{89}&{198} \\\ {89}&{198}&{440} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 2}&0&{ - 11} \\\ 4&9&{20} \\\ { - 1}&{ - 2}&{ - 5} \end{array}} \right]$$ Now we apply row transformation $${R_2} \to {R_2} + 2{R_1}$$ $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} {18}&{40}&{89} \\\ {40}&{89}&{198} \\\ {89}&{198}&{440} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 2}&0&{ - 11} \\\ 0&9&{ - 2} \\\ { - 1}&{ - 2}&{ - 5} \end{array}} \right]$$ Now we know that if a constant k is multiplied o all elements of the row, we can write the matrix as $$\left| {\begin{array}{*{20}{c}} a&b;&c; \\\ {kd}&{ke}&{kf} \\\ g&h;&i; \end{array}} \right| = k\left| {\begin{array}{*{20}{c}} a&b;&c; \\\ d&e;&f; \\\ g&h;&i; \end{array}} \right|$$ So we take the value of -1 from First and second row outside the matrix. $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} {18}&{40}&{89} \\\ {40}&{89}&{198} \\\ {89}&{198}&{440} \end{array}} \right] = ( - 1)( - 1)\left[ {\begin{array}{*{20}{c}} 2&0&{11} \\\ 0&9&{ - 2} \\\ 1&2&5 \end{array}} \right]$$ $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} {18}&{40}&{89} \\\ {40}&{89}&{198} \\\ {89}&{198}&{440} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2&0&{11} \\\ 0&9&{ - 2} \\\ 1&2&5 \end{array}} \right]$$ Now we apply row transformation $${R_1} \to {R_1} - 2{R_3}$$ $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} {18}&{40}&{89} \\\ {40}&{89}&{198} \\\ {89}&{198}&{440} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&{ - 4}&1 \\\ 0&9&{ - 2} \\\ 1&2&5 \end{array}} \right]$$ Now we find the determinant of the matrix on RHS. Since we know determinant of a matrix is same as determinant of transpose of a matrix. $$\therefore $$Det $$\left[ {\begin{array}{*{20}{c}} 0&{ - 4}&1 \\\ 0&9&{ - 2} \\\ 1&2&5 \end{array}} \right] = $$ Det $$\left[ {\begin{array}{*{20}{c}} 0&0&1 \\\ { - 4}&9&2 \\\ 1&{ - 2}&5 \end{array}} \right]$$ We know that determinant of a matrix can be calculated as: Det $$\left[ {\begin{array}{*{20}{c}} 0&0&1 \\\ { - 4}&9&2 \\\ 1&{ - 2}&5 \end{array}} \right] = 0[9 \times 5 - 2 \times - 2] - 0[ - 4 \times 5 - 1 \times 2] + 1[ - 4 \times - 2 - 1 \times 9]$$ Det $$\left[ {\begin{array}{*{20}{c}} 0&0&1 \\\ { - 4}&9&2 \\\ 1&{ - 2}&5 \end{array}} \right] = 1[8 - 9]$$ Det $$\left[ {\begin{array}{*{20}{c}} 0&0&1 \\\ { - 4}&9&2 \\\ 1&{ - 2}&5 \end{array}} \right] = - 1$$ **So, the correct option is C.** **Note:** Students many times make the mistake of writing negative signs after taking the transpose of the matrix which is not required because we use the property that transpose has the same determinant as that of matrix. Also, always try to apply row operations that make the matrix easier in the next step and not more complex.