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Question

Question: Find the determinant of the given matrix. \[\det \left( {\begin{array}{*{20}{c}} {18}&{40}&{89} ...

Find the determinant of the given matrix. \det \left( {\begin{array}{*{20}{c}} {18}&{40}&{89} \\\ {40}&{89}&{198} \\\ {89}&{198}&{440} \end{array}} \right) =
A) 8 - 8
B) 6 - 6
C) 1 - 1
D) 00

Explanation

Solution

In this question we have to choose the determinant value of the given matrix. First, we have to calculate the determinant of the given matrix. By using the relation of the matrix to the determinant value we are going to find the determinant.

Formula used:
Let us consider a matrix of 3×33 \times 3 as \left( {\begin{array}{*{20}{c}} a&b;&c; \\\ d&e;&f; \\\ g&h;&i; \end{array}} \right). The determinant of the same matrix is,

a&b;&c; \\\ d&e;&f; \\\ g&h;&i; \end{array}} \right) = a(ei - fh) - b(di - fg) + c(dh - eg)$$ Applying the formula we will find the value of the determinant of the given matrix. **Complete step by step answer:** We have to find the $$\det \left( {\begin{array}{*{20}{c}} {18}&{40}&{89} \\\ {40}&{89}&{198} \\\ {89}&{198}&{440} \end{array}} \right)$$ Let us consider a matrix of $$3 \times 3$$ as $$\left( {\begin{array}{*{20}{c}} a&b;&c; \\\ d&e;&f; \\\ g&h;&i; \end{array}} \right)$$. The determinant of the same matrix is, $$ \Rightarrow \det \left( {\begin{array}{*{20}{c}} a&b;&c; \\\ d&e;&f; \\\ g&h;&i; \end{array}} \right) = a(ei - fh) - b(di - fg) + c(dh - eg)$$ Applying the above formula, we get, $$ \Rightarrow \det \left( {\begin{array}{*{20}{c}} {18}&{40}&{89} \\\ {40}&{89}&{198} \\\ {89}&{198}&{440} \end{array}} \right) = 18(89 \times 440 - 198 \times 198) - 40(40 \times 440 - 198 \times 89) + 89(40 \times 198 - 89 \times 89)$$ Multiplying the terms that in the brackets we get, $$ \Rightarrow \det = 18(39160 - 39204) - 40(17600 - 17622) + 89(7920 - 7921)$$ Subtracting the terms inside the brackets we get, $$ \Rightarrow \det =18( - 44) - 40( - 22) + 89( - 1)$$ Simplifying we get, $$ \Rightarrow \det = - 792 + 880 - 89$$ Solving we get, the determinant as, $$ \Rightarrow \det = - 1$$ **$\therefore $ The determinant of the given matrix is $$ - 1$$. Hence, the correct option is c) $$ - 1$$.** **Note:** The determinant of a matrix is a number that is specially defined only for square matrices. Determinants are mathematical objects that are very useful in the analysis and solution of systems of linear equations. Every square matrix $$A = {[a]_{ij}}$$ of order n, we can associate a number (real or complex) called determinant of the square matrix $$A$$, where a = $$(i,j)$$ th element of $$A$$. This may be thought of as a function which associates each square matrix with a unique number (real or complex). If $$M$$ is the set of square matrices, $$K$$ is the set of numbers (real or complex) and $$f:M \to K$$ is defined by $$f(A) = k$$, where $$A \in M$$and $$k \in K$$, then $$f(A)$$ is called the determinant of $$A$$. It is also denoted by $$\left| A \right|$$ or $$\det A$$ or $\Delta $. Let us consider a matrix of $$3 \times 3$$ as $$\left( {\begin{array}{*{20}{c}} a&b;&c; \\\ d&e;&f; \\\ g&h;&i; \end{array}} \right)$$. The determinant of the same matrix is, $$\det \left( {\begin{array}{*{20}{c}} a&b;&c; \\\ d&e;&f; \\\ g&h;&i; \end{array}} \right) = a(ei - fh) - b(di - fg) + c(dh - eg)$$