Question

If $x,y,z$ are in A.P., then the value of the det$A$, where, A = \left( {\begin{array}{*{20}{c}} 4&5&6&x; \\\ 5&6&7&y; \\\ 6&7&8&z; \\\ x&y;&z;&0 \end{array}} \right)

Answer 1

Hint- Since, the given values are in AP, we will try to represent the elements in terms of that. As per our need we will rewrite the second column of the given matrix in terms of first and third columns.
Again, we know that, if any two rows or columns of a matrix are the same, then the value of the determinant is always zero.

Complete step by step answer:
It is given that,

4&5&6&x; \\\ 5&6&7&y; \\\ 6&7&8&z; \\\ x&y;&z;&0 \end{array}} \right)$$ Also, it is given, $$x,y,z$$ are in A.P. So, $$x + z = 2y$$ Now, the determinant of the given matrix we get, $$\det A = \left| {\begin{array}{*{20}{c}} 4&5&6&x; \\\ 5&6&7&y; \\\ 6&7&8&z; \\\ x&y;&z;&0 \end{array}} \right|$$ We can rewrite the determinant as, $$\det A = \left| {\begin{array}{*{20}{c}} 4&{\dfrac{{4 + 6}}{2}}&6&x; \\\ 5&{\dfrac{{5 + 7}}{2}}&7&y; \\\ 6&{\dfrac{{6 + 8}}{2}}&8&z; \\\ x&{\dfrac{{x + z}}{2}}&z;&0 \end{array}} \right|$$ From second column we can take $$\dfrac{1}{2}$$ as common and we have, $$\det A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}} 4&{4 + 6}&6&x; \\\ 5&{5 + 7}&7&y; \\\ 6&{6 + 8}&8&z; \\\ x&{x + z}&z;&0 \end{array}} \right|$$ We can rewrite the determinant as sum form. $$\det A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}} 4&4&6&x; \\\ 5&5&7&y; \\\ 6&6&8&z; \\\ x&x;&z;&0 \end{array}} \right| + \dfrac{1}{2}\left| {\begin{array}{*{20}{c}} 4&6&6&x; \\\ 5&7&7&y; \\\ 6&8&8&z; \\\ x&z;&z;&0 \end{array}} \right|$$ We know that, if two columns or rows are equal, then the value of the determinant is zero. Here, the first and second column of the first determinant and the second and third column of the second determinant are the same. So, the value of the determinant will be zero. Hence, $$\det A = 0 + 0 = 0$$ Therefore, $$\det A = 0$$ Note – In linear algebra, the determinant of a matrix is a scalar value that can be calculated from the elements of a square matrix. It is denoted by $$\det A$$or $$\left| A \right|$$. For $$2 \times 2$$ matrix, $$\left| A \right| = \left| {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right| = ad - bc$$ For $$3 \times 3$$matrix, $$\left| A \right| = \left( {\begin{array}{*{20}{c}} a&b;&c; \\\ d&e;&f; \\\ g&h;&i; \end{array}} \right) = a(ei - hf) - b(di - gf) + c(dh - eg)$$ For $$4 \times 4$$ and higher order matrices, it is quite difficult to find the determinant by applying the formula. For these types of sums, we will do rows or column operation as per the requirement and transform the matrix into simpler form. We can find the determinant for square matrices.