Question

Find the integral value of x, if \left| {\begin{array}{*{20}{c}} {{x^2}}&x&1 \\\ 0&2&1 \\\ 3&1&4 \end{array}} \right| = 28

Answer 1

Hint: The given problem consists of the determinant of a matrix which when solved is equal to the value 28. Thus it forms a quadratic equation of x. Now we just need to solve the quadratic equation. So we will get two values of x. The integer value between these two values will be the integral value of x.

Complete step-by-step answer:
Given \left| {\begin{array}{*{20}{c}} {{x^2}}&x&1 \\\ 0&2&1 \\\ 3&1&4 \end{array}} \right| = 28

A matrix is a rectangular array of numbers or other mathematical objects for which operations such as addition and multiplication are defined. It can also be defined as a matrix is a collection of numbers arranged into a fixed number of rows and columns.
Determinant of a scalar value that can be computed from the elements of a square matrix and encodes certain properties of the linear transformation described by the matrix. The determinant of a matrix A is denoted by det (A), det A or $\left| {\text{A}} \right|$ .
If {\text{A = }}\left| {\begin{array}{*{20}{c}} a&b&c \\\ d&{\text{e}}&f \\\ g&h&{\text{i}} \end{array}} \right|
Then $\left| {\text{A}} \right|$ $= a(ei - fh) - b(di - fg) + c(dh - eg)$
Therefore on comparing the values with the given question
$a = {x^2}$ , $b = x$ , $c = 1$
$d = 0$ , $e = 2$ , $f = 1$
$g = 3$ , $h = 1$ , $i = 4$
$\Rightarrow {x^2}\left( {8 - 1} \right) - x\left( {0 - 3} \right) + 1\left( {0 - 6} \right)$
$\Rightarrow 8{x^2} - {x^{^2}} + 3x - 6 = 28$
$\Rightarrow 7{x^2} + 3x - 6 = 28$
$\Rightarrow 7{x^2} + 3x - 34 = 0$
$\Rightarrow \left( {7x + 17} \right)\left( {x - 2} \right) = 0$
$\Rightarrow x = 2$ and $x = \dfrac{{ - 17}}{7}$
As, $x = \dfrac{{ - 17}}{7}$ is not an integer,
An integral value is the area or volume under or above a given mathematical function by an equation. It can be two dimensional or three dimensional. The greatest integer function is defined as $\left\lfloor {\text{x}} \right\rfloor$ is the largest integer that is less than or equal to x.
Therefore, Integral value of x is 2.

Note: In matrices, the determinant of the matrix must not be zero. This is instead of the real number not being zero to have an inverse, the determinant must not be zero to have an inverse. The matrix must be square (same number of rows and columns).