Which one of the following factors does the expansion of the determinant \(\left| {\begin{array}{c} x&y&3\\ {{x^2}}&{5{y^3}}&9\\ {{x^3}}&{10{y^3}}&{27} \end{array}} \right|\) Contain?
x - 3
Analyzing the Determinant Expansion and Finding a Factor
The question asks us to identify a factor that is contained within the expansion of the given 3x3 determinant:
\[ \left| {\begin{array}{c} x&y&3\\ {{x^2}}&{5{y^3}}&9\\ {{x^3}}&{10{y^3}}&{27} \end{array}} \right| \]We are given four options, and we need to determine which one is a factor of the polynomial expression resulting from the determinant's expansion.
Methods to Find a Factor of a Determinant Expansion
There are a few ways to approach this problem:
- Directly expand the determinant and then try to factor the resulting expression.
- Use properties of determinants to simplify the determinant before expansion.
- Use the Factor Theorem: If a polynomial \(P(x)\) has a factor \((x-a)\), then \(P(a) = 0\). We can extend this concept to multiple variables. If \((x-a)\) is a factor of a determinant expression \(\Delta(x, y)\), then substituting \(x=a\) should make the determinant zero. Similarly, if \((y-b)\) is a factor, substituting \(y=b\) should make the determinant zero. We can test each option using this substitution method.
Applying the Factor Theorem Method
The Factor Theorem method is often the quickest way to check potential linear factors like the ones given in the options. We will substitute the value of \(x\) or \(y\) that makes each potential factor zero into the original determinant and see if the determinant evaluates to zero.
Testing Option 1: \(x - 3\)
If \((x - 3)\) is a factor, then substituting \(x = 3\) into the determinant should result in a value of 0.
Substitute \(x = 3\) into the determinant:
\[ \left| {\begin{array}{c} 3&y&3\\ {{3^2}}&{5{y^3}}&9\\ {{3^3}}&{10{y^3}}&{27} \end{array}} \right| = \left| {\begin{array}{c} 3&y&3\\ {9}&{5{y^3}}&9\\ {27}&{10{y^3}}&{27} \end{array}} \right| \]Let's look at the columns of this new determinant:
| Column 1 (C1) | Column 2 (C2) | Column 3 (C3) |
|---|---|---|
| 3 | y | 3 |
| 9 | \(5y^3\) | 9 |
| 27 | \(10y^3\) | 27 |
Observe that Column 1 and Column 3 are identical. A property of determinants states that if any two rows or any two columns of a determinant are identical, the value of the determinant is zero.
Since C1 = C3, the value of the determinant is 0 when \(x=3\). Therefore, \((x - 3)\) is a factor of the determinant's expansion.
Testing Other Options
We can also quickly check the other options using the same method, although we've already found the factor.
- Option 2: \(x - y\)
Substitute \(x = y\) into the original determinant: \[ \left| {\begin{array}{c} y&y&3\\ {{y^2}}&{5{y^3}}&9\\ {{y^3}}&{10{y^3}}&{27} \end{array}} \right| \] Columns 1 and 2 are not identical (e.g., \(y^2\) vs \(5y^3\)). The determinant is not necessarily zero. - Option 3: \(y - 3\)
Substitute \(y = 3\) into the original determinant: \[ \left| {\begin{array}{c} x&3&3\\ {{x^2}}&{5 \cdot 3^3}&9\\ {{x^3}}&{10 \cdot 3^3}&{27} \end{array}} \right| = \left| {\begin{array}{c} x&3&3\\ {{x^2}}&{135}&9\\ {{x^3}}&{270}&{27} \end{array}} \right| \] Columns 2 and 3 are not identical. The determinant is not necessarily zero. - Option 4: \(x - 3y\)
Substitute \(x = 3y\) into the original determinant: \[ \left| {\begin{array}{c} 3y&y&3\\ {{(3y)}^2}&{5{y^3}}&9\\ {{(3y)}^3}&{10{y^3}}&{27} \end{array}} \right| = \left| {\begin{array}{c} 3y&y&3\\ {9y^2}&{5{y^3}}&9\\ {27y^3}&{10{y^3}}&{27} \end{array}} \right| \] No two columns are clearly identical or proportional for all \(y\). The determinant is not necessarily zero.
Based on the tests, only substituting \(x=3\) makes the determinant zero, confirming that \((x - 3)\) is a factor of its expansion.
Conclusion
By applying the Factor Theorem and substituting the root of each potential factor into the determinant, we found that substituting \(x=3\) (which is the root of the factor \(x-3\)) results in a determinant with identical columns, thus having a value of zero. This confirms that \((x - 3)\) is a factor of the determinant's expansion.
Revision Table: Key Concepts for Determinants
| Concept | Description | Relevance to Problem |
|---|---|---|
| Determinant | A scalar value calculated from the elements of a square matrix. | We need to find a factor of the expression resulting from its calculation (expansion). |
| Determinant Expansion (Cofactor Expansion) | A method to calculate the determinant by summing the products of elements of a row or column with their cofactors. | The determinant expands into a polynomial expression. |
| Properties of Determinants | Rules that govern how the determinant value changes under operations like row/column swaps, scaling, or adding multiples of rows/columns. Also includes properties like having a zero determinant if rows/columns are identical or linearly dependent. | The property that a determinant is zero if two columns are identical was crucial for the solution using substitution. |
| Factor Theorem | For a polynomial \(P(z)\), \((z-a)\) is a factor if and only if \(P(a)=0\). This extends to expressions involving multiple variables. | Used directly to test the options by substituting the root of each potential factor into the determinant expression. |
Additional Information: Understanding Factors and Roots
In algebra, a factor of a polynomial is an expression that divides the polynomial evenly, leaving no remainder. For example, \((x-2)\) is a factor of \(x^2 - 4\) because \(x^2 - 4 = (x-2)(x+2)\).
The roots or zeros of a polynomial are the values of the variable(s) that make the polynomial equal to zero. If \((x-a)\) is a factor of a polynomial in \(x\), then \(x=a\) is a root because substituting \(a\) for \(x\) makes the factor \((a-a)=0\), and thus the entire polynomial expression becomes zero.
When dealing with expressions derived from determinants, the expansion is a polynomial in the variables involved (in this case, \(x\) and \(y\)). If \((x-a)\) is a factor of this polynomial expansion, then substituting \(x=a\) into the determinant expression (which represents the polynomial) must result in a value of zero. This is the core idea behind using substitution to solve this problem.
The property of determinants that we used (determinant is zero if columns are identical) is a powerful shortcut that arises from the definition and properties of determinants. When we substituted \(x=3\), the first and third columns of the determinant became identical, making the determinant value zero without needing to calculate the full expansion first. This directly confirmed \((x-3)\) as a factor.
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2. \(\left| {\begin{array}{*{20}{c}} 1&a&{b + c}\\ 1&b&{c + a}\\ 1&c&{a + b} \end{array}} \right|\)
3. \(\left| {\begin{array}{*{20}{c}} 0&c&b\\ { - c}&0&a\\ { - b}&{ - a}&0 \end{array}} \right|\)
Select the correct answer using the code given below.
If A is an invertible matrix of order n and k is any positive real number, then the value of [det(kA)] -1 det A is
Consider the following statements in respect of the determinant \(\left| {\begin{array}{} {{{\cos }^2}\frac{\alpha }{2}}&{{{\sin }^2}\frac{\alpha }{2}}\\ {{{\sin }^2}\frac{\beta }{2}}&{{{\cos }^2}\frac{\beta }{2}} \end{array}} \right|\)
Where α, β are complementary angles
1. The value of the determinant is \(\frac{1}{{√ 2 }}\cos \left( {\frac{{\alpha - \beta }}{2}} \right)\;\)
2. The maximum value of the determinant is \(\frac{1}{\sqrt2}\)
Which of the above statements is/are correct?
Important Questions from Determinants
Let $A$ and $B$ be two invertible matrices of order $3 \times 3$. If $\det(A^2 B (A^T)^3) = 16$ and $\det(A^3 B^{-2}) = 32$, then $\det(B^2 A^{-1} (B^T)^3)$ is equal to:
The value of determinant \(\left| {\begin{array}{*{20}{c}} {a - b - c}&{2a}&{2a}\\ {2b}&{b - c - a}&{2b}\\ {2c}&{2c}&{c - a - b} \end{array}} \right|\) is:
If \(\left| {\begin{array}{*{20}{c}} 5&a\\ a&2 \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 2&1\\ 3&2 \end{array}} \right|\), then the values of a are:
The value of the determinant \(\left| {\begin{array}{*{20}{c}} {\sqrt {13} + \sqrt 3 }&{2\sqrt 5 }&{\sqrt 5 }\\ {\sqrt {15} + \sqrt {26} }&5&{\sqrt {10} }\\ {3 + \sqrt {65} }&{\sqrt {15} }&5 \end{array}} \right|\) is
The determinant \(\left| {\begin{array}{*{20}{c}} {xp + y}&x&y\\ {yp + z}&y&z\\ 0&{xp + y}&{yp + z} \end{array}} \right| = 0,\) if
