Which of the following determinants have value zero? 1. \(\left| {\begin{array}{*{20}{c}} {41}&1&5\\ {79}&7&9\\ {29}&5&3 \end{array}} \right|\) 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.
1, 2 and 3
Calculating Determinants with Zero Value
We are asked to identify which of the given determinants have a value of zero. We will evaluate each determinant one by one.
Evaluating Determinant 1
The first determinant is given by:
\[ \left| {\begin{array}{*{20}{c}} {41}&1&5\\ {79}&7&9\\ {29}&5&3 \end{array}} \right| \]To evaluate this determinant efficiently, we can use properties of determinants. Let's apply a column operation \(C_1 \to C_1 - 8C_3\). This operation does not change the value of the determinant.
Applying the operation:
- The first element of the new \(C_1\) is \(41 - 8 \times 5 = 41 - 40 = 1\).
- The second element of the new \(C_1\) is \(79 - 8 \times 9 = 79 - 72 = 7\).
- The third element of the new \(C_1\) is \(29 - 8 \times 3 = 29 - 24 = 5\).
The determinant transforms into:
\[ \left| {\begin{array}{*{20}{c}} 1&1&5\\ 7&7&9\\ 5&5&3 \end{array}} \right| \]Now, we observe that Column 1 and Column 2 are identical. A fundamental property of determinants states that if any two rows or any two columns of a determinant are identical or proportional, the value of the determinant is zero.
Since \(C_1 = C_2\), the value of the first determinant is 0.
Evaluating Determinant 2
The second determinant is given by:
\[ \left| {\begin{array}{*{20}{c}} 1&a&{b + c}\\ 1&b&{c + a}\\ 1&c&{a + b} \end{array}} \right| \]Let's apply a column operation \(C_3 \to C_3 + C_2\). This operation also does not change the value of the determinant.
Applying the operation:
- The first element of the new \(C_3\) is \((b + c) + a = a + b + c\).
- The second element of the new \(C_3\) is \((c + a) + b = a + b + c\).
- The third element of the new \(C_3\) is \((a + b) + c = a + b + c\).
The determinant becomes:
\[ \left| {\begin{array}{*{20}{c}} 1&a&{a + b + c}\\ 1&b&{a + b + c}\\ 1&c&{a + b + c} \end{array}} \right| \]Now, we can take out the common factor \( (a + b + c) \) from Column 3. Using the property that a common factor from any one row or column can be taken outside the determinant, we get:
\[ (a + b + c) \left| {\begin{array}{*{20}{c}} 1&a&1\\ 1&b&1\\ 1&c&1 \end{array}} \right| \]In the resulting determinant, we can see that Column 1 and Column 3 are identical (\(C_1 = C_3\)). According to the property mentioned earlier, if two columns are identical, the determinant's value is zero.
So, the value of the second determinant is \( (a + b + c) \times 0 = 0 \).
Evaluating Determinant 3
The third determinant is given by:
\[ \left| {\begin{array}{*{20}{c}} 0&c&b\\ { - c}&0&a\\ { - b}&{ - a}&0 \end{array}} \right| \]Let's expand this determinant along the first row to calculate its value:
\[ \text{Value} = 0 \times \text{cofactor of } a_{11} - c \times \text{cofactor of } a_{12} + b \times \text{cofactor of } a_{13} \] \[ = 0 \times \left| {\begin{array}{*{20}{c}} 0&a\\ { - a}&0 \end{array}} \right| - c \times \left| {\begin{array}{*{20}{c}} { - c}&a\\ { - b}&0 \end{array}} \right| + b \times \left| {\begin{array}{*{20}{c}} { - c}&0\\ { - b}&{ - a} \end{array}} \right| \]Calculate the 2x2 determinants:
\[ \left| {\begin{array}{*{20}{c}} 0&a\\ { - a}&0 \end{array}} \right| = (0)(0) - (a)(-a) = 0 - (-a^2) = a^2 \] \[ \left| {\begin{array}{*{20}{c}} { - c}&a\\ { - b}&0 \end{array}} \right| = (-c)(0) - (a)(-b) = 0 - (-ab) = ab \] \[ \left| {\begin{array}{*{20}{c}} { - c}&0\\ { - b}&{ - a} \end{array}} \right| = (-c)(-a) - (0)(-b) = ac - 0 = ac \]Substitute these values back:
\[ \text{Value} = 0 \times (a^2) - c \times (ab) + b \times (ac) \] \[ = 0 - abc + abc \] \[ = 0 \]Thus, the value of the third determinant is 0.
Alternatively, we can recognize this determinant as that of a skew-symmetric matrix of order 3. A skew-symmetric matrix \(A\) satisfies \(A^T = -A\), which means \(a_{ij} = -a_{ji}\) and \(a_{ii} = 0\). The given matrix fits this description:
| Col 1 | Col 2 | Col 3 | |
|---|---|---|---|
| Row 1 | 0 | c | b |
| Row 2 | -c | 0 | a |
| Row 3 | -b | -a | 0 |
For a skew-symmetric matrix of odd order \(n\), the determinant is always zero (\(\det(A) = 0\)). Since this is a 3x3 (odd order) skew-symmetric matrix, its determinant must be 0.
Summary of Determinant Results
After evaluating each determinant:
- Determinant 1 has a value of 0.
- Determinant 2 has a value of 0.
- Determinant 3 has a value of 0.
Therefore, all three determinants have a value of zero.
Revision Table: Key Determinant Properties
| Property | Description | Application |
|---|---|---|
| Identity Columns/Rows | If two columns or rows are identical, the determinant is zero. | Applied to Determinants 1 and 2. |
| Row/Column Operations | \(R_i \to R_i + k R_j\) or \(C_i \to C_i + k C_j\) does not change determinant value. | Used to simplify Determinants 1 and 2. |
| Common Factor | A scalar multiple in a row/column can be factored out. | Used in Determinant 2 evaluation. |
| Skew-symmetric Determinant | Determinant of an odd-order skew-symmetric matrix is zero. | Applied to Determinant 3. |
Additional Information: Determinant Calculation Methods
Apart from using properties, determinants can be calculated using various methods:
- Expansion by Cofactors: This involves expanding the determinant along any row or column. For a matrix \(A = [a_{ij}]\), the determinant is given by \( \det(A) = \sum_{j=1}^{n} a_{ij} C_{ij} \) (expanding along row \(i\)) or \( \det(A) = \sum_{i=1}^{n} a_{ij} C_{ij} \) (expanding along column \(j\)), where \(C_{ij} = (-1)^{i+j} M_{ij}\) is the cofactor and \(M_{ij}\) is the minor (determinant of the submatrix obtained by deleting row \(i\) and column \(j\)). This method was used for Determinant 3 calculation.
- Row/Column Reduction (Gaussian Elimination): Using row or column operations (like swapping rows, multiplying a row by a scalar, adding a multiple of one row to another) to transform the matrix into an upper or lower triangular form. The determinant of a triangular matrix is the product of its diagonal elements. While this method is powerful, using specific properties can sometimes be faster, as seen with Determinants 1 and 2.
Understanding the properties of determinants is crucial as it often simplifies complex calculations and provides shortcuts, especially in competitive exams.
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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:
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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
