If \(\rm \Delta = \begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i \end{vmatrix}\) then what is \(\rm \begin{vmatrix} 3d + 5g & 4a + 7g & 6g\\ 3e + 5h & 4b + 7h & 6h\\ 3f + 5i & 4c + 7i & 6i \end{vmatrix}\) equal to?
To relate \( \Delta' \) to \( \Delta \), we can use the properties of determinants. One useful property is that the determinant of a matrix is equal to the determinant of its transpose (\(\det(A) = \det(A^T)\)). Let's consider the transpose of \( \Delta' \):
Since \( \det(\Delta') = \det((\Delta')^T) \), we will work with the transpose matrix and apply row operations. Row operations on a determinant either leave its value unchanged (adding a multiple of one row to another), multiply it by a scalar (multiplying a row by a scalar), or multiply it by -1 (swapping two rows).
Step-by-Step Determinant Calculation
Let's manipulate \( (\Delta')^T \) using row operations to simplify it and make it look like \( \Delta \).
Step 1: Factor out common terms from rows.
Notice that the third row \((6g, 6h, 6i)\) has a common factor of 6. We can factor this out, multiplying the determinant by 6:
\[ (\Delta')^T = 6 \begin{vmatrix} 3d + 5g & 3e + 5h & 3f + 5i \\ 4a + 7g & 4b + 7h & 4c + 7i \\ g & h & i \end{vmatrix} \]
Step 2: Use row operations to eliminate terms.
Observe the term \(5g\) in the first row, first column element \(3d + 5g\). We can eliminate this term by subtracting 5 times the third row (\(5g, 5h, 5i\)) from the first row. This operation does not change the value of the determinant:
\( R_1 \leftarrow R_1 - 5R_3 \)
\[ (\Delta')^T = 6 \begin{vmatrix} (3d + 5g) - 5g & (3e + 5h) - 5h & (3f + 5i) - 5i \\ 4a + 7g & 4b + 7h & 4c + 7i \\ g & h & i \end{vmatrix} \]
\[ (\Delta')^T = 6 \begin{vmatrix} 3d & 3e & 3f \\ 4a + 7g & 4b + 7h & 4c + 7i \\ g & h & i \end{vmatrix} \]
Now, the first row \((3d, 3e, 3f)\) has a common factor of 3. Factor this out:
\[ (\Delta')^T = 6 \times 3 \begin{vmatrix} d & e & f \\ 4a + 7g & 4b + 7h & 4c + 7i \\ g & h & i \end{vmatrix} = 18 \begin{vmatrix} d & e & f \\ 4a + 7g & 4b + 7h & 4c + 7i \\ g & h & i \end{vmatrix} \]
Next, observe the term \(7g\) in the second row, first column element \(4a + 7g\). We can eliminate the terms involving \(g, h, i\) from the second row by subtracting 7 times the third row (\(7g, 7h, 7i\)) from the second row. This operation also does not change the determinant value:
\( R_2 \leftarrow R_2 - 7R_3 \)
\[ (\Delta')^T = 18 \begin{vmatrix} d & e & f \\ (4a + 7g) - 7g & (4b + 7h) - 7h & (4c + 7i) - 7i \\ g & h & i \end{vmatrix} \]
\[ (\Delta')^T = 18 \begin{vmatrix} d & e & f \\ 4a & 4b & 4c \\ g & h & i \end{vmatrix} \]
The second row \((4a, 4b, 4c)\) has a common factor of 4. Factor this out:
\[ (\Delta')^T = 18 \times 4 \begin{vmatrix} d & e & f \\ a & b & c \\ g & h & i \end{vmatrix} = 72 \begin{vmatrix} d & e & f \\ a & b & c \\ g & h & i \end{vmatrix} \]
Step 3: Compare the resulting determinant with the original \( \Delta \).
The determinant we have now is \( \begin{vmatrix} d & e & f \\ a & b & c \\ g & h & i \end{vmatrix} \). Let's compare this with the original \( \Delta = \begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i \end{vmatrix} \). The rows are in a different order. The first row of the new determinant is the second row of \( \Delta \), and the second row of the new determinant is the first row of \( \Delta \). The third row is the same.
Swapping two rows in a determinant multiplies its value by -1. If we swap R1 and R2 in \( \begin{vmatrix} d & e & f \\ a & b & c \\ g & h & i \end{vmatrix} \), we get \( \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \), which is \( \Delta \). Therefore:
\[ \begin{vmatrix} d & e & f \\ a & b & c \\ g & h & i \end{vmatrix} = - \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = -\Delta \]
Step 4: Substitute the result back to find \( \Delta' \).
We found that \( (\Delta')^T = 72 \begin{vmatrix} d & e & f \\ a & b & c \\ g & h & i \end{vmatrix} \). Substituting the relationship from Step 3:
\[ (\Delta')^T = 72 (-\Delta) = -72\Delta \]
Since \( \det(\Delta') = \det((\Delta')^T) \), we have:
\[ \Delta' = -72\Delta \]
Thus, the second determinant is equal to \( -72\Delta \).
Revision Table: Key Determinant Properties
Property
Description
Effect on Determinant
Transpose
\(\det(A) = \det(A^T)\)
Value unchanged
Row/Column Swap
Interchanging two rows or two columns
Multiplies the determinant by -1
Scalar Multiplication
Multiplying a row or column by a scalar \(k\)
Multiplies the determinant by \(k\)
Row/Column Addition
Adding a multiple of one row/column to another
Value unchanged
Additional Information on Determinants
A determinant is a scalar value that can be computed from the elements of a square matrix. It has many applications in linear algebra, including determining if a matrix is invertible (a matrix is invertible if and only if its determinant is non-zero), finding the area of a parallelogram or the volume of a parallelepiped defined by vectors forming the rows or columns of the matrix, and solving systems of linear equations using Cramer's rule.
Understanding determinant properties is crucial for simplifying calculations involving large matrices and for theoretical work. The properties used in this problem are fundamental. The ability to factor out scalars from rows or columns, perform row/column operations without changing the determinant's value (or changing it predictably), and knowing the effect of transposing or swapping rows/columns are essential skills for working with determinants.
Another important property is that the determinant of the product of two square matrices \(A\) and \(B\) of the same size is the product of their determinants: \( \det(AB) = \det(A)\det(B) \).
Was this answer helpful?
Similar Questions
The value of the determinant \(\left| {\begin{array}{*{20}{c}} {1 - {\rm{\alpha }}}&{{\rm{\alpha }} - {{\rm{\alpha }}^2}}&{{{\rm{\alpha }}^2}}\\ {1 - {\rm{\beta }}}&{{\rm{\beta }} - {{\rm{\beta }}^2}}&{{{\rm{\beta }}^2}}\\ {1 - {\rm{\gamma }}}&{{\rm{\gamma }} - {{\rm{\gamma }}^2}}&{{{\rm{\gamma }}^2}} \end{array}} \right|\) is equal to
Let p, q and r be three distinct positive real numbers. If \(\rm D = \left| {\begin{array}{*{20}{c}} \rm p&\rm q&\rm r\\ \rm q&\rm r&\rm p\\ \rm r&\rm p&\rm q \end{array}} \right|,\) then which one of the following is correct?
What is the value of the determinant \(\left| {\begin{array}{*{20}{c}} {\rm{i}}&{{{\rm{i}}^2}}&{{{\rm{i}}^3}}\\ {{{\rm{i}}^4}}&{{{\rm{i}}^6}}&{{{\rm{i}}^8}}\\ {{{\rm{i}}^9}}&{{{\rm{i}}^{12}}}&{{{\rm{i}}^{15}}} \end{array}} \right|\) where \(\rm i = \sqrt {-1}\) ?
If \(\left| {\begin{array}{*{20}{c}} {\rm{x}}&{\rm{y}}&0\\ 0&{\rm{x}}&{\rm{y}}\\ {\rm{y}}&0&{\rm{x}} \end{array}} \right| = 0\) , then which one of the following is correct?
If x, y, z are distinct real numbers and \(\left| {\begin{array}{*{20}{c}} x&{{x^2}}&{2 + {x^3}}\\ y&{{y^2}}&{2 + {y^3}}\\ z&{{z^2}}&{2 + {z^3}} \end{array}} \right| = 0\), then xyz =
If A + B + C = \(\pi \), then, the value of \(\left| {\begin{array}{*{20}{c}} {\sin \left( {A + B + C} \right)}&{\sin B}&{\cos C}\\ { - \sin B}&0&{\tan A}\\ {\cos \left( {A + B} \right)}&{ - \tan A}&0 \end{array}} \right|\) is
Let p, q and r be three distinct positive real numbers. If \(\rm D = \left| {\begin{array}{*{20}{c}} \rm p&\rm q&\rm r\\ \rm q&\rm r&\rm p\\ \rm r&\rm p&\rm q \end{array}} \right|,\) then which one of the following is correct?
