If \(\left|\begin{array}{ccc} 2 x-4 & 4 & 0 \\ 2 & x-1 & 1 \\ 2 & 2 & 0 \end{array}\right|=0\), then x = ?
4
Determinant Equation: Solving for x
To find the value of \(x\) in the given equation, we need to evaluate the determinant of the 3x3 matrix and set it equal to zero.
The given determinant equation is:
| \( \left|\begin{array}{ccc} 2 x-4 & 4 & 0 \\ 2 & x-1 & 1 \\ 2 & 2 & 0 \end{array}\right|=0 \) |
Expanding the Determinant
We can expand the determinant along any row or column. Observing the third column, we see that it contains two zero elements, which simplifies the calculation significantly. Let's expand along the third column (C3).
The formula for expanding a determinant along a column is:
\( \text{det}(A) = \sum_{i=1}^{n} a_{i j} C_{i j} \)
where \(a_{ij}\) is the element in the \(i\)-th row and \(j\)-th column, and \(C_{ij}\) is its cofactor. The cofactor \(C_{ij}\) is given by \( (-1)^{i+j} M_{i j} \), where \(M_{ij}\) is the minor (the determinant of the submatrix obtained by deleting the \(i\)-th row and \(j\)-th column).
Expanding along the third column (C3):
\( (2x-4) \cdot C_{13} + (4) \cdot C_{23} + (0) \cdot C_{33} \)
Wait, this is wrong. Let's restart the expansion. Elements are \(a_{13}=0, a_{23}=1, a_{33}=0\).
So, the expansion is:
\( a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33} = 0 \)
\( (0) \cdot C_{13} + (1) \cdot C_{23} + (0) \cdot C_{33} = 0 \)
This simplifies to:
\( 1 \cdot C_{23} = 0 \)
Therefore, we only need to calculate the cofactor \(C_{23}\).
Calculating the Cofactor \(C_{23}\)
The cofactor \(C_{23}\) is \( (-1)^{2+3} M_{23} \). The minor \(M_{23}\) is the determinant of the submatrix obtained by deleting the 2nd row and 3rd column from the original matrix:
| \( M_{23} = \left|\begin{array}{cc} 2x-4 & 4 \\ 2 & 2 \end{array}\right| \) |
Now, calculate this 2x2 determinant:
\( M_{23} = (2x-4)(2) - (4)(2) \)
\( M_{23} = 4x - 8 - 8 \)
\( M_{23} = 4x - 16 \)
Now, substitute \(M_{23}\) back into the cofactor formula:
\( C_{23} = (-1)^{5} (4x - 16) \)
\( C_{23} = -1 \cdot (4x - 16) \)
\( C_{23} = -4x + 16 \)
Solving for x
We found that \( 1 \cdot C_{23} = 0 \). Substituting the value of \(C_{23}\):
\( -4x + 16 = 0 \)
Now, we solve this linear equation for \(x\):
Subtract 16 from both sides:
\( -4x = -16 \)
Divide by -4:
\( x = \frac{-16}{-4} \)
\( x = 4 \)
Conclusion
The value of \(x\) that satisfies the given determinant equation is 4.
Important Questions from Determinants
If \(A=\left[\begin{array}{rrr} 2 & -1 & 0 \\ -1 & 3 & 0 \\ 1 & 0 & 1 \end{array}\right]\), then what is the value of det[adj(adjA)] ?
If A, B and C are square matrices of order 3 and det(BC) = 2 det(A), then what is the value of det(2A-1BC)?
If \(A=\left[\begin{array}{rrr} 0 & 3 & 4 \\ -3 & 0 & 5 \\ -4 & -5 & 0 \end{array}\right]\), then which one of the following statements is correct?
If \(\left|\begin{array}{ccc} x^2+3 x & x-1 & x+3 \\ x+1 & -2 x & x-4 \\ x-3 & x+4 & 3 x \end{array}\right|\) = ax4 + bx3 + cx2 + dx + e, then what is the value of e?"
If all elements of a third order determinant are equal to 1 or -1, then the value of the determinant is:
