Evaluate the determinant $\Delta=\left|\begin{array}{rrr}1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0\end{array}\right|$.

Step 1: Choose a row or column for expansion

To evaluate a determinant, we can use the cofactor expansion method. It's often easiest to expand along a row or column that contains the most zeros, as this simplifies calculations. In this matrix, the third column has two zeros, so we will expand along the third column.

Step 2: Apply cofactor expansion along the third column

The determinant is calculated as the sum of the products of each element in the chosen column (or row) with its corresponding cofactor. The cofactor $C_{ij}$ is given by $(-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor determinant obtained by removing the $i$ -th row and $j$ -th column. Since the elements in the second and third rows of the third column are zero, their contributions to the determinant will be zero.

Step 3: Calculate the cofactor $C_{13}$

For the element $a_{13} = 4$, its cofactor $C_{13}$ is $(-1)^{1+3}$ times the determinant of the $2 \times 2$ matrix obtained by removing the first row and third column. This $2 \times 2$ matrix is $\begin{pmatrix}-1 & 3 \\ 4 & 1\end{pmatrix}$.

Step 4: Evaluate the $2 \times 2$ minor determinant

The determinant of a $2 \times 2$ matrix $\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ is $ad - bc$. Applying this formula to our minor determinant, we get $(-1)(1) - (3)(4)$, which simplifies to $-1 - 12 = -13$.

Step 5: Substitute the minor determinant back into the cofactor

Now we substitute the value of the $2 \times 2$ determinant back into the cofactor formula. Since $(-1)^{1+3} = (-1)^4 = 1$, the cofactor $C_{13}$ is $1 \times (-13) = -13$.

Step 6: Calculate the final determinant

Finally, we substitute the value of $C_{13}$ back into the expansion formula: $\Delta = 4 \cdot C_{13}$. This gives us $4 \cdot (-13) = -52$.

More from this chapter

• Evaluate the determinant 1 & 0 & 4 3 & 5 & -1 0 & 1 & 2 (i) with the help of second row. (ii) with the help of third column.
• If A = [[1, 2], [3, 4]] find det(A) and adj(A).
• The value of 1 & 2 & 3 4 & 5 & 6 7 & 8 & 9 is:
• Using Cramer's rule, show that the following system of linear equations is consistent and hence solve it: x + y + z = 1, 2x + 3y + 2z = 2, x + y + 2z = 4.
• Let a R and A be a matrix of order 3 3 such that det(A)=-4 and A+I= 1 & a & 1 2 & 1 & 0 a & 1 & 2, where I is the 3 3 identity. If det ((a+1)adj((a-1)A))=2^m3^n
• Let A be a 3 × 3 matrix such that |adj (adj(adj A))| = 81. If S = n Z: |adj(adj A)|^(((n-1)²)/(2)) = |A|³n² - 5n - 4, then Σ_n S | A^(n² + n) | is equal to:
• Find the equation of the line joining A(1,3) and B(0,0) using determinants and find k if D(k, 0) is a point such that area of triangle ABD is 3 sq units.
• If the system of linear equations 3x + y + β z = 3, 2x + α y - z = -3, x + 2y + z = 4 has infinitely many solutions, then the value of 22β - 9α is:

All Determinants solutions →