The value of $\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$ is :

Step 1: Define the Determinant

To find the determinant of a $3 \times 3$ matrix, we use the cofactor expansion method. We multiply each element of the first row by the determinant of its corresponding $2 \times 2$ submatrix, alternating signs.

Step 2: Substitute the values

Now, we substitute the given values into the determinant formula. We take the first element, 1, and multiply it by the determinant of the submatrix formed by removing its row and column. We do the same for 2 and 3, remembering to alternate the signs.

Step 3: Calculate the $2 \times 2$ determinants

Next, we calculate the determinants of the $2 \times 2$ submatrices. The determinant of a $2 \times 2$ matrix $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ is $ad - bc$.

Step 4: Simplify the expression

We perform the multiplications within each parenthesis to simplify the expression further.

Step 5: Perform the final calculations

Finally, we complete the arithmetic operations: multiplication, then addition and subtraction, to arrive at the determinant's value.

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).
• 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:
• Let A = 2 & 2 + p & 2 + p + q 4 & 6 + 2p & 8 + 3p + 2q 6 & 12 + 3p & 20 + 6p + 3q. If det(adj(adj(3A))) = 2^m · 3^n, m, n N, then m + n is equal to:

All Determinants solutions →