Determinant of 3×3 Matrix (Practice Questions)

The determinant of a 3×3 matrix is a single number calculated from its elements.

Solved Examples

Example 1: Find the determinant of matrix A \begin{vmatrix}2 & 3 & 1 \\0 & 4 & 5 \\1 & 6 & 2 \\\end{vmatrix}

Determinant of A = 2 (4×2 - 5×6) - 3(0×2 - 5×1) + 1(0×6 - 4×1)
⇒ Determinant of A = 2(8-30) - 3(0-5) +1(0-4)
⇒ Determinant of A =2(-22) - 3(-5) +1(-4)
⇒ Determinant of A = (-44) +15 - 4
⇒ Determinant of A =-44+11

∴ Determinant of A i.e., |A| = (-33)

Example 2: Find the determinant of matrix B =\begin{vmatrix}1 & 2 & 1 \\0 & 3 & 0 \\4 & 1 & 2 \\\end{vmatrix}

Determinant of B = 1(3×2 - 0×1) - 2(0×2 - 0×4) + 1(0×1 - 3×4)
⇒ Determinant of B = 1(6-0) - 2(0) + 1(-12)
⇒ Determinant of B = 1(6) - 0 - 12
⇒ Determinant of B =6-12
⇒ Determinant of B = (-6)

∴ Determinant of B i.e., |B| = 6

Example 3: Find the Determinant of matrix C \begin{vmatrix}3 & 1 & 2 \\0 & 2 & 5 \\2 & 0 & 4 \\\end{vmatrix}

Determinant of matrix C = 3(2×4 - 5×0) - 1(0×4 - 5×2) + 2(0×0 - 2×2)
⇒ Determinant of C = 3(8-0) - 1(0-10) + 2(0-4)
⇒ Determinant of C =3(8) - 1(-10) + 2(-4)
⇒ Determinant of C = 24 + 10 -8
⇒ Determinant of C = 26

∴ Determinant of C i.e., |C| = 26

Example 4: Solve the given system of Equations using Cramer's Rule.

2x + 3y - z = 7
4x - 2y + 3z = 8
x + y + 2z = 10

Solution:

Step1: First, find the Determinant D of coefficient matrix.

D = \begin{vmatrix}2 & 3 & -1 \\4 & -2 & 3 \\1 & 1 & 2\end{vmatrix}

On Solving this determinant D

D= 2(-2×2-3×1) - 3(4×2-1×3) - (-1)(4×1-(-2)×1)
⇒ D= 2(-4-3) - 3(8-3) - 1(4+2)
⇒ D= 2(-7) - 3(5) - 1(6)
⇒ D= -14 - 15 - 6
⇒ D= -35

Step2: Now, find the determinants of D_x, D_y and D_z

For D_x, we replace the coefficients of x with the constants on the right-hand side:

Dx = \begin{vmatrix}7 & 3 & -1 \\8 & -2 & 3 \\10 & 1 & 2\end{vmatrix}

For D_y, we replace the coefficients of y with the constants:

Dy = \begin{vmatrix}2 & 7 & -1 \\4 & 8 & 3 \\1 & 10 & 2\end{vmatrix}

For D_z, we replace the coefficients of z with the constants:

Dz = \begin{vmatrix}2 & 3 & 7 \\4 & -2 & 8 \\1 & 1 & 10\end{vmatrix}

On Solving the determinant D_x

D_x = 7(-2×2 - 3×1) - 3(8×2 - 3×10) - (-1)(8×1 - (-2×10)
⇒ D_x = 7(-4 - 3) - 3(16 - 30) - -1(8 + 20)
⇒ D_x = 7(-7) - 3(-14) - 28
⇒ D_x = -49 + 42 - 28

Thus, D_x = - 35

On Solving the determinant D_y

D_y = 2(8 × 2 - 3×10) - 7(4×2 - 3×1) - (-1)(4×10 - 8×1)

⇒ D_y = 2(16 - 30) - 7(8 - 3) - 1(40 - 8)
⇒ D_y = 2(-14) - 7(8 - 3) + 1(32)
⇒ D_y = -28 - 35 -32
⇒ D_y = - 95

On Solving the determinant D_z

D_z = 2(-2×(10) - 8×(1)) - 3(4×(10) - 8×(1)) - 7(4×1 - (-2×1)

⇒ D_z = 2(-20 - 8) - 3(40 - 8) - 7(4 + 2)
⇒ D_z = 2(-28) - 3(32) - 7(6)
⇒ D_z = -56- 96 + 42
⇒ D_z = - 110

Step 3: Now putting the values of D, D_x, D_y and D_z in the Carmer's Rule Formula to find the values of x,y and z.

x = D_x/D = (-35)/(-35 )= 1
y = D_y/D = (-95)/(-35) = 19 /7
z = D_z/D = (-110)/(-35) = 22/7

Practice Questions

Question 1: Calculate the determinant of the identity matrix:

\begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{bmatrix}

Question 2: Find the determinant of the matrix:

\begin{bmatrix}3 & 2 & 0 \\0 & 4 & -1 \\2 & 1 & 5\end{bmatrix}

Question 3: Determine the determinant of the matrix:

\begin{bmatrix}2 & 1 & 1 \\1 & 2 & 1 \\1 & 1 & 2\end{bmatrix}

Question 4: Calculate the determinant of the matrix:

\begin{bmatrix}-1 & 0 & 0 \\0 & 2 & 0 \\0 & 0 & -3\end{bmatrix}

Question 5: Find the determinant of the matrix:

\begin{bmatrix}4 & 3 & 2 \\1 & 0 & 1 \\2 & 1 & 4\end{bmatrix}

Question 6: Determine the determinant of the matrix:

\begin{bmatrix}0 & 1 & 2 \\2 & -1 & 3 \\1 & 0 & -2\end{bmatrix}

Answer key:

1. 1
2. 59
3. 4
4. 6
5. -8
6. 9