Determinant of a 3×3 Matrix – Examples with Answers

To find the determinant of the matrix $latex B$, we can use the cofactor expansion method along the first row.

Then, if we denote the determinant by $latex |B|$ or $latex \det(B)$.

$latex |B| = 2 \cdot C_{11} – 5 \cdot C_{12} + 3 \cdot C_{13} $

Where $latex C_{11}$, $latex C_{12}$, and $latex C_{13}$ are the cofactors of the elements of the first row.

To find the cofactors, we need to calculate the determinants of the 2×2 matrices obtained by eliminating the row and column of each element of the first row:

$$ C_{11} = \begin{vmatrix} 4 & 6 \\ 8 & 9 \end{vmatrix} $$

$$= (4 \cdot 9) – (6 \cdot 8)$$

$$= 36 – 48 = -12$$

$$ C_{12} = \begin{vmatrix} 1 & 6 \\ 7 & 9 \end{vmatrix}$$

$$ = (1 \cdot 9) – (6 \cdot 7) $$

$$= 9 – 42 = -33 $$

$$ C_{13} = \begin{vmatrix} 1 & 4 \\ 7 & 8 \end{vmatrix} $$

$$= (1 \cdot 8) – (4 \cdot 7) $$

$$= 8 – 28 = -20$$

Now, we substitute the values of the cofactors into the determinant formula:

$$ |B| = 2 \cdot (-12) – 5 \cdot (-33) + 3 \cdot (-20) $$

$$= -24 + 165 – 60 = 81 $$

Therefore, the determinant of the 3×3 matrix is 81.

We can take the first row and form the following formula. Remember that we multiply each element by the determinant of the corresponding minor.

$$ |B| = b_{11}(b_{22}b_{33} – b_{23}b_{32}) – b_{12}(b_{21}b_{33} – b_{23}b_{31}) + b_{13}(b_{21}b_{32} – b_{22}b_{31}) $$

Substituting the values of matrix B, we have:

$$ |B| = 1(5 \cdot 9 – 8 \cdot 6) – 4(2 \cdot 9 – 8 \cdot 3) + 7(2 \cdot 6 – 5 \cdot 3) $$

Simplifying, we have:

$$ |B| = 1(-3) – 4(-6) + 7(-3) $$

$$ |B| = -3 + 24 – 21 $$

$latex |B| = 0 $

Similar to the previous examples, we will obtain a formula using the cofactor expansion method in the first row:

$$|C| = c_{11}(c_{22}c_{33} – c_{23}c_{32}) – c_{12}(c_{21}c_{33} – c_{23}c_{31}) + c_{13}(c_{21}c_{32} – c_{22}c_{31})$$

Using the values of the elements of the matrix C, we have:

$$|C| = 3(1 \cdot 9 – 5 \cdot 8) – 0(4 \cdot 9 – 5 \cdot 7) + 2(4 \cdot 8 – 1 \cdot 7)$$

Performing the operations, we have:

$latex |C| = 3(-31) – 0(-1) + 2(25)$

$latex |C| = -93 + 0 + 50$

$latex |C| = -43$

Using the first row of the matrix, we form the following formula:

$$|D| = d_{11}(d_{22}d_{33} – d_{23}d_{32}) – d_{12}(d_{21}d_{33} – d_{23}d_{31}) + d_{13}(d_{21}d_{32} – d_{22}d_{31})$$

Substituting the values of the elements of the matrix D in this formula, we have:

$$|D| = -1(-5 \cdot -9 – 6 \cdot 8) – 2(4 \cdot -9 – 6 \cdot -7) + (-3)(4 \cdot 8 – (-5) \cdot -7)$$

Solving the operations and simplifying, we have:

$$|D| = -1(45 – 48) – 2(-36 + 42) + (-3)(32 + 35)$$

$$|D| = -1(-3) – 2(-6) + (-3)(67)$$

$latex |D| = 3 + 12 – 201$

$latex |D| = -186$

Similar to the previous exercises, we will select row 1 to apply the cofactor method:

$$|E| = e_{11}(e_{22}e_{33} – e_{23}e_{32}) – e_{12}(e_{21}e_{33} – e_{23}e_{31}) + e_{13}(e_{21}e_{32} – e_{22}e_{31})$$

Using the values of the elements of the matrix E in the formula, we have:

$$|E| = 2(3 \cdot -6 – (-7) \cdot 5) – (-4)(-1 \cdot -6 – (-7) \cdot 4) + 6(-1 \cdot 5 – 3 \cdot 4)$$

Solving the operations, we have:

$$|E| = 2(-18 + 35) – (-4)(6 + 28) + 6(-5 – 12)$$

$$|E| = 2(17) – (-4)(34) + 6(-17)$$

$$|E| = 34 + 136 – 102$$

$latex |E| = 68$

Using the cofactor expansion method, we can form the following formula using the first row of the matrix:

$$|F| = f_{11}(f_{22}f_{33} – f_{23}f_{32}) – f_{12}(f_{21}f_{33} – f_{23}f_{31}) + f_{13}(f_{21}f_{32} – f_{22}f_{31})$$

Using the values of the elements of the matrix F, we have:

$$|F| = 3(4 \cdot -3 – (-5) \cdot 6) – (-2)(-1 \cdot -3 – (-5) \cdot 2) + 1(-1 \cdot 6 – 4 \cdot 2)$$

Solving the operations, we have:

$$|F| = 3(-12 + 30) – (-2)(3 + 10) + 1(-6 – 8)$$

$$|F| = 3(18) – (-2)(13) + 1(-14)$$

$latex |F| = 54 + 26 – 14$

$latex |F| = 66$

If we take the first row of the matrix, we can form the following formula:

$$|G| = g_{11}(g_{22}g_{33} – g_{23}g_{32}) – g_{12}(g_{21}g_{33} – g_{23}g_{31}) + g_{13}(g_{21}g_{32} – g_{22}g_{31})$$

Using this formula with the values of the elements of the matrix G, we have:

$$|G| = -4(-2 \cdot 8 – 6 \cdot 1) – 7(5 \cdot 8 – 6 \cdot -3) + (-1)(5 \cdot 1 – (-2) \cdot -3)$$

Simplifying this expression, we have:

$$|G| = -4(-16 – 6) – 7(40 + 18) + (-1)(5 – 6)$$

$$|G| = -4(-22) – 7(58) + (-1)(-1)$$

$latex |G| = 88 – 406 + 1$

$latex |G| = -317$

By taking the first row of the matrix, we can multiply each element by the determinant of the smallest to form the following formula:

$$|H| = h_{11}(h_{22}h_{33} – h_{23}h_{32}) – h_{12}(h_{21}h_{33} – h_{23}h_{31}) + h_{13}(h_{21}h_{32} – h_{22}h_{31})$$

Substituting the values of the elements of the matrix H, we have:

$$|H| = 1(1 \cdot 1 – (-4) \cdot (-1)) – 2(-2 \cdot 1 – (-4) \cdot 3) + 3(-2 \cdot (-1) – 1 \cdot 3)$$

Now, we solve the operations and simplify:

$$|H| = 1(1 – 4) – 2(-2 + 12) + 3(2 – 3)$$

$latex |H| = 1(-3) – 2(10) + 3(-1)$

$latex |H| = -3 – 20 – 3$

$latex |H| = -26$

We form the following formula by taking the first row of the I matrix:

$$|I| = i_{11}(i_{22}i_{33} – i_{23}i_{32}) – i_{12}(i_{21}i_{33} – i_{23}i_{31}) + i_{13}(i_{21}i_{32} – i_{22}i_{31})$$

Substituting the values of matrix I, we have:

$$|I| = 2(4 \cdot 1 – 2 \cdot -2) – 3(-1 \cdot 1 – 2 \cdot 3) + (-1)(-1 \cdot -2 – 4 \cdot 3)$$

Solving operations and simplifying:

$$|I| = 2(4 + 4) – 3(-1 – 6) + (-1)(2 – 12)$$

$latex |I| = 2(8) – 3(-7) + (-1)(-10)$

$latex |I| = 16 + 21 + 10$

$latex |I| = 47$

Using the first row of the matrix, we form the following formula:

$$|J| = j_{11}(j_{22}j_{33} – j_{23}j_{32}) – j_{12}(j_{21}j_{33} – j_{23}j_{31}) + j_{13}(j_{21}j_{32} – j_{22}j_{31})$$

Now, we use the values of the matrix elements:

$$|J| = -1(2 \cdot -2 – (-1) \cdot 1) – 3(3 \cdot -2 – (-1) \cdot 1) + 5(3 \cdot 1 – 2 \cdot 1)$$

Solving the operations, we have

$$|J| = -1(-4 – 1) – 3(-6 + 1) + 5(3 – 2)$$

$latex |J| = -1(-5) – 3(-5) + 5(1)$

$latex |J| = 5 + 15 + 5$

$latex |J| = 25$