We compute the 2 × 2 determinants: M_21 = (0)(2) - (4)(1) = -4, M_22 = (1)(2) - (4)(0) = 2, and M_23 = (1)(1) - (0)(0) = 1.

Evaluating the 2 × 2 determinants: M_13 = (3)(1) - (5)(0) = 3, M_23 = (1)(1) - (0)(0) = 1, and M_33 = (1)(5) - (0)(3) = 5.

Evaluate the determinant $\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}$ (i) with the help of second row. (ii) with the help of third column.

Answer: The value of the determinant is $23$ by both methods: (i) expansion along the second row gives $23$ , and (ii) expansion along the third column gives $23$ .

Step-by-step solution

Step 1: Set up expansion along the second row

To expand along row 2, we use the alternating sign pattern $(-1)^{i+j}$ for each entry. The signs for the second row are negative, positive, and negative, corresponding to positions $(2{,}1)$ , $(2{,}2)$ , and $(2{,}3)$ .

Step 2: Evaluate minors for row 2 expansion

We compute the $2 \times 2$ determinants: $M_{21} = (0)(2) - (4)(1) = -4$ , $M_{22} = (1)(2) - (4)(0) = 2$ , and $M_{23} = (1)(1) - (0)(0) = 1$ .

Step 3: Calculate the determinant value via row 2

Multiplying each element by its signed minor yields $12 + 10 + 1$ , which equals $23$ .

Step 4: Set up expansion along the third column

For the third column, the signs determined by $(-1)^{i+j}$ are positive for $(1{,}3)$ , negative for $(2{,}3)$ , and positive for $(3{,}3)$ . The elements in this column are $4$ , $-1$ , and $2$ .

Step 5: Evaluate minors for column 3 expansion

Evaluating the $2 \times 2$ determinants: $M_{13} = (3)(1) - (5)(0) = 3$ , $M_{23} = (1)(1) - (0)(0) = 1$ , and $M_{33} = (1)(5) - (0)(3) = 5$ .

Step 6: Calculate the determinant value via column 3

Summing the products gives $12 + 1 + 10 = 23$ , which matches the result obtained from row 2 expansion as expected.

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Evaluate the determinant ∣10435−1012∣\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}​130​051​4−12​​ (i) with the help of second row. (ii) with the help of third column.