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Exercise 4.4 · Q1

Q.Find the value of the following: [1234]\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}

Sikkim CbseNCERTSubjective· 2mImportance★★★★★
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✓ Free question

The determinant of a 2×22 \times 2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} is ad−bcad - bc. For [1234]\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, this gives 1⋅4−2⋅3=4−6=−21 \cdot 4 - 2 \cdot 3 = 4 - 6 = -2.

The determinant is a single number that captures key information about a matrix: whether it is invertible, the scaling factor of the area it transforms, and more. For a 2×22 \times 2 matrix, the formula is straightforward, but it’s worth understanding why it works.

Think of the rows as vectors: (1,2)(1, 2) and (3,4)(3, 4). The determinant measures the signed area of the parallelogram they span. The formula ad−bcad - bc comes from subtracting the product of the “wrong” diagonal from the product of the “main” diagonal. Here’s how we apply it.

  1. Identify the entries. In the matrix [1234]\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, we have:

    • a=1a = 1 (top-left)
    • b=2b = 2 (top-right)
    • c=3c = 3 (bottom-left)
    • d=4d = 4 (bottom-right)
  2. Apply the determinant formula. For any 2×22 \times 2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}, the determinant is:

det⁡=ad−bc\det = ad - bc

Substitute the values:

det⁡=(1)(4)−(2)(3)\det = (1)(4) - (2)(3)

  1. Compute the products. First, 1×4=41 \times 4 = 4. Then, 2×3=62 \times 3 = 6.

  2. Subtract. 4−6=−24 - 6 = -2.

Watch out

A common mistake is to compute ac−bdac - bd or to mix up the order. Always remember: it’s the product of the main diagonal (top-left to bottom-right) minus the product of the other diagonal (top-right to bottom-left). The order matters — swapping gives the wrong sign.

The result is −2-2. This negative value tells us that the orientation of the vectors is reversed relative to the standard basis — the area is still 2 square units, but with a flipped direction.

Tip

If you ever forget the formula, think of the cross product of the row vectors: (1,2)×(3,4)=1⋅4−2⋅3=−2(1,2) \times (3,4) = 1 \cdot 4 - 2 \cdot 3 = -2. The determinant is essentially the 2D version of that.

✓Final answer

The value is −2\boxed{-2}.

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