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Worked Examples · Example 19

Q.Given A=[35−46]A = \begin{bmatrix} 3 & 5 \\ -4 & 6 \end{bmatrix} and B=[−921−7]B = \begin{bmatrix} -9 & 2 \\ 1 & -7 \end{bmatrix}, find the following

(i) ∣A∣|A|
(ii) ∣B∣|B|
(iii) 2∣A∣2|A|
(iv) ∣2A∣|2A|
(v) ∣A∣∣B∣|A||B|
(vi) ∣AB∣|AB|.
CBSENCERTSubjective· 3mImportance★★★★★
24% · 19/80 Questions
✓ Free question

Evaluate each 2×22\times2 determinant with det⁡=ad−bc\det = ad-bc; note 2∣A∣2|A| multiplies the value by 22, while ∣2A∣=22∣A∣|2A| = 2^2|A|, and ∣AB∣=∣A∣∣B∣|AB| = |A||B|.

∣abcd∣=ad−bc,∣kA∣=kn∣A∣ (n=order=2),∣AB∣=∣A∣ ∣B∣.\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc,\qquad |kA| = k^n|A|\ (n=\text{order}=2),\qquad |AB| = |A|\,|B|.

Given A=[35−46]A = \begin{bmatrix} 3 & 5 \\ -4 & 6 \end{bmatrix}, B=[−921−7]B = \begin{bmatrix} -9 & 2 \\ 1 & -7 \end{bmatrix}.

  1. (i) ∣A∣=(3)(6)−(5)(−4)=18+20=38.|A| = (3)(6) - (5)(-4) = 18 + 20 = 38.
  2. (ii) ∣B∣=(−9)(−7)−(2)(1)=63−2=61.|B| = (-9)(-7) - (2)(1) = 63 - 2 = 61.
  3. (iii) 2∣A∣=2×38=76.2|A| = 2 \times 38 = 76.
  4. (iv) 2A=[610−812]2A = \begin{bmatrix} 6 & 10 \\ -8 & 12 \end{bmatrix}, so ∣2A∣=(6)(12)−(10)(−8)=72+80=152.|2A| = (6)(12) - (10)(-8) = 72 + 80 = 152. (Check: 22∣A∣=4×38=1522^2|A| = 4\times 38 = 152.) ✓
  5. (v) ∣A∣∣B∣=38×61=2318.|A||B| = 38 \times 61 = 2318.
  6. (vi) AB=[35−46][−921−7]=[−22−2942−50]AB = \begin{bmatrix} 3 & 5 \\ -4 & 6 \end{bmatrix}\begin{bmatrix} -9 & 2 \\ 1 & -7 \end{bmatrix} = \begin{bmatrix} -22 & -29 \\ 42 & -50 \end{bmatrix}, so ∣AB∣=(−22)(−50)−(−29)(42)=1100+1218=2318.|AB| = (-22)(-50) - (-29)(42) = 1100 + 1218 = 2318. This equals ∣A∣∣B∣|A||B|. ✓
✓Final answer

  1. ∣A∣=38|A| = 38;
  2. ∣B∣=61|B| = 61;
  3. 2∣A∣=762|A| = 76;
  4. ∣2A∣=152|2A| = 152;
  5. ∣A∣∣B∣=2318|A||B| = 2318;
  6. ∣AB∣=2318|AB| = 2318.

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