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Exercise B · Q2

Q.For A=[6−5−74]A = \begin{bmatrix} 6 & -5 \\ -7 & 4 \end{bmatrix}, B=[1−3−24]B = \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix} and C=[−213−1]C = \begin{bmatrix} -2 & 1 \\ 3 & -1 \end{bmatrix}, show that

(i) Commutative property does not hold true for multiplication of matrices AA and BB i.e. AB≠BAAB \ne BA
(ii) Associative property holds true for multiplication of three matrices, i.e. A(BC)=(AB)CA(BC) = (AB)C.
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✓ Free question

Computing the products shows AB≠BAAB\ne BA (non-commutative) while A(BC)=(AB)CA(BC)=(AB)C (associative).

For 2×22\times2 matrices, (XY)ij=∑kXikYkj(XY)_{ij}=\sum_k X_{ik}Y_{kj}. Commutativity would require AB=BAAB=BA; associativity requires A(BC)=(AB)CA(BC)=(AB)C.

Part (i): show AB≠BAAB\ne BA

  1. Compute ABAB with A=[6−5−74], B=[1−3−24]A=\begin{bmatrix}6&-5\\-7&4\end{bmatrix},\ B=\begin{bmatrix}1&-3\\-2&4\end{bmatrix}:

AB=[6(1)+(−5)(−2)6(−3)+(−5)(4)(−7)(1)+4(−2)(−7)(−3)+4(4)]=[16−38−1537]AB=\begin{bmatrix}6(1)+(-5)(-2)&6(-3)+(-5)(4)\\(-7)(1)+4(-2)&(-7)(-3)+4(4)\end{bmatrix}=\begin{bmatrix}16&-38\\-15&37\end{bmatrix}

  1. Compute BABA:

BA=[1(6)+(−3)(−7)1(−5)+(−3)(4)(−2)(6)+4(−7)(−2)(−5)+4(4)]=[27−17−4026]BA=\begin{bmatrix}1(6)+(-3)(-7)&1(-5)+(-3)(4)\\(-2)(6)+4(-7)&(-2)(-5)+4(4)\end{bmatrix}=\begin{bmatrix}27&-17\\-40&26\end{bmatrix}

  1. Since [16−38−1537]≠[27−17−4026]\begin{bmatrix}16&-38\\-15&37\end{bmatrix}\ne\begin{bmatrix}27&-17\\-40&26\end{bmatrix}, we get AB≠BAAB\ne BA.

Part (ii): show A(BC)=(AB)CA(BC)=(AB)C

  1. Compute BCBC with C=[−213−1]C=\begin{bmatrix}-2&1\\3&-1\end{bmatrix}:

BC=[1(−2)+(−3)(3)1(1)+(−3)(−1)(−2)(−2)+4(3)(−2)(1)+4(−1)]=[−11416−6]BC=\begin{bmatrix}1(-2)+(-3)(3)&1(1)+(-3)(-1)\\(-2)(-2)+4(3)&(-2)(1)+4(-1)\end{bmatrix}=\begin{bmatrix}-11&4\\16&-6\end{bmatrix}

  1. Compute A(BC)A(BC):

A(BC)=[6(−11)+(−5)(16)6(4)+(−5)(−6)(−7)(−11)+4(16)(−7)(4)+4(−6)]=[−14654141−52]A(BC)=\begin{bmatrix}6(-11)+(-5)(16)&6(4)+(-5)(-6)\\(-7)(-11)+4(16)&(-7)(4)+4(-6)\end{bmatrix}=\begin{bmatrix}-146&54\\141&-52\end{bmatrix}

  1. Compute (AB)C(AB)C using AB=[16−38−1537]AB=\begin{bmatrix}16&-38\\-15&37\end{bmatrix}:

(AB)C=[16(−2)+(−38)(3)16(1)+(−38)(−1)(−15)(−2)+37(3)(−15)(1)+37(−1)]=[−14654141−52](AB)C=\begin{bmatrix}16(-2)+(-38)(3)&16(1)+(-38)(-1)\\(-15)(-2)+37(3)&(-15)(1)+37(-1)\end{bmatrix}=\begin{bmatrix}-146&54\\141&-52\end{bmatrix}

  1. Both equal [−14654141−52]\begin{bmatrix}-146&54\\141&-52\end{bmatrix}, hence A(BC)=(AB)CA(BC)=(AB)C.
✓Final answer

  1. AB=[16−38−1537]≠[27−17−4026]=BAAB=\begin{bmatrix}16&-38\\-15&37\end{bmatrix}\ne\begin{bmatrix}27&-17\\-40&26\end{bmatrix}=BA — not commutative.
  2. A(BC)=(AB)C=[−14654141−52]A(BC)=(AB)C=\begin{bmatrix}-146&54\\141&-52\end{bmatrix} — associativity holds.

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