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NCERT Exemplar · Q32

Q.Let A=[12−13]A = \begin{bmatrix}1 & 2\\ -1 & 3\end{bmatrix}, B=[4015]B = \begin{bmatrix}4 & 0\\ 1 & 5\end{bmatrix}, C=[201−2]C = \begin{bmatrix}2 & 0\\ 1 & -2\end{bmatrix} and a=4a = 4, b=−2b = -2. Show that:

(a) A+(B+C)=(A+B)+CA + (B + C) = (A + B) + C
(b) A(BC)=(AB)CA(BC) = (AB)C
(c) (a+b)B=aB+bB(a + b)B = aB + bB
(d) a(C−A)=aC−aAa(C - A) = aC - aA
(e) (AT)T=A(A^T)^T = A
(f) (bA)T=b AT(bA)^T = b\,A^T
(g) (AB)T=BTAT(AB)^T = B^T A^T
(h) (A−B)C=AC−BC(A - B)C = AC - BC
(i) (A−B)T=AT−BT(A - B)^T = A^T - B^T.
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All nine identities (a)–(i) are true. We verify each by direct computation with A=[12−13]A=\begin{bmatrix}1&2\\-1&3\end{bmatrix}, B=[4015]B=\begin{bmatrix}4&0\\1&5\end{bmatrix}, C=[201−2]C=\begin{bmatrix}2&0\\1&-2\end{bmatrix}, a=4a=4, b=−2b=-2.

The plan is simple: matrices obey the same associative, distributive and transpose laws as ordinary numbers (the one thing you cannot do is swap the order in a product). To verify each law here, we compute the left side and the right side separately and check they are the identical matrix.

(a) A+(B+C)=(A+B)+CA+(B+C)=(A+B)+C — addition is associative

B+C=[4+20+01+15−2]=[6023],A+(B+C)=[7216].B+C=\begin{bmatrix}4+2&0+0\\1+1&5-2\end{bmatrix}=\begin{bmatrix}6&0\\2&3\end{bmatrix},\quad A+(B+C)=\begin{bmatrix}7&2\\1&6\end{bmatrix}.

A+B=[5208],(A+B)+C=[7216].A+B=\begin{bmatrix}5&2\\0&8\end{bmatrix},\quad (A+B)+C=\begin{bmatrix}7&2\\1&6\end{bmatrix}.

Both equal [7216]\begin{bmatrix}7&2\\1&6\end{bmatrix}.

(b) A(BC)=(AB)CA(BC)=(AB)C — multiplication is associative

BC=[807−10],A(BC)=[1⋅8+2⋅71⋅0+2(−10)−1⋅8+3⋅70+3(−10)]=[22−2013−30].BC=\begin{bmatrix}8&0\\7&-10\end{bmatrix},\quad A(BC)=\begin{bmatrix}1\cdot8+2\cdot7&1\cdot0+2(-10)\\-1\cdot8+3\cdot7&0+3(-10)\end{bmatrix}=\begin{bmatrix}22&-20\\13&-30\end{bmatrix}.

AB=[610−115],(AB)C=[12+10−20−2+15−30]=[22−2013−30].AB=\begin{bmatrix}6&10\\-1&15\end{bmatrix},\quad (AB)C=\begin{bmatrix}12+10&-20\\-2+15&-30\end{bmatrix}=\begin{bmatrix}22&-20\\13&-30\end{bmatrix}.

Both equal [22−2013−30]\begin{bmatrix}22&-20\\13&-30\end{bmatrix}.

(c) (a+b)B=aB+bB(a+b)B=aB+bB

a+b=2a+b=2, so (a+b)B=[80210](a+b)B=\begin{bmatrix}8&0\\2&10\end{bmatrix}, and aB+bB=[160420]+[−80−2−10]=[80210]aB+bB=\begin{bmatrix}16&0\\4&20\end{bmatrix}+\begin{bmatrix}-8&0\\-2&-10\end{bmatrix}=\begin{bmatrix}8&0\\2&10\end{bmatrix}.

(d) a(C−A)=aC−aAa(C-A)=aC-aA

C−A=[1−22−5]C-A=\begin{bmatrix}1&-2\\2&-5\end{bmatrix}, so a(C−A)=[4−88−20]a(C-A)=\begin{bmatrix}4&-8\\8&-20\end{bmatrix}, and aC−aA=[804−8]−[48−412]=[4−88−20]aC-aA=\begin{bmatrix}8&0\\4&-8\end{bmatrix}-\begin{bmatrix}4&8\\-4&12\end{bmatrix}=\begin{bmatrix}4&-8\\8&-20\end{bmatrix}.

(e) (AT)T=A(A^{T})^{T}=A

AT=[1−123]A^{T}=\begin{bmatrix}1&-1\\2&3\end{bmatrix}; transposing again gives [12−13]=A\begin{bmatrix}1&2\\-1&3\end{bmatrix}=A.

(f) (bA)T=b AT(bA)^{T}=b\,A^{T}

bA=[−2−42−6]bA=\begin{bmatrix}-2&-4\\2&-6\end{bmatrix}, so (bA)T=[−22−4−6](bA)^{T}=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}, and b AT=−2[1−123]=[−22−4−6]b\,A^{T}=-2\begin{bmatrix}1&-1\\2&3\end{bmatrix}=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}.

(g) (AB)T=BTAT(AB)^{T}=B^{T}A^{T} — the order reverses …

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