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Class 12 RD Sharma Solutions – Chapter 5 Algebra of Matrices – Exercise 5.3 | Set 1

  • Last Updated : 21 Jul, 2021
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Question 1. Compute the indicated products:

(i) \begin{bmatrix}a & b\\-b & a\end{bmatrix}\begin{bmatrix}a & -b\\b & a\end{bmatrix}

Solution:

We have,

\begin{bmatrix}a & b\\-b & a\end{bmatrix}\begin{bmatrix}a & -b\\b & a\end{bmatrix}=\begin{bmatrix}a(a) +b(b) & a(-b) +b(a) \\-b(a) +a(b) & (-b) (-b) +a(a) \end{bmatrix}

=\begin{bmatrix}a^2+b^2& -ab+ab\\-ab+ab & a^2-b^2\end{bmatrix}



=\begin{bmatrix}a^2+b^2& 0\\0 & a^2-b^2\end{bmatrix}

(ii) \begin{bmatrix}1 & -2\\2 & 3\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\-3 & 2 & -1\end{bmatrix}

Solution:

We have,

\begin{bmatrix}1 & -2\\2 & 3\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\-3 & 2 & -1\end{bmatrix}=\begin{bmatrix}1(1) +(-2) (-3) & 1(2) +(-2) (2) & 1(3) +(-2) (-1) \\2(1) +3(-3) & 2(2) +3(2) & 2(3) +3(-1) \end{bmatrix}

=\begin{bmatrix}1+6 & 2-4 & 3+2\\2-9 & 4+6 & 6-3\end{bmatrix}

=\begin{bmatrix}7 & -2 & 5\\-7 & 10 & 3\end{bmatrix}

(iii) \begin{bmatrix}2 & 3 & 4\\3 & 4 & 5\\4 & 5 & 6\end{bmatrix}\begin{bmatrix}1 & -3 & 5\\0 & 2 & 4\\3 & 0 & 5\end{bmatrix}



Solution:

We have,

\begin{bmatrix}2 & 3 & 4\\3 & 4 & 5\\4 & 5 & 6\end{bmatrix}\begin{bmatrix}1 & -3 & 5\\0 & 2 & 4\\3 & 0 & 5\end{bmatrix}=\begin{bmatrix}2(1) +3(0) +4(3) & 2(-3) +3(2) +4(0) & 2(5) +3(4) +4(5) \\3(1) +4(0) +5(3) & 3(-3) +4(2) +5(0) & 3(5) +4(4) +5(5) \\4(1) +5(0) +6(3) & 4(-3) +5(2) +6(0) & 4(5) +5(4) +6(5) \end{bmatrix}

=\begin{bmatrix}2+0+12 & -6+6+0 & 10+12+20\\3+0+15 & -9+8+0 & 15+16+25\\4+0+18 & -12+10+0 & 20+20+30\end{bmatrix}

=\begin{bmatrix}14 & 0 & 42\\18 & -1 & 56\\22 & -2 & 70\end{bmatrix}

Question 2. Show that AB ≠ BA in each of the following cases:

(i) A=\begin{bmatrix}5 & -1\\6 & 7\end{bmatrix}   and B=\begin{bmatrix}2 & 1\\3 & 4\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}5 & -1\\6 & 7\end{bmatrix}   and B =\begin{bmatrix}2 & 1\\3 & 4\end{bmatrix}

AB =\begin{bmatrix}5 & -1\\6 & 7\end{bmatrix}\begin{bmatrix}2 & 1\\3 & 4\end{bmatrix}



=\begin{bmatrix}10-3 & 5-4\\12+21 & 6+28\end{bmatrix}

=\begin{bmatrix}7 & 1\\33 & 34\end{bmatrix}

And we have,

BA =\begin{bmatrix}2 & 1\\3 & 4\end{bmatrix}\begin{bmatrix}5 & -1\\6 & 7\end{bmatrix}

=\begin{bmatrix}10+6 & -2+7\\15+24 & -3+28\end{bmatrix}

=\begin{bmatrix}16 & 5\\39 & 25\end{bmatrix}

Therefore, AB ≠ BA.

Hence, proved.

(ii) A=\begin{bmatrix}-1 & 1 & 0\\0 & -1 & 1\\2 & 3 & 4\end{bmatrix}   and B=\begin{bmatrix}1 & 2 & 3\\0 & 1 & 0\\1 & 1 & 0\end{bmatrix}

Solution:



We have,

A =\begin{bmatrix}-1 & 1 & 0\\0 & -1 & 1\\2 & 3 & 4\end{bmatrix}   and B =\begin{bmatrix}1 & 2 & 3\\0 & 1 & 0\\1 & 1 & 0\end{bmatrix}

AB =\begin{bmatrix}-1 & 1 & 0\\0 & -1 & 1\\2 & 3 & 4\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\0 & 1 & 0\\1 & 1 & 0\end{bmatrix}

=\begin{bmatrix}-1+0+0 & -2+1+0 & -3+0+0\\0+0+1 & 0-1+1 & 0+0+0\\2+0+4 & 4+3+4 & 6+0+0\end{bmatrix}

=\begin{bmatrix}-1 & -1 & 3\\0 & 1 & 0\\1 & 1 & 0\end{bmatrix}

And we have,

BA =\begin{bmatrix}1 & 2 & 3\\0 & 1 & 0\\1 & 1 & 0\end{bmatrix}\begin{bmatrix}-1 & 1 & 0\\0 & -1 & 1\\2 & 3 & 4\end{bmatrix}

=\begin{bmatrix}-1+0+6 & 1-2+9 & 0+2+12\\0+0+0 & 0-1+0 & 0+1+0\\-1+0+0 & 1-1+0 & 0+1+0\end{bmatrix}

=\begin{bmatrix}5 & 8 & 14\\0 & -1 & 1\\-1 & 0 & 1\end{bmatrix}

Therefore, AB ≠ BA.



Hence proved.

(iii) A=\begin{bmatrix}1 & 3 & 0\\1 & 1 & 0\\4 & 1 & 0\end{bmatrix}   and B=\begin{bmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & 5 & 1\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}1 & 3 & 0\\1 & 1 & 0\\4 & 1 & 0\end{bmatrix}   and B =\begin{bmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & 5 & 1\end{bmatrix}

AB =\begin{bmatrix}1 & 3 & 0\\1 & 1 & 0\\4 & 1 & 0\end{bmatrix}\begin{bmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & 5 & 1\end{bmatrix}

=\begin{bmatrix}0+3+0 & 1+0+0 & 0+0+0\\0+1+0 & 1+0+0 & 0+0+0\\0+1+0 & 4+0+0 & 0+0+0\end{bmatrix}

=\begin{bmatrix}3 & 1 & 0\\1 & 1 & 0\\1 & 4 & 0\end{bmatrix}

And we have,

BA =\begin{bmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & 5 & 1\end{bmatrix}\begin{bmatrix}1 & 3 & 0\\1 & 1 & 0\\4 & 1 & 0\end{bmatrix}



=\begin{bmatrix}0+1+0 & 0+1+0 & 0+0+0\\1+0+0 & 3+0+0 & 0+0+0\\0+5+4 & 0+5+1 & 0+0+0\end{bmatrix}

=\begin{bmatrix}1 & 1 & 0\\1 & 3 & 0\\9 & 6 & 0\end{bmatrix}

Therefore, AB ≠ BA.

Hence proved.

Question 3. Compute the products AB and BA whichever exists in each of the following cases:

(i) A=\begin{bmatrix}1 & -2\\2 & 3\end{bmatrix}   and B=\begin{bmatrix}1 & 2 & 3\\2 & 3 & 1\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}1 & -2\\2 & 3\end{bmatrix}   and B =\begin{bmatrix}1 & 2 & 3\\2 & 3 & 1\end{bmatrix}

As A is of order 2 × 2 and B is of order 2 × 3, AB is possible but BA is not possible.

So, we get



AB =\begin{bmatrix}1 & -2\\2 & 3\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\2 & 3 & 1\end{bmatrix}

=\begin{bmatrix}1-4 & 2-6 & 3-2\\2+6 & 4+9 & 6+3\end{bmatrix}

=\begin{bmatrix}-3 & -4 & 1\\8 & 13 & 9\end{bmatrix}

(ii) A=\begin{bmatrix}3 & 2\\-1 & 0\\-1 & 1\end{bmatrix}   and B=\begin{bmatrix}4 & 5 & 6\\0 & 1 & 2\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}3 & 2\\-1 & 0\\-1 & 1\end{bmatrix}   and B =\begin{bmatrix}4 & 5 & 6\\0 & 1 & 2\end{bmatrix}

As A is of order 3 × 2 and B is of order 2 × 3, AB and BA both are possible.

So, we get,

AB =\begin{bmatrix}3 & 2\\-1 & 0\\-1 & 1\end{bmatrix}\begin{bmatrix}4 & 5 & 6\\0 & 1 & 2\end{bmatrix}



=\begin{bmatrix}12+0 & 15+2 & 18+4\\-4+0 & -5+0 & -6+0\\-4+0 & -5+0 & -6+2\end{bmatrix}

=\begin{bmatrix}12 & 17 & 22\\-4 & -5 & -6\\-4 & -5 & -4\end{bmatrix}

Also we have,

BA =\begin{bmatrix}4 & 5 & 6\\0 & 1 & 2\end{bmatrix}\begin{bmatrix}3 & 2\\-1 & 0\\-1 & 1\end{bmatrix}

=\begin{bmatrix}12-5-6 & 8+0+6\\0-1-2 & 0+0+2\end{bmatrix}

=\begin{bmatrix}1 & 14\\-3 & 2\end{bmatrix}

(iii) A=\begin{bmatrix}1 & -1 & 2 & 3\end{bmatrix}   and B=\begin{bmatrix}0\\1\\3\\2\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}1 & -1 & 2 & 3\end{bmatrix}   and B =\begin{bmatrix}0\\1\\3\\2\end{bmatrix}



As A is of order 1 × 4 and B is of order 4 × 1, AB and BA both are possible.

So, we get,

AB =\begin{bmatrix}1 & -1 & 2 & 3\end{bmatrix}\begin{bmatrix}0\\1\\3\\2\end{bmatrix}

=\begin{bmatrix}1(0) +(-1) (1) +2(3) +3(2) \end{bmatrix}

=\begin{bmatrix}0-1+6+6\end{bmatrix}

=\begin{bmatrix}11\end{bmatrix}

Also, we have,

BA =\begin{bmatrix}0\\1\\3\\2\end{bmatrix}\begin{bmatrix}1 & -1 & 2 & 3\end{bmatrix}

=\begin{bmatrix}0(1) & 0(-1) & 0(2) & 0(3) \\1(1) & 1(-1) & 1(2) & 1(3) \\3(1) & 3(-1) & 3(2) & 3(3) \\2(1) & 2(-1) & 3(2) & 2(3) \end{bmatrix}

=\begin{bmatrix}0 & 0 & 0 & 0\\1 & -1 & 2 & 3\\3 & -3 & 6 & 9\\2 & -2 & 6 & 6\end{bmatrix}



(iv) \begin{bmatrix}a & b\end{bmatrix}\begin{bmatrix}c\\d\end{bmatrix}+\begin{bmatrix}a & b & c & d\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\end{bmatrix}

Solution:

We have,

=\begin{bmatrix}a & b\end{bmatrix}\begin{bmatrix}c\\d\end{bmatrix}+\begin{bmatrix}a & b & c & d\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\end{bmatrix}

=\begin{bmatrix}ac+bd\end{bmatrix}\begin{bmatrix}a^2+b^2+c^2+d^2\end{bmatrix}

=\begin{bmatrix}ac+bd+a^2+b^2+c^2+d^2\end{bmatrix}

Question 4. Show that AB ≠ BA in each of the following cases:

(i) A=\begin{bmatrix}1 & 3 & -1\\2 & -1 & -1\\3 & 0 & -1\end{bmatrix}   and B=\begin{bmatrix}-2 & 3 & -1\\1 & 2 & -1\\-6 & 9 & -4\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}1 & 3 & -1\\2 & -1 & -1\\3 & 0 & -1\end{bmatrix}   and B =\begin{bmatrix}-2 & 3 & -1\\1 & 2 & -1\\-6 & 9 & -4\end{bmatrix}



AB =\begin{bmatrix}1 & 3 & -1\\2 & -1 & -1\\3 & 0 & -1\end{bmatrix}\begin{bmatrix}-2 & 3 & -1\\1 & 2 & -1\\-6 & 9 & -4\end{bmatrix}

=\begin{bmatrix}-2-3+6 & 3+6-9 & -1-3+4\\-4+1+6 & 6-2-9 & -2+1+4\\-6+0+6 & 9+0-9 & -3+0+4\end{bmatrix}

=\begin{bmatrix}1 & 0 & 0\\3 & -5 & 3\\0 & 0 & 1\end{bmatrix}

And we have,

BA =\begin{bmatrix}-2 & 3 & -1\\1 & 2 & -1\\-6 & 9 & -4\end{bmatrix}\begin{bmatrix}1 & 3 & -1\\2 & -1 & -1\\3 & 0 & -1\end{bmatrix}

=\begin{bmatrix}-2+6-3 & -6-3+0 & 2-3+1\\-1+4-3 & -3-2+0 & 1-2+1\\-6+18-12 & -18-9+0 & 6-9+4\end{bmatrix}

=\begin{bmatrix}1 & -9 & 0\\0 & -5 & 0\\0 & -27 & 1\end{bmatrix}

Therefore, AB ≠ BA.

Hence proved.

(ii) A=\begin{bmatrix}10 & -4 & -1\\-11 & 5 & 0\\9 & -5 & 1\end{bmatrix}   and B=\begin{bmatrix}1 & 2 & 1\\3 & 4 & 2\\1 & 3 & 2\end{bmatrix}



Solution:

We have,

A =\begin{bmatrix}10 & -4 & -1\\-11 & 5 & 0\\9 & -5 & 1\end{bmatrix}   and B =\begin{bmatrix}1 & 2 & 1\\3 & 4 & 2\\1 & 3 & 2\end{bmatrix}

AB =\begin{bmatrix}10 & -4 & -1\\-11 & 5 & 0\\9 & -5 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 1\\3 & 4 & 2\\1 & 3 & 2\end{bmatrix}

=\begin{bmatrix}10-12-1 & 20-16-3 & 10-8-2\\-11+15+0 & -22+20+0 & -11+10+0\\9-15+1 & 18-20+3 & 9-10+2\end{bmatrix}

=\begin{bmatrix}-3 & 1 & 0\\4 & -2 & -1\\-5 & 1 & 1\end{bmatrix}

And we have,

BA =\begin{bmatrix}1 & 2 & 1\\3 & 4 & 2\\1 & 3 & 2\end{bmatrix}\begin{bmatrix}10 & -4 & -1\\-11 & 5 & 0\\9 & -5 & 1\end{bmatrix}

=\begin{bmatrix}10-22+9 & -4+10-5 & -9+0+1\\30-44+10 & -12+20-10 & -3+0+2\\10-33+18 & -4+15-10 & -1+0+2\end{bmatrix}

=\begin{bmatrix}-3 & 1 & 0\\4 & -2 & -1\\-5 & 1 & 1\end{bmatrix}



Therefore, AB ≠ BA.

Hence proved.

Question 5. Evaluate the following:

(i) \left(\begin{bmatrix}1 & 3\\-1 & -4\end{bmatrix}+\begin{bmatrix}3 & -2\\-1 & 1\end{bmatrix}\right) \begin{bmatrix}1 & 3 & 5\\2 & 4 & 6\end{bmatrix}

Solution:

We have,

=\left(\begin{bmatrix}1 & 3\\-1 & -4\end{bmatrix}+\begin{bmatrix}3 & -2\\-1 & 1\end{bmatrix}\right) \begin{bmatrix}1 & 3 & 5\\2 & 4 & 6\end{bmatrix}

=\left(\begin{bmatrix}1+3 & 3-2\\-1+1 & -4+1\end{bmatrix}\right) \begin{bmatrix}1 & 3 & 5\\2 & 4 & 6\end{bmatrix}

=\begin{bmatrix}4 & 1\\0 & -3\end{bmatrix}\begin{bmatrix}1 & 3 & 5\\2 & 4 & 6\end{bmatrix}

=\begin{bmatrix}4+2 & 12+4 & 20+6\\-2-6 & -6-12 & -10-18\\\end{bmatrix}

=\begin{bmatrix}6 & 16 & 26\\-8 & -18 & -28\\\end{bmatrix}

(ii) \begin{bmatrix}1 & 2 & 3\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\2 & 0 & 1\\0 & 1 & 2\end{bmatrix}\begin{bmatrix}2\\4\\6\end{bmatrix}

Solution:

We have,

=\begin{bmatrix}1 & 2 & 3\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\2 & 0 & 1\\0 & 1 & 2\end{bmatrix}\begin{bmatrix}2\\4\\6\end{bmatrix}

=\begin{bmatrix}1+4+0 & 0+0+3 & 2+0+6\end{bmatrix}\begin{bmatrix}2\\4\\6\end{bmatrix}

=\begin{bmatrix}5 & 3 & 8\end{bmatrix}\begin{bmatrix}2\\4\\6\end{bmatrix}

=\begin{bmatrix}10+12+60\end{bmatrix}

=\begin{bmatrix}82\end{bmatrix}

(iii) \begin{bmatrix}1 & -1\\0 & 2\\2 & 3\end{bmatrix}\left(\begin{bmatrix}1 & 0 & 2\\2 & 0 & 1\end{bmatrix}-\begin{bmatrix}0 & 1 & 2\\1 & 0 & 2\end{bmatrix}\right)

Solution:



We have,

=\begin{bmatrix}1 & -1\\0 & 2\\2 & 3\end{bmatrix}\left(\begin{bmatrix}1 & 0 & 2\\2 & 0 & 1\end{bmatrix}-\begin{bmatrix}0 & 1 & 2\\1 & 0 & 2\end{bmatrix}\right)

=\begin{bmatrix}1 & -1\\0 & 2\\2 & 3\end{bmatrix}\begin{bmatrix}1-0 & 0-1 & 2-2\\2-1 & 0-0 & 1-2\end{bmatrix}

=\begin{bmatrix}1 & -1\\0 & 2\\2 & 3\end{bmatrix}\begin{bmatrix}1 & -1 & 0\\1 & 0 & -1\end{bmatrix}

=\begin{bmatrix}1-1 & -1+0 & 0+1\\0+2 & 0+0 & 0-2\\2+3 & -2+0 & 0-3\\\end{bmatrix}

=\begin{bmatrix}0 & -1 & 1\\2 & 0 & -2\\5 & -2 & -3\\\end{bmatrix}

Question 6. If A =\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}   , B =\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}   and C =\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}   , then show that A2 = B2 = C2 = I2.

Solution:

We have,

A =\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}   , B =\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}   and C =

\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}



A2 =\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}1+0 & 0+0\\0+0 & 0+1\end{bmatrix}

=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

Therefore, A2 = I2

B2 =\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}

=\begin{bmatrix}1+0 & 0+0\\0+0 & 0+1\end{bmatrix}

=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

Therefore, B2 = I2

C2 =\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}

=\begin{bmatrix}1+0 & 0+0\\0+0 & 0+1\end{bmatrix}

=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

Therefore, C2 = I2

So, we get A2 = B2 = C2 = I2

Hence proved.

Question 7. If A =\begin{bmatrix}2 & -1\\3 & 2\end{bmatrix}   and B =\begin{bmatrix}0 & 4\\-1 & 7\end{bmatrix}   , find 3A2 – 2B + I.

Solution:

We are given,

A =\begin{bmatrix}2 & -1\\3 & 2\end{bmatrix}   and B =\begin{bmatrix}0 & 4\\-1 & 7\end{bmatrix}

So, we get,

3A2 – 2B + I =3\begin{bmatrix}2 & -1\\3 & 2\end{bmatrix}\begin{bmatrix}2 & -1\\3 & 2\end{bmatrix}-2\begin{bmatrix}0 & 4\\-1 & 7\end{bmatrix}+\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=3\begin{bmatrix}4-3 & -2-2\\6+6 & -3+4\end{bmatrix}-\begin{bmatrix}0 & 8\\-2 & 14\end{bmatrix}+\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}



=3\begin{bmatrix}1 & -4\\12 & 1\end{bmatrix}-\begin{bmatrix}0 & 8\\-2 & 14\end{bmatrix}+\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}3 & -12\\36 & 3\end{bmatrix}-\begin{bmatrix}0 & 8\\-2 & 14\end{bmatrix}+\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}3-0+1 & -12-8+0\\36+2+0 & 3-14+1\end{bmatrix}

=\begin{bmatrix}4 & -20\\38 & -10\end{bmatrix}

Question 8. If A =\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix}   , prove that (A – 2I) (A – 3I) = 0.

Solution:

We are given,

A =\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix}

L.H.S. = (A – 2I) (A – 3I)

=\left(\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix}-2\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\right) \left(\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix}-3\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\right)

=\left(\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix}-\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}\right) \left(\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix}-\begin{bmatrix}3 & 0\\0 & 3\end{bmatrix}\right)

=\begin{bmatrix}4-2 & 2-0\\-1-0 & 1-2\end{bmatrix}\begin{bmatrix}4-3 & 2-0\\-1-0 & 1-3\end{bmatrix}

=\begin{bmatrix}2 & 2\\-1 & -1\end{bmatrix}\begin{bmatrix}1 & 2\\-1 & -2\end{bmatrix}

=\begin{bmatrix}2-2 & 4-4\\-1+1 & -2+2\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 9. If A =\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}   , show that A2 =\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}   and A3 =\begin{bmatrix}1 & 3\\0 & 1\end{bmatrix}   .

Solution:

We have,

A =\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}



So, A2 =\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}

=\begin{bmatrix}1+0 & 1+1\\0+0 & 0+1\end{bmatrix}

=\begin{bmatrix}1 & 2\\0 & 1\end{bmatrix}

Hence, A3 = A2 . A

=\begin{bmatrix}1 & 2\\0 & 1\end{bmatrix}\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}

=\begin{bmatrix}1+0 & 1+2\\0+0 & 0+1\end{bmatrix}

=\begin{bmatrix}1 & 3\\0 & 1\end{bmatrix}

Hence proved.

Question 10. If A =\begin{bmatrix}ab & b^2\\-a^2 & -ab\end{bmatrix}   , show that A2 = 0.

Solution:

We have,

A =\begin{bmatrix}ab & b^2\\-a^2 & -ab\end{bmatrix}

So, we get

L.H.S. = A2 =\begin{bmatrix}ab & b^2\\-a^2 & -ab\end{bmatrix}\begin{bmatrix}ab & b^2\\-a^2 & -ab\end{bmatrix}

=\begin{bmatrix}a^2b^2-a^2b^2 & ab^3-ab^3\\-a^3b+a^3b & -a^2b^2+a^2b^2\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 11. If A =\begin{bmatrix}\cos2\theta & \sin2\theta\\-\sin2\theta & \cos2\theta\end{bmatrix}   , find A2.

Solution:

We have,

A =\begin{bmatrix}\cos2\theta & \sin2\theta\\-\sin2\theta & \cos2\theta\end{bmatrix}

So, we get

A2 =\begin{bmatrix}\cos2\theta & \sin2\theta\\-\sin2\theta & \cos2\theta\end{bmatrix}\begin{bmatrix}\cos2\theta & \sin2\theta\\-\sin2\theta & \cos2\theta\end{bmatrix}

=\begin{bmatrix}\cos^22\theta-\sin^22\theta & \cos2\theta\sin2\theta+\cos2\theta\sin2\theta\\-\cos2\theta\sin2\theta-\cos2\theta\sin2\theta & \cos^22\theta-\sin^22\theta\end{bmatrix}

=\begin{bmatrix}\cos4\theta & 2\cos2\theta\sin2\theta\\-2\cos2\theta\sin2\theta & \cos4\theta\end{bmatrix}

=\begin{bmatrix}\cos4\theta & \sin4\theta\\-\sin4\theta & \cos4\theta\end{bmatrix}

Question 12. If A =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}   and B =\begin{bmatrix}-1 & 3 & 5\\1 & -3 & -5\\-1 & 3 & 5\end{bmatrix}   , show that AB = BA = O3×3.

Solution:

We have,

A =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}   and B =\begin{bmatrix}-1 & 3 & 5\\1 & -3 & -5\\-1 & 3 & 5\end{bmatrix}

So, we get



AB =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}\begin{bmatrix}-1 & 3 & 5\\1 & -3 & -5\\-1 & 3 & 5\end{bmatrix}

=\begin{bmatrix}-2-3+5 & 6+9-15 & 10+15+25\\1+4-5 & -3-12+15 & -5-20+25\\-1-3+4 & 3+9-12 & 5+15-20\end{bmatrix}

=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

= O3×3

And we have,

BA =\begin{bmatrix}-1 & 3 & 5\\1 & -3 & -5\\-1 & 3 & 5\end{bmatrix}\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}

=\begin{bmatrix}-2-3+5 & 3+12-15 & 5+15-20\\2+3-5 & -3-12+15 & -5-15+20\\-2-3+5 & 3+12-15 & 5+15-20\end{bmatrix}

=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

= O3×3

Therefore, AB = BA = O3×3.

Hence proved.

Question 13. If A =\begin{bmatrix}0 & c & -b\\-c & 0 & a\\b & -a & 0\end{bmatrix}   and B =\begin{bmatrix}a^2 & ab & ac\\ab & b^2 & bc\\ac & bc & c^2\end{bmatrix}   , show that AB = BA = O3×3.

Solution:

We have,

A =\begin{bmatrix}0 & c & -b\\-c & 0 & a\\b & -a & 0\end{bmatrix}   and B =\begin{bmatrix}a^2 & ab & ac\\ab & b^2 & bc\\ac & bc & c^2\end{bmatrix}

So, we have,

AB =\begin{bmatrix}0 & c & -b\\-c & 0 & a\\b & -a & 0\end{bmatrix}\begin{bmatrix}a^2 & ab & ac\\ab & b^2 & bc\\ac & bc & c^2\end{bmatrix}

=\begin{bmatrix}0+abc-abc & 0+b^2c-b^2c & 0+bc^2-bc^2\\-a^2c+0+a^2c & -abc+0+abc & -ac^2+0+ac^2\\a^2b-a^2b+0 & ab^2-ab^2+0 & abc-abc+0\end{bmatrix}

=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

And we have,

BA =\begin{bmatrix}a^2 & ab & ac\\ab & b^2 & bc\\ac & bc & c^2\end{bmatrix}\begin{bmatrix}0 & c & -b\\-c & 0 & a\\b & -a & 0\end{bmatrix}



=\begin{bmatrix}0+abc-abc & 0+b^2c-b^2c & 0+bc^2-bc^2\\-a^2c+0+a^2c & -abc+0+abc & -ac^2+0+ac^2\\a^2b-a^2b+0 & ab^2-ab^2+0 & abc-abc+0\end{bmatrix}

=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

= O3×3

Therefore, AB = BA = O3×3.

Hence proved.

Question 14. If A =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}   and B =\begin{bmatrix}2 & -2 & -4\\-1 & 3 & 4\\1 & -2 & -3\end{bmatrix}   , show that AB = A and BA = B.

Solution:

We have,

A =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}   and B =\begin{bmatrix}2 & -2 & -4\\-1 & 3 & 4\\1 & -2 & -3\end{bmatrix}

AB =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}\begin{bmatrix}2 & -2 & -4\\-1 & 3 & 4\\1 & -2 & -3\end{bmatrix}

=\begin{bmatrix}4+3-5 & -4-9+10 & -8-12+15\\-2-4+5 & 2+12-10 & 4+16-15\\2+3-4 & -2-9+18 & -4-12+12\end{bmatrix}

=\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}

= A

And we have,

BA =\begin{bmatrix}2 & -2 & -4\\-1 & 3 & 4\\1 & -2 & -3\end{bmatrix}\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}

=\begin{bmatrix}4+2-4 & -6-8+12 & -10-10+16\\-2-3+4 & 3+12-12 & 5+15-16\\2+2-3 & -3-8+9 & -5-10+12\end{bmatrix}

=\begin{bmatrix}2 & -2 & -4\\-1 & 3 & 4\\1 & -2 & -3\end{bmatrix}

= B

Hence proved.

Question 15. If A =\begin{bmatrix}-1 & 1 & -1\\3 & -3 & 3\\5 & 5 & 5\end{bmatrix}   and B =\begin{bmatrix}0 & 4 & 3\\1 & -3 & -3\\-1 & 4 & 4\end{bmatrix}   , compute A2 – B2.

Solution:

We have,



A =\begin{bmatrix}-1 & 1 & -1\\3 & -3 & 3\\5 & 5 & 5\end{bmatrix}   and B =\begin{bmatrix}0 & 4 & 3\\1 & -3 & -3\\-1 & 4 & 4\end{bmatrix}

A2 =\begin{bmatrix}-1 & 1 & -1\\3 & -3 & 3\\5 & 5 & 5\end{bmatrix}\begin{bmatrix}-1 & 1 & -1\\3 & -3 & 3\\5 & 5 & 5\end{bmatrix}

=\begin{bmatrix}4+2-4 & -6-8+12 & -10-10+16\\-2-3+4 & 3+12-12 & 5+15-16\\2+2-3 & -3-8+9 & -5-10+12\end{bmatrix}

=\begin{bmatrix}-1 & -9 & -1\\3 & 27 & 3\\35 & 15 & 35\end{bmatrix}

And we have,

B2 =\begin{bmatrix}0 & 4 & 3\\1 & -3 & -3\\-1 & 4 & 4\end{bmatrix}\begin{bmatrix}0 & 4 & 3\\1 & -3 & -3\\-1 & 4 & 4\end{bmatrix}

=\begin{bmatrix}-2-3+5 & 3+12-15 & 5+15-20\\2+3-5 & -3-12+15 & -5-15+20\\-2-3+5 & 3+12-15 & 5+15-20\end{bmatrix}

=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}

So, we get

A2 – B2 =\begin{bmatrix}-1 & -9 & -1\\3 & 27 & 3\\35 & 15 & 35\end{bmatrix}-\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}

=\begin{bmatrix}-1-1 & -9-0 & -1-0\\3-0 & 27-1 & 3-0\\35-0 & 15-0 & 35-1\end{bmatrix}

=\begin{bmatrix}-2 & -9 & -1\\3 & 26 & 3\\35 & 15 & 34\end{bmatrix}

Question 16. For the following matrices verify the associativity of matrix multiplication i.e. (AB) C = A (BC).

(i) A =\begin{bmatrix}1 & 2 & 0\\-1 & 0 & 1\end{bmatrix}   , B =\begin{bmatrix}1 & 0\\-1 & 2\\0 & 3\end{bmatrix}   , C =\begin{bmatrix}1\\-1\end{bmatrix}

Solution:

We are given,

A =\begin{bmatrix}1 & 2 & 0\\-1 & 0 & 1\end{bmatrix}   , B =\begin{bmatrix}1 & 0\\-1 & 2\\0 & 3\end{bmatrix}   , C =\begin{bmatrix}1\\-1\end{bmatrix}

L.H.S. = (AB) C

=\left(\begin{bmatrix}1 & 2 & 0\\-1 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 0\\-1 & 2\\0 & 3\end{bmatrix}\right) \begin{bmatrix}1\\-1\end{bmatrix}

=\begin{bmatrix}1-2+0 & 0+4+0\\-1+0+0 & 0+0+3\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}

=\begin{bmatrix}-1 & 4\\-1 & 3\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}



=\begin{bmatrix}-5\\-4\end{bmatrix}

And R.H.S. = A (BC)

=\begin{bmatrix}1 & 2 & 0\\-1 & 0 & 1\end{bmatrix}\left(\begin{bmatrix}1 & 0\\-1 & 2\\0 & 3\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}\right)

=\begin{bmatrix}1 & 2 & 0\\-1 & 0 & 1\end{bmatrix}\begin{bmatrix}1+0\\-1-2\\0-3\end{bmatrix}

=\begin{bmatrix}1 & 2 & 0\\-1 & 0 & 1\end{bmatrix}\begin{bmatrix}1\\-3\\-3\end{bmatrix}

=\begin{bmatrix}1-6+0\\-1+0-3\end{bmatrix}

=\begin{bmatrix}-5\\-4\end{bmatrix}

= L.H.S.

Hence proved.

(ii) A =\begin{bmatrix}4 & 2 & 3\\1 & 1 & 2\\3 & 0 & 1\end{bmatrix}   , B =\begin{bmatrix}1 & -1 & 1\\0 & 1 & 2\\2 & -1 & 1\end{bmatrix}   , C =\begin{bmatrix}1 & 2 & -1\\3 & 0 & 1\\0 & 0 & 1\end{bmatrix}

Solution:

We are given,

A =\begin{bmatrix}4 & 2 & 3\\1 & 1 & 2\\3 & 0 & 1\end{bmatrix}   , B =\begin{bmatrix}1 & -1 & 1\\0 & 1 & 2\\2 & -1 & 1\end{bmatrix}   , C =\begin{bmatrix}1 & 2 & -1\\3 & 0 & 1\\0 & 0 & 1\end{bmatrix}

L.H.S. = (AB) C

=\left(\begin{bmatrix}4 & 2 & 3\\1 & 1 & 2\\3 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & -1 & 1\\0 & 1 & 2\\2 & -1 & 1\end{bmatrix}\right) \begin{bmatrix}1 & 2 & -1\\3 & 0 & 1\\0 & 0 & 1\end{bmatrix}

=\begin{bmatrix}4+0+6 & -4+2-3 & 4+4+3\\1+0+4 & -1+1-2 & 1+2+2\\3+0+2 & -3+0-1 & 3+0+1\end{bmatrix}\begin{bmatrix}1 & 2 & -1\\3 & 0 & 1\\0 & 0 & 1\end{bmatrix}

=\begin{bmatrix}10 & -5 & 11\\5 & -2 & 5\\5 & -4 & 4\end{bmatrix}\begin{bmatrix}1 & 2 & -1\\3 & 0 & 1\\0 & 0 & 1\end{bmatrix}

=\begin{bmatrix}10-15+0 & 20+0+0 & -10+5+11\\5-6+0 & 10+0+0 & -5-2+5\\5-12+0 & 10+0+0 & -5-4+4\end{bmatrix}

=\begin{bmatrix}-5 & 20 & -4\\-1 & 10 & -2\\-7 & 10 & -5\end{bmatrix}

And R.H.S. = A (BC)



=\begin{bmatrix}4 & 2 & 3\\1 & 1 & 2\\3 & 0 & 1\end{bmatrix} \left(\begin{bmatrix}1 & -1 & 1\\0 & 1 & 2\\2 & -1 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & -1\\3 & 0 & 1\\0 & 0 & 1\end{bmatrix}\right)

=\begin{bmatrix}4 & 2 & 3\\1 & 1 & 2\\3 & 0 & 1\end{bmatrix}\begin{bmatrix}1-3+0 & 2+0+0 & -1-1+1\\0+3+0 & 0+0+0 & 0+1+2\\2-3+0 & 4+0+0 & -2-1+1\end{bmatrix}

=\begin{bmatrix}4 & 2 & 3\\1 & 1 & 2\\3 & 0 & 1\end{bmatrix}\begin{bmatrix}-2 & 2 & -1\\3 & 0 & 3\\-1 & 4 & -2\end{bmatrix}

=\begin{bmatrix}-8+6-3 & 8+12+0 & -4+6-6\\-2+3-2 & 2+0+8 & -1+3-4\\-6+0-1 & 6+0+4 & -3-0-2\end{bmatrix}

=\begin{bmatrix}-5 & 20 & -4\\-1 & 10 & -2\\-7 & 10 & -5\end{bmatrix}

= L.H.S.

Hence proved.

Question 17. For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A (B + C) = AB + AC.

(i) A =\begin{bmatrix}1 & -1\\0 & 2\end{bmatrix}   , B =\begin{bmatrix}-1 & 0\\2 & 1\end{bmatrix}   , C =\begin{bmatrix}0 & 1\\1 & -1\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}1 & -1\\0 & 2\end{bmatrix}   , B =\begin{bmatrix}-1 & 0\\2 & 1\end{bmatrix}   , C =\begin{bmatrix}0 & 1\\1 & -1\end{bmatrix}

L.H.S. = A (B + C)

=\begin{bmatrix}1 & -1\\0 & 2\end{bmatrix} \left(\begin{bmatrix}-1 & 0\\2 & 1\end{bmatrix}+\begin{bmatrix}0 & 1\\1 & -1\end{bmatrix}\right)

=\begin{bmatrix}1 & -1\\0 & 2\end{bmatrix}\begin{bmatrix}-1+0 & 0+1\\2+1 & 1-1\end{bmatrix}

=\begin{bmatrix}1 & -1\\0 & 2\end{bmatrix}\begin{bmatrix}-1 & 1\\3 & 0\end{bmatrix}

=\begin{bmatrix}-1-3 & 1+0\\0+6 & 0+0\end{bmatrix}

=\begin{bmatrix}-4 & 1\\6 & 0\end{bmatrix}

R.H.S. = AB + AC

=\begin{bmatrix}1 & -1\\0 & 2\end{bmatrix}\begin{bmatrix}-1 & 0\\2 & 1\end{bmatrix}+\begin{bmatrix}1 & -1\\0 & 2\end{bmatrix}\begin{bmatrix}0 & 1\\1 & -1\end{bmatrix}

=\begin{bmatrix}-1-2 & 0-1\\0+4 & 0+2\end{bmatrix}+\begin{bmatrix}0-1 & 1+1\\0+2 & 0-2\end{bmatrix}



=\begin{bmatrix}-3 & -1\\4 & 2\end{bmatrix}+\begin{bmatrix}-1 & 2\\2 & -2\end{bmatrix}

=\begin{bmatrix}-3-1 & -1+2\\4+2 & 2-2\end{bmatrix}

=\begin{bmatrix}-4 & 1\\6 & 0\end{bmatrix}

= L.H.S.

Hence proved.

(ii) A =\begin{bmatrix}2 & -1\\1 & 1\\-1 & 2\end{bmatrix}   , B =\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}   , C =\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}2 & -1\\1 & 1\\-1 & 2\end{bmatrix}   , B =\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}   , C =\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}

L.H.S. = A (B + C)

=\begin{bmatrix}2 & -1\\1 & 1\\-1 & 2\end{bmatrix} \left(\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}+\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}\right)

=\begin{bmatrix}2 & -1\\1 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}0+1 & 1-1\\1+0 & 1+1\end{bmatrix}

=\begin{bmatrix}2 & -1\\1 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}1 & 0\\1 & 2\end{bmatrix}

=\begin{bmatrix}2-1 & 0+2\\1+1 & 0+2\\-1+2 & 0+4\end{bmatrix}

=\begin{bmatrix}1 & 2\\2 & 2\\1 & 4\end{bmatrix}

R.H.S. = AB + AC

=\begin{bmatrix}2 & -1\\1 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}+\begin{bmatrix}2 & -1\\1 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}

=\begin{bmatrix}0+1 & 2-1\\0+1 & 1+1\\0+2 & -1+2\end{bmatrix}+\begin{bmatrix}2+0 & -2-1\\1+0 & -1+1\\-1+0 & 1+2\end{bmatrix}

=\begin{bmatrix}-1 & 1\\1 & 2\\2 & 1\end{bmatrix}+\begin{bmatrix}2 & -3\\1 & 0\\-1 & 3\end{bmatrix}

=\begin{bmatrix}-1+2 & 1-3\\1+1 & 2+0\\2-1 & 1+3\end{bmatrix}



=\begin{bmatrix}1 & -2\\2 & 2\\1 & 4\end{bmatrix}

= L.H.S.

Hence proved.

Question 18. If A =\begin{bmatrix}1 & 0 & -2\\3 & -1 & 0\\-2 & 1 & 1\end{bmatrix}   , B =\begin{bmatrix}0 & 5 & -4\\-2 & 1 & 3\\-1 & 0 & 2\end{bmatrix}   and C =\begin{bmatrix}1 & 5 & 2\\-1 & 1 & 0\\0 & -1 & 1\end{bmatrix}   , show that A (B – C) = AB – AC.

Solution:

We have,

A =\begin{bmatrix}1 & 0 & -2\\3 & -1 & 0\\-2 & 1 & 1\end{bmatrix}   , B =\begin{bmatrix}0 & 5 & -4\\-2 & 1 & 3\\-1 & 0 & 2\end{bmatrix}   and C =\begin{bmatrix}1 & 5 & 2\\-1 & 1 & 0\\0 & -1 & 1\end{bmatrix}

L.H.S. = A (B – C)

=\begin{bmatrix}1 & 0 & -2\\3 & -1 & 0\\-2 & 1 & 1\end{bmatrix}\left(\begin{bmatrix}0 & 5 & -4\\-2 & 1 & 3\\-1 & 0 & 2\end{bmatrix}-\begin{bmatrix}1 & 5 & 2\\-1 & 1 & 0\\0 & -1 & 1\end{bmatrix}\right)

=\begin{bmatrix}1 & 0 & -2\\3 & -1 & 0\\-2 & 1 & 1\end{bmatrix}\begin{bmatrix}0-1 & 5-5 & -4-2\\-2+1 & 1-1 & 3-0\\-1-0 & 0+1 & 2-1\end{bmatrix}

=\begin{bmatrix}1 & 0 & -2\\3 & -1 & 0\\-2 & 1 & 1\end{bmatrix}\begin{bmatrix}-1 & 0 & -6\\-1 & 0 & 3\\-1 & 1 & 1\end{bmatrix}

=\begin{bmatrix}-1+0+2 & 0-0-2 & -6+0-2\\-3+1+0 & 0+0+0 & -18-3+0\\2-1-1 & 0+0+1 & 12+3+1\end{bmatrix}

=\begin{bmatrix}1 & -2 & -8\\-2 & 0 & -21\\0 & 1 & 16\end{bmatrix}

R.H.S. = AB – AC

=\begin{bmatrix}1 & 0 & -2\\3 & -1 & 0\\-2 & 1 & 1\end{bmatrix}\begin{bmatrix}0 & 5 & -4\\-2 & 1 & 3\\-1 & 0 & 2\end{bmatrix}-\begin{bmatrix}1 & 0 & -2\\3 & -1 & 0\\-2 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 5 & 2\\-1 & 1 & 0\\0 & -1 & 1\end{bmatrix}

=\begin{bmatrix}0+0+2 & 5+0+0 & -4+0-4\\0+2+0 & 15-1+0 & -12-3+0\\0-2-1 & -10+1+0 & 8+3+2\end{bmatrix}-\begin{bmatrix}1+0+0 & -5+0+2 & 2+0-2\\3+1+0 & 15-1+0 & 6+0+0\\0-2+1 & -10+1+1 & -4+0+1\end{bmatrix}

=\begin{bmatrix}2 & 5 & -8\\2 & 14 & -15\\-3 & -9 & 13\end{bmatrix}-\begin{bmatrix}1 & 7 & 0\\4 & 14 & 6\\-3 & -10 & -3\end{bmatrix}

=\begin{bmatrix}2-1 & 5-7 & -8-0\\2-4 & 14-14 & -15-6\\-3+3 & -9+10 & 13+3\end{bmatrix}

=\begin{bmatrix}1 & -2 & -8\\-2 & 0 & -21\\0 & 1 & 16\end{bmatrix}

Question 19. Compute the elements a43 and a22 of the matrix:

A =\begin{bmatrix}0 & 1 & 0\\2 & 0 & 2\\0 & 3 & 2\\4 & 0 & 4\end{bmatrix}\begin{bmatrix}2 & -1\\-3 & 2\\4 & 3\end{bmatrix}\begin{bmatrix}0 & 1 & -1 & 2 & -2\\3 & -3 & 4 & -4 & 0\end{bmatrix}

Solution:



We are given,

A =\begin{bmatrix}0 & 1 & 0\\2 & 0 & 2\\0 & 3 & 2\\4 & 0 & 4\end{bmatrix}\begin{bmatrix}2 & -1\\-3 & 2\\4 & 3\end{bmatrix}\begin{bmatrix}0 & 1 & -1 & 2 & -2\\3 & -3 & 4 & -4 & 0\end{bmatrix}

=\begin{bmatrix}0-3+0 & 0+2+0\\4+0+8 & -2+0+6\\0-9+8 & 0+6+6\\8+0+16 & -4+0+12\end{bmatrix}\begin{bmatrix}0 & 1 & -1 & 2 & -2\\3 & -3 & 4 & -4 & 0\end{bmatrix}

=\begin{bmatrix}-3 & 2\\12 & 4\\-1 & 12\\24 & 8\end{bmatrix}\begin{bmatrix}0 & 1 & -1 & 2 & -2\\3 & -3 & 4 & -4 & 0\end{bmatrix}

=\begin{bmatrix}0+6 & -3-6 & 3+8 & -6-8 & 6+0\\0+12 & 12-12 & -12+16 & 24-16 & 24+0\\0+36 & -1-36 & 1+48 & -2-48 & 2+0\\0+24 & -24+24 & -24+34 & 48-32 & -48+0\end{bmatrix}

=\begin{bmatrix}6 & -9 & 11 & -14 & 6\\12 & 0 & 4 & 8 & -24\\36 & -37 & 49 & -50 & 2\\24 & 0 & 8 & 16 & -48\end{bmatrix}

Therefore, a43 = 8 and a22 = 0.

Question 20. If A =\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}   and I is the identity matrix of order 3, show that A3 = pI + qA + rA2.

Solution:

We have,

A =\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}

L.H.S. = A3

=\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}

=\begin{bmatrix}0+0+0 & 0+0+0 & 0+1+0\\0+0+p & 0+0+q & 0+0+r\\0+0+pr & p+0+qr & 0+q+r^2\end{bmatrix}\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}

=\begin{bmatrix}0 & 0 & 1\\p & q & r\\pr & p+qr & q+r^2\end{bmatrix}\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}

=\begin{bmatrix}0+0+p & 0+0+q & 0+0+r\\0+0+pr & p+0+qr & 0+q+r^2\\0+0+pq+pr^2 & pr+0+q^2+qr^2 & 0+p+qr+qr+r^2\end{bmatrix}

=\begin{bmatrix}p & q & r\\pr & p+qr & q+r^2\\pq+pr^2 & pr+q^2+qr^2 & p+2qr+r^2\end{bmatrix}

And R.H.S. = pI + qA + rA2

=p\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}+q\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}+r\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}

=\begin{bmatrix}p & 0 & 0\\0 & p & 0\\0 & 0 & p\end{bmatrix}+\begin{bmatrix}0 & q & 0\\0 & 0 & q\\pq & q^2 & qr\end{bmatrix}+r\begin{bmatrix}0+0+0 & 0+0+0 & 0+1+0\\0+0+p & 0+0+q & 0+0+r\\0+0+pr & p+0+qr & 0+q+r^2\end{bmatrix}

=\begin{bmatrix}p & 0 & 0\\0 & p & 0\\0 & 0 & p\end{bmatrix}+\begin{bmatrix}0 & q & 0\\0 & 0 & q\\pq & q^2 & qr\end{bmatrix}+r\begin{bmatrix}0 & 0 & 1\\p & q & r\\pr & p+qr & q+r^2\end{bmatrix}



=\begin{bmatrix}p & 0 & 0\\0 & p & 0\\0 & 0 & p\end{bmatrix}+\begin{bmatrix}0 & q & 0\\0 & 0 & q\\pq & q^2 & qr\end{bmatrix}+\begin{bmatrix}0 & 0 & r\\pr & qr & r^2\\pr^2 & pr+qr^2 & qr+r^3\end{bmatrix}

=\begin{bmatrix}0+0+p & 0+0+q & 0+0+r\\0+0+pr & p+0+qr & 0+q+r^2\\0+0+pq+pr^2 & pr+0+q^2+qr^2 & 0+p+qr+qr+r^2\end{bmatrix}

=\begin{bmatrix}p & q & r\\pr & p+qr & q+r^2\\pq+pr^2 & pr+q^2+qr^2 & p+2qr+r^2\end{bmatrix}

= L.H.S.

Hence proved.

Question 21. If ω is a complex cube root of unity, show that

\left(\begin{bmatrix}1 & ω & ω^2\\ω & ω^2 & 1\\ω^2 & 1 & ω\end{bmatrix}+\begin{bmatrix}ω & ω^2 & 1\\ω^2 & 1 & ω\\ω & ω^2 & 1\end{bmatrix}\right) \begin{bmatrix}1\\ω\\ω^2\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}

Solution:

We have,

L.H.S. =\left(\begin{bmatrix}1 & ω & ω^2\\ω & ω^2 & 1\\ω^2 & 1 & ω\end{bmatrix}+\begin{bmatrix}ω & ω^2 & 1\\ω^2 & 1 & ω\\ω & ω^2 & 1\end{bmatrix}\right) \begin{bmatrix}1\\ω\\ω^2\end{bmatrix}

=\begin{bmatrix}1+ω & ω+ω^2 & ω^2+1\\ω+ω^2 & ω^2+1 & 1+ω\\ω^2+ω & 1+ω^2 & ω+1\end{bmatrix}\begin{bmatrix}1\\ω\\ω^2\end{bmatrix}

=\begin{bmatrix}-ω^2 & -1 & -ω\\-1 & -ω & ω^2\\-1 & -ω & -ω^2\end{bmatrix}\begin{bmatrix}1\\ω\\ω^2\end{bmatrix}

=\begin{bmatrix}-ω^2-ω-ω^3\\-1-ω^2-ω^4\\-1-ω^2-ω^4\end{bmatrix}

=\begin{bmatrix}-ω(1+ω+ω^2) \\-1-ω^2-ω\\-1-ω^2-ω\end{bmatrix}

=\begin{bmatrix}-ω(0) \\-(1+ω^2+ω) \\-(1+ω^2+ω) \end{bmatrix}

=\begin{bmatrix}0\\0\\0\end{bmatrix}

= R.H.S.

Hence proved.

Question 22. If A =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}   , prove that A2 = A.

Solution:

We have,

A =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}

So, A2 =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}

=\begin{bmatrix}4+3-5 & -6-12+15 & -10-15+20\\-2-4+5 & 3+16-15 & 5+20-20\\2+3-4 & -3-12+12 & -5-15+16\end{bmatrix}

=\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}

= A

Hence proved.

Question 23. If A =\begin{bmatrix}4 & -1 & -4\\3 & 0 & -4\\3 & -1 & -3\end{bmatrix}   , show that A2 = I3.

Solution:

We have,

A =\begin{bmatrix}4 & -1 & -4\\3 & 0 & -4\\3 & -1 & -3\end{bmatrix}

So, A2 =\begin{bmatrix}4 & -1 & -4\\3 & 0 & -4\\3 & -1 & -3\end{bmatrix}\begin{bmatrix}4 & -1 & -4\\3 & 0 & -4\\3 & -1 & -3\end{bmatrix}

=\begin{bmatrix}16-3-12 & -4+4+0 & -16+4+12\\12+0-12 & -3+0+4 & -12+0+12\\12-3-9 & -3+3+0 & -12+4+9\end{bmatrix}



=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}

= I3

Hence proved.

Question 24.

(i) If\begin{bmatrix}1 & 1 & x\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2 & 1 & 0\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}   = 0, find x.

Solution:

We have,

=>\begin{bmatrix}1 & 1 & x\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2 & 1 & 0\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}   = 0

=>\begin{bmatrix}1+0+2x & 0+2+x & 2+1+0\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}   = 0

=>\begin{bmatrix}1+2x & 2+x & 3\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}   = 0

=>\begin{bmatrix}1+2x+2+x+3\end{bmatrix}   = 0

=> [3x + 6] = 0

=> 3x = –6

=> x = –6/3

=> x = –2

Therefore, the value of x is –2.

(ii) If\begin{bmatrix}2 & 3\\5 & 7\end{bmatrix}\begin{bmatrix}1 & -3\\-2 & 4\end{bmatrix}=\begin{bmatrix}-4 & 6\\-9 & x\end{bmatrix}   = 0, find x.

Solution:

We have,

=>\begin{bmatrix}2 & 3\\5 & 7\end{bmatrix}\begin{bmatrix}1 & -3\\-2 & 4\end{bmatrix}=\begin{bmatrix}-4 & 6\\-9 & x\end{bmatrix}

=>\begin{bmatrix}2-6 & -6+12\\5-14 & -15+28\end{bmatrix}=\begin{bmatrix}-4 & 6\\-9 & x\end{bmatrix}

=>\begin{bmatrix}-4 & 6\\-9 & 13\end{bmatrix}=\begin{bmatrix}-4 & 6\\-9 & x\end{bmatrix}

On comparing the above matrix we get,

x = 13

Therefore, the value of x is –13.

Question 25. If\begin{bmatrix}x & 4 & 1\end{bmatrix}\begin{bmatrix}2 & 1 & 2\\1 & 0 & 2\\0 & 2 & -4\end{bmatrix}\begin{bmatrix}x\\4\\-1\end{bmatrix}=0   , find x.

Solution:

We have,

=>\begin{bmatrix}x & 4 & 1\end{bmatrix}\begin{bmatrix}2 & 1 & 2\\1 & 0 & 2\\0 & 2 & -4\end{bmatrix}\begin{bmatrix}x\\4\\-1\end{bmatrix}=0

=>\begin{bmatrix}2x+4+0 & x+0+2 & 2x+8-4\end{bmatrix}\begin{bmatrix}x\\4\\-1\end{bmatrix}=0

=>\begin{bmatrix}2x+4 & x+2 & 2x+4\end{bmatrix}\begin{bmatrix}x\\4\\-1\end{bmatrix}=0

=>\begin{bmatrix}2x^2+4x+4x+8-2x-4\end{bmatrix}=0



=> 2x2 + 4x + 4x + 8 – 2x – 4 = 0

=> 2x2 + 6x + 4 = 0

=> 2x2 + 2x + 4x + 4 = 0

=> 2x (x + 1) + 4 (x + 1) = 0

=> (x + 1) (2x + 4) = 0

=> x = –1 or x = –2

Therefore, the value of x is –1 or –2.

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