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Class 12 RD Sharma Solutions – Chapter 5 Algebra of Matrices – Exercise 5.2 | Set 1

  • Last Updated : 03 Mar, 2021
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Question 1(i): Compute the following sum:\begin{bmatrix} 3 & -2\\ 1 & 4 \end{bmatrix} + \begin{bmatrix} -2 & 4\\ 1 & 3 \end{bmatrix} .

Solution:

As the matrices are of the same dimensions, we can add them to get a matrix of the same dimensions which is 2×2.

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

=>\begin{bmatrix} 1 & 2 \\ 2 & 7 \end{bmatrix}

Question 1(ii): Compute the following sum:\begin{bmatrix} 2 & 1 & 3 \\ 0 & 3 & 5 \\ -1 & 2 & 5 \\ \end{bmatrix} + \begin{bmatrix} 1 & -2 & 3 \\ 2 & 6 & 1 \\ 0 & -3 & 1 \\ \end{bmatrix} .

Solution:



As the matrices are of the same dimensions, we can add them to get a matrix of the same dimensions which is 3×3.

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

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

Question 2: Let A =\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} , B =\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} and C =\begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} . Find each of the following:

(i): 2A – 3B

Solution:

Both the matrices A and B are of the same order which is 2×2, hence the operation can be performed.

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

=> 3B =3\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 3 & 9 \\ -6 & 15 \end{bmatrix}

=> 2A – 3B =\begin{bmatrix} 4-3 & 8-9 \\ 6+6 & 4-15 \end{bmatrix}



=> 2A – 3B =\begin{bmatrix} 1 & -1 \\ 12 & -11 \end{bmatrix}

(ii): B – 4C

Solution:

Both the matrices B and C are of the same order which is 2×2, hence the operation can be performed.

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

=> 4C =4\begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} -8 & 20 \\ 12 & 16 \end{bmatrix}

=> B – 4C =\begin{bmatrix} 1+8 & 3-20 \\ -2-12 & 5-16 \end{bmatrix}

=> B – 4C =\begin{bmatrix} 9 & -17 \\ -14 & -11 \end{bmatrix}

(iii): 3A – C

Solution:

Both the matrices A and C are of the same order which is 2×2, hence the operation can be performed.

=> 3A =3\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix}



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

=> 3A – C =\begin{bmatrix} 6+2 & 12-5 \\ 9-3 & 6-4 \end{bmatrix}

=> 3A – C =\begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}

(iv): 3A -2B + 3C

Solution:

The matrices A, B and C are of the same order which is 2×2, hence the operation can be performed.

=> 3A =3\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix}

=> 2B =2\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 2 & 6 \\ -4 & 10 \end{bmatrix}

=> 3C =3\begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} -6 & 15 \\ 9 & 12 \end{bmatrix}

=> 3A – 2B + 3C =\begin{bmatrix} 6-2-6 & 12-6+15 \\ 9+4+9 & 6-10+12 \end{bmatrix}

=> 3A – 2B + 3C =\begin{bmatrix} -2 & 21 \\ 22 & 8 \end{bmatrix}



Question 3: If A =\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} , B =\begin{bmatrix} -1 & 0 & 2\\ 3 & 4 & 1 \end{bmatrix} , C =\begin{bmatrix} -1 & 2 & 3\\ 2 & 1 & 0 \end{bmatrix} , find:

(i): A + B and B + C

Solution:

A and B can not be added since A’s order is 2×2 which is different from B’s order which is 2×3.

B+C can be computed and is solved as follows:

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

=> B + C =\begin{bmatrix} -2 & 2 & 5 \\ 5 & 5 & 1 \end{bmatrix}

(ii): 2B + 3A and 3C – 4B

Solution:

A and B can not be added since A’s order is 2×2 which is different from B’s order which is 2×3, and thus 2B + 3A can not be calculated.

3C – 4B can be computed and is solved as follows:

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

=> 4B =4\begin{bmatrix} -1 & 0 & 2\\ 3 & 4 & 1 \end{bmatrix} = \begin{bmatrix} -4 & 0 & 8\\ 12 & 16 & 4 \end{bmatrix}



=> 3C – 4B =\begin{bmatrix} -3+4 & 6+0 & 9-8 \\ 6-12 & 3-16 & 10-4 \end{bmatrix}

=> 3C – 4B =\begin{bmatrix} 1 & 6 & 1 \\ -6 & -13 & 6 \end{bmatrix}

Question 4: Let A =\begin{bmatrix} -1 & 0 & 2 \\ 3 & 1 & 4 \end{bmatrix} , B =\begin{bmatrix} 0 & -2 & 5 \\ 1 & -3 & 1 \end{bmatrix} and C =\begin{bmatrix} 1 & -5 & 2 \\ 6 & 0 & -4 \end{bmatrix} . Compute 2A – 3B + 4C.

Solution:

The result can be computed since A, B and C are of the same order which is 2×3.

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

=> 3B =3\begin{bmatrix} 0 & -2 & 5 \\ 1 & -3 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -6 & 15 \\ 3 & -9 & 3 \end{bmatrix}

=> 4C =4\begin{bmatrix} 1 & -5 & 2 \\ 6 & 0 & -4 \end{bmatrix} = \begin{bmatrix} 4 & -20 & 8 \\ 24 & 0 & -16 \end{bmatrix}

=> 2A – 3B + 4C =\begin{bmatrix} -2-0+4 & 0+6-20 & 4-15+8 \\ 6-3+24 & 2+9+0 & 8-3-16 \end{bmatrix}

=> 2A – 3B + 4C =\begin{bmatrix} 2 & -14 & -3 \\ 27 & 11 & -11 \end{bmatrix}

Question 5: If A =diag(2, -5, 9), B = diag(1, 1, -4) and C = diag(-6, 3, 4), find:

(i): A – 2B

Solution:



In the given question A and B are diagonal matrices of the order 3×3, thus the only non-zero elements are present in the diagonal.

=> A = diag(2, -5, 9)

=> 2B = 2. diag(1, 1, -4) = diag(2, 2, -8)

=> A – 2B = diag(2-2, -5-2, 9+8)

=> A – 2B = diag(0, -7, 17)

(ii): B + C – 2A

Solution:

In the given question A, B and C are diagonal matrices of the order 3×3, thus the only non-zero elements are present in the diagonal.

=> B = diag(1, 1, -4)

=> C = diag(-6, 3, 4)

=> 2A = 2. diag(2, -5, 9) = diag(4, -10, 18)

=> B + C – 2A = diag(1-6-4, 1+3+10, -4+4-18)

=> B + C – 2A = diag(-9, 14, -18)

(iii): 2A + 3B – 5C

Solution:

In the given question A, B and C are diagonal matrices of the order 3×3, thus the only non-zero elements are present in the diagonal.

=> 2A = 2. diag(2, -5, 9) = diag(4, -10, 18)

=> 3B = 3. diag(1, 1, -4) = diag(3, 3, -12)

=> 5C = 5. diag(-6, 3, 4) = diag(-30, 15, 20)

=> 2A + 3B – 5C = diag(4+3+30, -10+3-15, 18-12-20)

=> 2A + 3B – 5C = diag(37, -22, -14)

Question 6: Given the matrices A =\begin{bmatrix} 2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4 \\ \end{bmatrix} , B =\begin{bmatrix} 9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6 \\ \end{bmatrix} and C =\begin{bmatrix} 2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5 \\ \end{bmatrix} . Verify that (A + B) + C = A + (B + C).

Solution:



Given L.H.S :

=> (A + B) = \begin{bmatrix} 2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4 \\ \end{bmatrix} + \begin{bmatrix} 9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6 \\ \end{bmatrix}

=> (A + B) =\begin{bmatrix} 2+9 & 1+7 & 1-1 \\ 3+3 & -1+5 & 0+4 \\ 0+2 & 2+1 & 4+6 \\ \end{bmatrix}

=> (A + B) =\begin{bmatrix} 11 & 8 & 0 \\ 6 & 4 & 4 \\ 2 & 3 & 10 \\ \end{bmatrix}

=> (A + B) + C =\begin{bmatrix} 11 & 8 & 0 \\ 6 & 4 & 4 \\ 2 & 3 & 10 \\ \end{bmatrix} + \begin{bmatrix} 2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5 \\ \end{bmatrix}

=> (A + B) + C =\begin{bmatrix} 11+2 & 8-4 & 0+3 \\ 6+1 & 4-1 & 4+0 \\ 2+9 & 3+4 & 10+5 \\ \end{bmatrix}

=> (A + B) + C =\begin{bmatrix} 13 & 4 & 3 \\ 7 & 3 & 4 \\ 11 & 7 & 15 \\ \end{bmatrix}

Given R.H.S :

=> (B + C) =\begin{bmatrix} 9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6 \\ \end{bmatrix} + \begin{bmatrix} 2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5 \\ \end{bmatrix}

=> (B + C) =\begin{bmatrix} 9+2 & 7-4 & -1+3 \\ 3+1 & 5-1 & 4+0 \\ 2+9 & 1+4 & 6+5 \\ \end{bmatrix}



=> (B + C) =\begin{bmatrix} 11 & 3 & 2 \\ 4 & 4 & 4 \\ 11 & 5 & 11 \\ \end{bmatrix}

=> A + (B + C) =\begin{bmatrix} 2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4 \\ \end{bmatrix} + \begin{bmatrix} 11 & 3 & 2 \\ 4 & 4 & 4 \\ 11 & 5 & 11 \\ \end{bmatrix}

=> A + (B + C) =\begin{bmatrix} 2+11 & 1+3 & 1+2 \\ 3+4 & -1+4 & 0+4 \\ 0+11 & 2+5 & 4+11 \\ \end{bmatrix}

=> A + (B + C) =\begin{bmatrix} 13 & 4 & 3 \\ 7 & 3 & 4 \\ 11 & 7 & 15 \\ \end{bmatrix}

Hence R.H.S = L.H.S has been verified.

Question 7: Find the matrices X and Y, if X + Y =\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} and X – Y =\begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix} .

Solution:

We know that (X + Y) + (X – Y) = 2X.

=> (X + Y) + (X – Y) =\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} + \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}

=> 2X =\begin{bmatrix} 5+3 & 2+6 \\ 0+0 & 9-1 \end{bmatrix}

=> 2X =\begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}



=> X =\dfrac {1}{2}\begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}

=> X =\begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}

Now Y = (X + Y) – X

=> Y =\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}

=> Y =\begin{bmatrix} 5-4 & 2-4 \\ 0-0 & 9-4 \end{bmatrix}

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

Question 8: Find X, if Y =\begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} and 2X + Y =\begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} .

Solution:

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

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

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



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

=> 2X =\begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix}

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

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

Question 9: Find matrices X and Y, if 2X – Y =\begin{bmatrix} 6 & -6 & 0\\ -4 & 2 & 1\end{bmatrix} and X + 2Y =\begin{bmatrix} 3 & 2 & 5\\ -2 & 1 & -7\end{bmatrix} .

Solution:

We know that 2 (2X – Y) + (X + 2Y) = 4X – 2Y + X + 2Y = 5X .

=> 2 (2X – Y) =2\begin{bmatrix} 6 & -6 & 0\\ -4 & 2 & 1\end{bmatrix}

=> 2 (2X -Y) =\begin{bmatrix} 12 & -12 & 0\\ -8 & 4 & 2\end{bmatrix}

=> 2 (2X – Y) + (X + 2Y) =\begin{bmatrix} 12 & -12 & 0\\ -8 & 4 & 2\end{bmatrix} + \begin{bmatrix} 3 & 2 & 5\\ -2 & 1 & -7\end{bmatrix}

=> 5X =\begin{bmatrix} 12+3 & -12+2 & 0+5\\ -8-2 & 4+1 & 2-7\end{bmatrix}



=> 5X =\begin{bmatrix} 15 & -10 & 5\\ -10 & 5 & -5\end{bmatrix}

=> X =\dfrac{1}{5}\begin{bmatrix} 15 & -10 & 5\\ -10 & 5 & -5\end{bmatrix}

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

As (X + 2Y) =\begin{bmatrix} 3 & 2 & 5\\ -2 & 1 & -7\end{bmatrix}

=>\begin{bmatrix} 3 & -2 & 1\\ -2 & 1 & -1\end{bmatrix} + 2Y = \begin{bmatrix} 3 & 2 & 5\\ -2 & 1 & -7\end{bmatrix}

=> 2Y =\begin{bmatrix} 3 & 2 & 5\\ -2 & 1 & -7\end{bmatrix} -\begin{bmatrix} 3 & -2 & 1\\ -2 & 1 & -1\end{bmatrix}

=> 2Y =\begin{bmatrix} 0 & 4 & 4\\ 0 & 0 & -6\end{bmatrix}

=> Y =\dfrac{1}{2}\begin{bmatrix} 0 & 4 & 4\\ 0 & 0 & -6\end{bmatrix}

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

Question 10: If X – Y =\begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 0 \\1 & 0 & 0 \end{bmatrix} and X + Y =\begin{bmatrix} 3 & 5 & 1\\ -1 & 1 & 4 \\11 & 8 & 0 \end{bmatrix} , find X and Y.

Solution:



We know that (X + Y) + (X – Y) = 2X.

=> 2X =\begin{bmatrix} 3 & 5 & 1\\ -1 & 1 & 4 \\11 & 8 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 0 \\1 & 0 & 0 \end{bmatrix}

=> 2X =\begin{bmatrix} 1+3 & 1+5 & 1+1\\ 1-1 & 1+1 & 0+4 \\1+11 & 0+8 & 0+0 \end{bmatrix}

=> 2X =\begin{bmatrix} 4 & 6 & 2\\ 0 & 2 & 4 \\12 & 8 & 0 \end{bmatrix}

=> X =\dfrac {1}{2}\begin{bmatrix} 4 & 6 & 2\\ 0 & 2 & 4 \\12 & 8 & 0 \end{bmatrix}

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

Also (X + Y) – (X -Y) = 2Y.

=> 2Y =\begin{bmatrix} 3 & 5 & 1\\ -1 & 1 & 4 \\11 & 8 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 0 \\1 & 0 & 0 \end{bmatrix}

=> 2Y =\begin{bmatrix} 3-1 & 5-1 & 1-1\\ -1-1 & 1-1 & 4-0 \\11-1 & 8-0 & 0-0 \end{bmatrix}

=> 2Y =\begin{bmatrix} 2 & 4 & 0\\ -2 & 0 & 4 \\10 & 8 & 0 \end{bmatrix}

=> Y =\dfrac{1}{2}\begin{bmatrix} 2 & 4 & 0\\ -2 & 0 & 4 \\10 & 8 & 0 \end{bmatrix}

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

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