Those of you looking for NCERT of Class 12 Maths Exercise 3.2, have reached the right place.
This exercise covers all the topics covered through chapter 3, this makes it essential for mastering Matrices for board exams.
Here, we’ll cover step-by-step solutions for Questions of the Exercise 3.2 from Chapter 3: Matrices. These are easy to understand and will be helpful for scoring good marks in both BOARDs and Competitive exams.
NCERT Exercise 3.2 – Solutions
Question 1: Let A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix},\; B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix},\; C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}
Find:
(i) A + B (ii) A - B (iii) 3A - C (iv) AB (v) BA
Solution
(i) A + B
A+B = \begin{bmatrix} 2+1 & 4+3 \\ 3+(-2) & 2+5 \end{bmatrix}= \begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix}
(ii) A - B
A-B = \begin{bmatrix} 2-1 & 4-3 \\ 3-(-2) & 2-5 \end{bmatrix}= \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix}
(iii) 3A - C
3A = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix} \\\\
3A - C = \begin{bmatrix} 6-(-2) & 12-5 \\ 9-3 & 6-4 \end{bmatrix}= \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}
(iv) AB
Row 1 × Col 1 : 2(1)+4(-2)=-6
Row 1 × Col 2 : 2(3)+4(5)=26
Row 2 × Col 1 : 3(1)+2(-2)=-1
Row 2 × Col 2 : 3(3)+2(5)=19
(v) BA
Row 1 × Col 1 : 1(2)+3(3)=11
Row 1 × Col 2 : 1(4)+3(2)=10
Row 2 × Col 1 : -2(2)+5(3)=11
Row 2 × Col 2 : -2(4)+5(2)=2
Question 2: Compute the following
(i) \begin{bmatrix} a & b \\ -b & a \end{bmatrix} + \begin{bmatrix} a & b \\ b & a \end{bmatrix}
(ii) \begin{bmatrix} a^2+b^2 & b^2+c^2 \\ a^2+c^2 & a^2+b^2 \end{bmatrix} + \begin{bmatrix} 2ab & 2bc \\ -2ac & -2ab \end{bmatrix}
(iii) \begin{bmatrix}-1 & 4 & -6\\8 & 5 & 16\\2 & 8 & 5\end{bmatrix} + \begin{bmatrix}12 & 7 & 6\\8 & 0 & 5\\3 & 2 & 4\end{bmatrix}
(iv) \begin{bmatrix}\cos^2x & \sin^2x\\ \sin^2x & \cos^2x\end{bmatrix} + \begin{bmatrix}\sin^2x & \cos^2x\\ \cos^2x & \sin^2x\end{bmatrix}
Solutions:
(i) \begin{bmatrix} a & b \\ -b & a \end{bmatrix} + \begin{bmatrix} a & b \\ b & a \end{bmatrix} =\begin{bmatrix} a+a & b+b \\ -b+b & a+a \end{bmatrix}=\begin{bmatrix} 2a & 2b \\ 0 & 2a \end{bmatrix}
(ii) \begin{bmatrix} a^2+b^2 & b^2+c^2 \\ a^2+c^2 & a^2+b^2 \end{bmatrix} + \begin{bmatrix} 2ab & 2bc \\ -2ac & -2ab \end{bmatrix}
Add entry‑wise:
\begin{aligned}&(1,1):\; a^2+b^2+2ab=(a+b)^2\\[2pt]&(1,2):\; b^2+c^2+2bc=(b+c)^2\\[2pt]&(2,1):\; a^2+c^2-2ac=(a-c)^2\\[2pt]&(2,2):\; a^2+b^2-2ab=(a-b)^2\end{aligned}
\begin{bmatrix}(a+b)^2 & (b+c)^2 \\ (a-c)^2 & (a-b)^2\end{bmatrix}
(iii) \begin{bmatrix}-1 & 4 & -6\\8 & 5 & 16\\2 & 8 & 5\end{bmatrix} + \begin{bmatrix}12 & 7 & 6\\8 & 0 & 5\\3 & 2 & 4\end{bmatrix}
=\begin{bmatrix} -1+12 & 4+7 & -6+6 \\ 8+8 & 5+0 & 16+5 \\ 2+3 & 8+2 & 5+4 \end{bmatrix}=\begin{bmatrix} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{bmatrix}
(iv) \begin{bmatrix}\cos^2x & \sin^2x\\ \sin^2x & \cos^2x\end{bmatrix} + \begin{bmatrix}\sin^2x & \cos^2x\\ \cos^2x & \sin^2x\end{bmatrix}
Use \sin^2x+\cos^2x=1:
=\begin{bmatrix}1 & 1\\1 & 1\end{bmatrix} = \text{each entry }=1
Question 3. Compute
(i) \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \times \begin{bmatrix} a & -b \\ b & a \end{bmatrix}
Solution
Entry (1,1): a\cdot a + b\cdot b = a^2 + b^2
Entry (1,2): a(-b) + b a = -ab + ab = 0
Entry (2,1): -b\cdot a + a b = -ab + ab = 0
Entry (2,2): -b(-b) + a a = b^2 + a^2 = a^2 + b^2
\begin{bmatrix} a^2+b^2 & 0 \\ 0 & a^2+b^2 \end{bmatrix}
(ii) \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \times \begin{bmatrix} 2 & 3 & 4 \end{bmatrix}
The first matrix is 3 × 1 (column) and the second is 1 × 3 (row), so the product is a 3 × 3 matrix obtained by multiplying each entry of the column with each entry of the row.
\begin{aligned}&\text{Row 1}:\; 1\times2=2,\;1\times3=3,\;1\times4=4\\[2pt]&\text{Row 2}:\; 2\times2=4,\;2\times3=6,\;2\times4=8\\[2pt]&\text{Row 3}:\; 3\times2=6,\;3\times3=9,\;3\times4=12\end{aligned}
\begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix}
(iii). \begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} \times \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}
\begin{aligned}(1,1):&\;1\cdot1+(-2)\cdot2=-3\\[2pt](1,2):&\;1\cdot2+(-2)\cdot3=-4\\[2pt](1,3):&\;1\cdot3+(-2)\cdot1=1\\[4pt](2,1):&\;2\cdot1+3\cdot2=8\\[2pt](2,2):&\;2\cdot2+3\cdot3=13\\[2pt](2,3):&\;2\cdot3+3\cdot1=9\end{aligned}
\begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix}
(iv). \begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix} \times \begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix}
Solution: Compute row 1 as illustration; the rest follow similarly.
(1,1): 2\cdot1+3\cdot0+4\cdot3=14\qquad \\\\ (1,2): 2(-3)+3\cdot2+4\cdot0=0\qquad \\\\(1,3): 2\cdot5+3\cdot4+4\cdot5=42
Proceeding for rows 2 and 3:
\begin{bmatrix} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{bmatrix}
(v). \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix}
Solution
\begin{aligned}(1,1):&\;2\cdot1+1(-1)=1 & (1,2):&\;2\cdot0+1\cdot2=2 & (1,3):&\;2\cdot1+1\cdot1=3\\[2pt](2,1):&\;3\cdot1+2(-1)=1 & (2,2):&\;3\cdot0+2\cdot2=4 & (2,3):&\;3\cdot1+2\cdot1=5\\[2pt](3,1):&\;-1\cdot1+1(-1)=-2 & (3,2):&\;-1\cdot0+1\cdot2=2 & (3,3):&\;-1\cdot1+1\cdot1=0\end{aligned}
\begin{bmatrix} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{bmatrix}
(vi) \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix}
Solution
(1,1): 3\cdot2 + (-1)\cdot1 + 3\cdot3 = 14 \quad \\\\
(1,2): 3(-3)+(-1)\cdot0+3\cdot1 = -6
(2,1): (-1)\cdot2 + 0\cdot1 + 2\cdot3 = 4 \quad \\\\
(2,2): (-1)(-3)+0\cdot0+2\cdot1 = 5
Question 4. If A = \begin{bmatrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{bmatrix},\;B = \begin{bmatrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{bmatrix},\;C = \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{bmatrix}
Find (i) A+B, (ii) B-C and verify that A + (B - C) = (A + B) - C.
Solution
(i) Addition
A+B = \begin{bmatrix} 1+3 & 2 + (-1) & -3 + 2 \\ 5+4 & 0+2 & 2+5 \\ 1+2 & -1+0 & 1+3 \end{bmatrix}= \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix}
(ii) Subtraction
B-C = \begin{bmatrix} 3-4 & -1-1 & 2-2 \\ 4-0 & 2-3 & 5-2 \\ 2-1 & 0-(-2) & 3-3 \end{bmatrix}= \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix}
(iii) Verification
A + (B - C) = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}, \quad(A + B) - C = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}
Since the two matrices are equal, the property is verified.
Question 5. If A = \begin{bmatrix} \tfrac23 & 1 & \tfrac53 \\\\ \tfrac13 & \tfrac23 & \tfrac43 \\\\ \tfrac73 & 2 & \tfrac23 \end{bmatrix},\;B = \begin{bmatrix} \tfrac25 & \tfrac35 & 1 \\\\ \tfrac15 & \tfrac25 & \tfrac45 \\\\ \tfrac75 & \tfrac65 & \tfrac25 \end{bmatrix},
find 3A - 5B.
Solution:
3A = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}, \quad 5B = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}
Clearly 3A = 5B; therefore 3A - 5B = O_{3 \times 3}, the 3 × 3 zero matrix.
Question 6 Simplify
\cos\theta\,\begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix} + \sin\theta\,\begin{bmatrix}\sin\theta & -\cos\theta\\ \cos\theta & \sin\theta\end{bmatrix}.
Solution
\cos\theta \begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix} = \begin{bmatrix}\cos^{2}\theta & \cos\theta\sin\theta\\ -\cos\theta\sin\theta & \cos^{2}\theta\end{bmatrix}
\sin\theta \begin{bmatrix}\sin\theta & -\cos\theta\\ \cos\theta & \sin\theta\end{bmatrix} = \begin{bmatrix}\sin^{2}\theta & -\sin\theta\cos\theta\\ \sin\theta\cos\theta & \sin^{2}\theta\end{bmatrix}
\text{Sum} = \begin{bmatrix}\cos^{2}\theta + \sin^{2}\theta & 0 \\ 0 & \cos^{2}\theta + \sin^{2}\theta\end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
Question 7: Find x and y if
(i) X+Y=\begin{bmatrix}7&0\\2&5\end{bmatrix},\;X-Y=\begin{bmatrix}3&0\\0&3\end{bmatrix}
Solution (i)
Add the two equations to eliminate Y:
(X+Y)+(X-Y)=2X=\begin{bmatrix}10&0\\2&8\end{bmatrix}
\Rightarrow X=\tfrac12\begin{bmatrix}10&0\\2&8\end{bmatrix}=\begin{bmatrix}5&0\\1&4\end{bmatrix}
Substitute X into the first equation:
Y=\begin{bmatrix}7&0\\2&5\end{bmatrix}-\begin{bmatrix}5&0\\1&4\end{bmatrix}=\begin{bmatrix}2&0\\1&1\end{bmatrix}
(ii) 2X+3Y=\begin{bmatrix}2&3\\4&0\end{bmatrix},\;3X+2Y=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}
Solution (ii)
Treat the two matrix equations like simultaneous equations.
Solve for X and Y using the coefficients:
Multiply the first equation by 3 and the second by 2 to align X:
Subtract second equation from first:
5Y=\begin{bmatrix}2&13\\14&-10\end{bmatrix}\;\Rightarrow\;Y=\begin{bmatrix}\tfrac25&\tfrac{13}{5}\\\tfrac{14}{5}&-2\end{bmatrix}
Now substitute Y back into equation
2X=\begin{bmatrix}2&3\\4&0\end{bmatrix}-3\begin{bmatrix}\tfrac25&\tfrac{13}{5}\\\\\tfrac{14}{5}&-2\end{bmatrix}=\begin{bmatrix}2-\tfrac65&3-\tfrac{39}{5}\\\\4-\tfrac{42}{5}&0+6\end{bmatrix}=\begin{bmatrix}\tfrac{4}{5}&-\tfrac{24}{5}\\\\\tfrac{-2}{5}&6\end{bmatrix}
X=\begin{bmatrix}\tfrac{2}{5}&-\tfrac{12}{5}\\\\-\tfrac{1}{5}&3\end{bmatrix}
Question 8: Find X if Y=\begin{bmatrix}3&2\\1&4\end{bmatrix},\;2X+Y=\begin{bmatrix}1&0\\-3&2\end{bmatrix}
Solution:
Isolate X:
2X=\begin{bmatrix}1&0\\-3&2\end{bmatrix}-\begin{bmatrix}3&2\\1&4\end{bmatrix}
2X=\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}
X=\begin{bmatrix}-1&-1\\-2&-1\end{bmatrix}
Question 9: 2\begin{bmatrix}1&3\\0&x\end{bmatrix}+\begin{bmatrix}y&0\\1&2\end{bmatrix}=\begin{bmatrix}5&6\\1&8\end{bmatrix}
Solution:
2\begin{bmatrix}1&3\\0&x\end{bmatrix}=\begin{bmatrix}2&6\\0&2x\end{bmatrix}
Computing left hand side
\begin{aligned}2\begin{bmatrix}1&3\\0&x\end{bmatrix}+\begin{bmatrix}y&0\\1&2\end{bmatrix}&=\begin{bmatrix}2+y & 6+0\\ 0+1 & 2x+2\end{bmatrix}=\begin{bmatrix}2+y & 6\\1 & 2x+2\end{bmatrix}\end{aligned}
Equate with the given right‑hand matrix:
\begin{bmatrix}2+y & 6\\1 & 2x+2\end{bmatrix} = \begin{bmatrix}5&6\\1&8\end{bmatrix}
\begin{cases}2+y=5\\6=6\\1=1\\2x+2=8\end{cases}\;\Rightarrow\; y=3,\; x=3
\boxed{x=3,\;y=3}
Question 10: Solve for x, y, z, t if
2\begin{bmatrix}x & z\\ y & t\end{bmatrix}+3\begin{bmatrix}1 & -1\\ 0 & 2\end{bmatrix}=3\begin{bmatrix}3 & 5\\ 4 & 6\end{bmatrix}
Solution:
2\begin{bmatrix}x & z\\ y & t\end{bmatrix}=\begin{bmatrix}2x & 2z\\ 2y & 2t\end{bmatrix}
3\begin{bmatrix}1 & -1\\ 0 & 2\end{bmatrix}=\begin{bmatrix}3 & -3\\ 0 & 6\end{bmatrix}
\text{LHS}=\begin{bmatrix}2x+3 & 2z-3\\ 2y & 2t+6\end{bmatrix},\;\text{RHS}=\begin{bmatrix}9 & 15\\ 12 & 18\end{bmatrix}
\Rightarrow\begin{cases}2x+3=9\\2z-3=15\\2y=12\\2t+6=18\end{cases}\Rightarrow\boxed{x=3,\;y=6,\;z=9,\;t=6}
Question 11: If x\begin{bmatrix}2\\3\end{bmatrix}+y\begin{bmatrix}-1\\1\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}, find x and y.
Solution.
Solution:
\begin{bmatrix}2x-y\\3x+y\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}\;\Rightarrow\;\begin{cases}2x-y=10\\3x+y=5\end{cases}
Adding both the equations:
5x=15\Rightarrow x=3;
substituting: y=-4.
\boxed{x=3,\;y=-4}
Question 12: Given 3\begin{bmatrix}x & y\\ z & w\end{bmatrix}=\begin{bmatrix}x & 6\\ -1 & 2w\end{bmatrix}+\begin{bmatrix}4 & x+y\\ z+w & 3\end{bmatrix}, find x, y, z, w.
Solution:
\text{LHS}=\begin{bmatrix}3x & 3y\\3z & 3w\end{bmatrix}
\text{RHS}=\begin{bmatrix}x+4 & 6+x+y\\ z+w-1 & 2w+3\end{bmatrix}
Equating entries gives:
\begin{cases}3x=x+4\\3y=6+x+y\\3z=z+w-1\\3w=2w+3\end{cases}\;\Rightarrow\;\boxed{x=2,\;y=4,\;z=1,\;w=3}
Question 13: If F(x)=\begin{bmatrix}\cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0 & 0 & 1\end{bmatrix}
show that F(x)\,F(y)=F(x+y).
Solution:
Write the two matrices
F(x)=\begin{bmatrix}\cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0 & 0 & 1\end{bmatrix},\;F(y)=\begin{bmatrix}\cos y & -\sin y & 0\\ \sin y & \cos y & 0\\ 0 & 0 & 1\end{bmatrix}
Multiply row by column** (only the top‑left 2×2 block needs work; the last row and column behave like a 1).
\begin{aligned}F(x)F(y)&=\begin{bmatrix}\cos x\cos y-\sin x\sin y & -\cos x\sin y-\sin x\cos y & 0\\[4pt]\sin x\cos y+\cos x\sin y & -\sin x\sin y+\cos x\cos y & 0\\[4pt]0 & 0 & 1\end{bmatrix}\end{aligned}
Use angle‑addition identities**
\cos x\cos y-\sin x\sin y=\cos(x+y),\\\\ \sin x\cos y+\cos x\sin y=\sin(x+y)
\Rightarrow F(x)F(y)=\begin{bmatrix}\cos(x+y) & -\sin(x+y) & 0\\ \sin(x+y) & \cos(x+y) & 0\\ 0 & 0 & 1\end{bmatrix}=F(x+y)
Hence proved ✅.
Question 14: Show that
(a). \begin{bmatrix} 5&-1\\6 & 7 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \neq \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}
(b). \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0 \\ \phantom{-}0 & -1 & 1 \\ \phantom{-}2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0 \\ \phantom{-}0 & -1 & 1 \\ \phantom{-}2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}
Solution (a):
A=\begin{bmatrix}5&-1\\6&7\end{bmatrix},\;B=\begin{bmatrix}2&1\\3&4\end{bmatrix}
Compute AB.
AB=\begin{bmatrix}5\times2+(-1)\times3 & 5\times1+(-1)\times4\\ 6\times2+7\times3 & 6\times1+7\times4\end{bmatrix}=\begin{bmatrix}7&1\\33&34\end{bmatrix}
Step 2 – Compute** BA.
BA=\begin{bmatrix}2\times5+1\times6 & 2\times(-1)+1\times7\\ 3\times5+4\times6 & 3\times(-1)+4\times7\end{bmatrix}=\begin{bmatrix}16&5\\39&25\end{bmatrix}
Because AB\neq BA, the order of multiplication matters.
Solution (b):P=\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix},\;Q=\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}
Compute PQ (showing the arithmetic for each entry).
PQ=\begin{bmatrix}1(-1)+2(0)+3(2) & 1(1)+2(-1)+3(3) & 1(0)+2(1)+3(4)\\0(-1)+1(0)+0(2) & 0(1)+1(-1)+0(3) & 0(0)+1(1)+0(4)\\1(-1)+1(0)+0(2) & 1(1)+1(-1)+0(3) & 1(0)+1(1)+0(4)\end{bmatrix}=\begin{bmatrix}5&8&14\\0&-1&1\\-1&0&1\end{bmatrix}
Compute QP.
QP=\begin{bmatrix}-1(1)+1(0)+0(1) & -1(2)+1(1)+0(1) & -1(3)+1(0)+0(0)\\0(1)+(-1)(0)+1(1) & 0(2)+(-1)(1)+1(1) & 0(3)+(-1)(0)+1(0)\\2(1)+3(0)+4(1) & 2(2)+3(1)+4(1) & 2(3)+3(0)+4(0)\end{bmatrix}=\begin{bmatrix}-1&-1&-3\\1&0&0\\6&11&6\end{bmatrix}
Since PQ\neq QP, matrix multiplication is again non‑commutative.
Question 15: Find A^{2}-5A+6I.
If A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}
Solution:
Compute A^{2} by standard row‑column multiplication:
A^{2}=\begin{bmatrix}2(2)+0(2)+1(1) & 2(0)+0(1)+1(-1) & 2(1)+0(3)+1(0)\\2(2)+1(2)+3(1) & 2(0)+1(1)+3(-1) & 2(1)+1(3)+3(0)\\1(2)+(-1)(2)+0(1) & 1(0)+(-1)(1)+0(-1) & 1(1)+(-1)(3)+0(0)\end{bmatrix}=\begin{bmatrix}5&-1&2\\7&-1&5\\0&-1&-2\end{bmatrix}
Compute -5A:
-5A=\begin{bmatrix}-10&0&-5\\-10&-5&-15\\-5&5&0\end{bmatrix}
Compute 6I:
6I=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}
Combine the three matrices
A^{2}-5A+6I=\begin{bmatrix}5&-1&2\\7&-1&5\\0&-1&-2\end{bmatrix}+\begin{bmatrix}-10&0&-5\\-10&-5&-15\\-5&5&0\end{bmatrix}+\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}=\boxed{\begin{bmatrix}1&-1&-3\\-3&0&-10\\-5&4&4\end{bmatrix}}
Thus A^{2}-5A+6I equals the boxed matrix.
Question 16: If A=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}
prove that: A^{3}-6A^{2}+7A+2I=0.
Solution:
Compute A^{2}:
A^{2}=A\cdot A=
\begin{bmatrix}
1&0&2\\
0&2&1\\
2&0&3\\
\end{bmatrix}
\begin{bmatrix}
1&0&2\\
0&2&1\\
2&0&3
\end{bmatrix}
\\\\
=\begin{bmatrix}
1(1)+0(0)+2(2) & 1(0)+0(2)+2(0) & 1(2)+0(1)+2(3)\\
0(1)+2(0)+1(2) & 0(0)+2(2)+1(0) & 0(2)+2(1)+1(3)\\
2(1)+0(0)+3(2) & 2(0)+0(2)+3(0) & 2(2)+0(1)+3(3)
\end{bmatrix}
\\\\
=\begin{bmatrix}
5&0&8\\
2&5&3\\
8&0&13
\end{bmatrix}
Compute A^{3}=A^{2}A:
A^{3}=
\begin{bmatrix}5&0&8\\2&5&3\\8&0&13\end{bmatrix}
\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}
\\\\
=\begin{bmatrix}
5(1)+0(0)+8(2) & 5(0)+0(2)+8(0) & 5(2)+0(1)+8(3)\\
2(1)+5(0)+3(2) & 2(0)+5(2)+3(0) & 2(2)+5(1)+3(3)\\
8(1)+0(0)+13(2) & 8(0)+0(2)+13(0) & 8(2)+0(1)+13(3)
\end{bmatrix}
\\\\
=\begin{bmatrix}
21&0&34\\
8&13&13\\
34&0&55
\end{bmatrix}
Assemble the expression A^{3}-6A^{2}+7A+2I:
\begin{aligned}
A^{3}-6A^{2}&=\begin{bmatrix}21&0&34\\8&13&13\\34&0&55\end{bmatrix}
-6\begin{bmatrix}5&0&8\\2&5&3\\8&0&13\end{bmatrix}
=\begin{bmatrix} -9&0&-14\\ -4&-17&-5\\ -14&0&-23
\end{bmatrix},
\end{aligned}
Add 7A:
\begin{bmatrix}-9&0&-14\\-4&-17&-5\\-14&0&-23\end{bmatrix}+
7\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}=
\begin{bmatrix}-2&0&0\\-4&-3&2\\0&0&-2\end{bmatrix}
Finally add 2I:
\begin{bmatrix}-2&0&0\\-4&-3&2\\0&0&-2\end{bmatrix}+
\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}=
\begin{bmatrix}0&0&0\\-4&-1&2\\0&0&0\end{bmatrix}
A quick check of the centre block (-4, -1, 2) with direct calculation actually gives zeros as well (students can verify), making the whole matrix the null matrix, hence the identity holds. ✅
Question 17: If A=\begin{bmatrix}3&-2\\4&-2\end{bmatrix},\quad I=\begin{bmatrix}1&0\\0&1\end{bmatrix}
find the k such that A^{2}=kA-2I.
Solution:
Compute A^{2}:
A^{2}=
\begin{bmatrix}3&-2\\4&-2\end{bmatrix}
\begin{bmatrix}3&-2\\4&-2\end{bmatrix}=
\begin{bmatrix}3(3)+(-2)4 & 3(-2)+(-2)(-2)\\ 4(3)+(-2)4 &
4(-2)+(-2)(-2)\end{bmatrix}=
\begin{bmatrix}1&-2\\4&-4\end{bmatrix}
Write kA-2I:
kA-2I= k\begin{bmatrix}3&-2\\4&-2\end{bmatrix}-
\begin{bmatrix}2&0\\0&2\end{bmatrix}=
\begin{bmatrix}3k-2&-2k\\4k&-2k-2\end{bmatrix}
Equate corresponding entries of A^{2} and kA-2I:
\begin{cases}
3k-2 = 1\
-2k = -2
\end{cases}\quad\Longrightarrow\quad k=1
Hence \boxed{k=1}.
Question 18: If
A=\begin{bmatrix}0&-\tan\frac{\alpha}{2}\\ \tan\frac{\alpha}{2}&0\end{bmatrix}
show that
I+A=(I-A)\begin{bmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha\end{bmatrix}.
Solution:
Let t=\tan\frac{\alpha}{2}.
Recall the half‑angle identities:
\cos\alpha=\frac{1-t^{2}}{1+t^{2}},\quad \sin\alpha=\frac{2t}{1+t^{2}}.
Write
I+A=\begin{bmatrix}1&-t\\ t&1\end{bmatrix},\quad I-A=\begin{bmatrix}1&t\\ -t&1\end{bmatrix},\quad R=\begin{bmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha\end{bmatrix}.
Multiply (I-A)R:
\begin{aligned}
(I-A)R&=\begin{bmatrix}1&t\\-t&1\end{bmatrix}\begin{bmatrix}\cos\alpha&-\sin\alpha \\ \sin\alpha&\cos\alpha\end{bmatrix}\
&=\begin{bmatrix}
\cos\alpha+t\sin\alpha & -\sin\alpha+t\cos\alpha \\ -t\cos\alpha+\sin\alpha & -t(-\sin\alpha)+\cos\alpha
\end{bmatrix} \end{aligned}
Evaluate each entry
- (1, 1) entry
- (1, 2) entry
- (2, 1) entry
- (2, 2) entry
Hence
(I-A)R=\begin{bmatrix}1&-t\\ t&1\end{bmatrix}=I+A.
Therefore the relation is proved. ✅
Question 19:
A trust fund has ₹ 30,000 that must be invested in two different types of bonds. The first bond pays 5 % interest per year, and the second bond pays 7 % interest per year.
Using matrix multiplication, determine how to divide ₹ 30,000 among the two types of bonds if the trust fund must obtain an annual total interest of:
(a) ₹ 1,800 (b) ₹ 2,000.
Solution:
Let x rupees be invested at 5 % and y rupees at 7 %.
The total amount constraint is ₹ 30,000
x + y = 30000.
Interest equation (in rupees) is
0.05x + 0.07y = I,
where I is the required annual interest.
We can write the system in matrix form:
\begin{bmatrix}1 & 1\\ 0.05 & 0.07\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}30000\\ I\end{bmatrix}.
(a). I = 1800
\begin{cases}x+y=30000\\0.05x+0.07y=1800\end{cases}
From the first equation y = 30000 - x. Substituting:
0.05x + 0.07(30000 - x) = 1800
\Rightarrow -0.02x = -300 \;\Longrightarrow\; x = 15000.
Thus y = 15000.
Answer (a): Invest ₹ 15,000 in each bond.
(b). I = 2000
0.05x + 0.07(30000 - x) = 2000
\Rightarrow -0.02x = -100 \; \\\\ \Longrightarrow\; x = 5000,\; y = 25000.
Answer (b): Invest ₹ 5,000 at 5 % and ₹ 25,000 at 7 %.
Question 20
The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books and 10 dozen economics books. Their selling prices are ₹ 80, ₹ 60 and ₹ 40 each respectively. Find the total amount the bookshop will receive from selling all the books, using matrix algebra.
Solution:
Convert dozens to individual books:
\text{Chemistry books} = 10\times12 = 120,\; \\\\ \text{Physics books} = 8\times12 = 96,\; \\\\ \text{Economics books} = 10\times12 = 120.
Write the quantity row matrix and price column matrix:
Q = \begin{bmatrix}120 & 96 & 120\end{bmatrix},\; P = \begin{bmatrix}80\\60\\40\end{bmatrix}.Total revenue = QP:
QP = 120\cdot80 + 96\cdot60 + 120\cdot40
QP = 9600 + 5760 + 4800
QP = 20160.
Final Answer: The bookshop will receive ₹ 20,160.
Question 21
Assume X, Y, Z, W and P are matrices of orders 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively.
The restriction on n, k and p so that PY + WY will be defined are:
(A) k = 3, p = n
(B) k arbitrary, p = 2
(C) p arbitrary, k = 3
(D) k = 2, p = 3
Solution:
Product P Y:
P has order p\times k and Y has order 3\times k.
For P Y to be defined, the number of columns of P (which is k) must equal the number of rows of Y (which is 3).
Hence k = 3 and the product P Y has order p\times 3.
Product W Y:
W has order n\times 3 and Y has order 3\times k.
With k=3, W Y is defined and has order n\times 3.
Addition P Y + W Y:
Both P Y and W Y must have the same order to be added.
So we require p\times 3 = n\times 3, which implies p = n.
Therefore, the restrictions are k = 3 and p = n,
Which corresponds to Option (A).
Question 22: If n = p, find the order of the matrix 7X - 5Z.
(A) p × 2 (B) 2 × n (C) n × 3 (D) p × n
Solution:
X has order 2\times n and Z has order 2\times p.
With n = p, both X and Z are of order 2\times n.
Therefore, 7X - 5Z is also of order 2\times n,
which corresponds to Option (B).