MATHEMATICS CLASS- 12
CHAPTER-3
(MATRICES)
Exercise 3.2
Question 1 (i)
Given
$$ A= \begin{bmatrix} 2 & 4\\ 3 & 2 \end{bmatrix}, \qquad B= \begin{bmatrix} 1 & 3\\ -2 & 5 \end{bmatrix}, \qquad C= \begin{bmatrix} -2 & 5\\ 3 & 4 \end{bmatrix}. $$Find
$$ A+B. $$Solution
Add the corresponding elements of the two matrices.
$$ \begin{aligned} A+B &= \begin{bmatrix} 2 & 4\\ 3 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 3\\ -2 & 5 \end{bmatrix}\\[4mm] &= \begin{bmatrix} 2+1 & 4+3\\ 3+(-2) & 2+5 \end{bmatrix}\\[4mm] &= \begin{bmatrix} 3 & 7\\ 1 & 7 \end{bmatrix}. \end{aligned} $$Answer:
$$ \boxed{ A+B= \begin{bmatrix} 3 & 7\\ 1 & 7 \end{bmatrix} } $$Question 1 (ii)
Find
$$ A-B. $$Solution
Subtract the corresponding elements of
$$ B $$from
$$ A. $$ $$ \begin{aligned} A-B &= \begin{bmatrix} 2 & 4\\ 3 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 3\\ -2 & 5 \end{bmatrix}\\[4mm] &= \begin{bmatrix} 2-1 & 4-3\\ 3-(-2) & 2-5 \end{bmatrix}\\[4mm] &= \begin{bmatrix} 1 & 1\\ 5 & -3 \end{bmatrix}. \end{aligned} $$Answer:
$$ \boxed{ A-B= \begin{bmatrix} 1 & 1\\ 5 & -3 \end{bmatrix} } $$Question 1 (iii)
Find
$$ 3A-C. $$Solution
First, multiply every element of
$$ A $$by
$$ 3. $$ $$ 3A= 3 \begin{bmatrix} 2 & 4\\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 12\\ 9 & 6 \end{bmatrix}. $$Now subtract the corresponding elements of
$$ C. $$ $$ \begin{aligned} 3A-C &= \begin{bmatrix} 6 & 12\\ 9 & 6 \end{bmatrix} - \begin{bmatrix} -2 & 5\\ 3 & 4 \end{bmatrix}\\[4mm] &= \begin{bmatrix} 6-(-2) & 12-5\\ 9-3 & 6-4 \end{bmatrix}\\[4mm] &= \begin{bmatrix} 8 & 7\\ 6 & 2 \end{bmatrix}. \end{aligned} $$Answer:
$$ \boxed{ 3A-C= \begin{bmatrix} 8 & 7\\ 6 & 2 \end{bmatrix} } $$Question 1 (iv)
Given
$$ A= \begin{bmatrix} 2 & 4\\ 3 & 2 \end{bmatrix}, \qquad B= \begin{bmatrix} 1 & 3\\ -2 & 5 \end{bmatrix}. $$Find
$$ AB. $$Solution
Multiply the rows of
$$ A $$by the columns of
$$ B. $$ $$ \begin{aligned} AB &= \begin{bmatrix} 2 & 4\\ 3 & 2 \end{bmatrix} \begin{bmatrix} 1 & 3\\ -2 & 5 \end{bmatrix}. \end{aligned} $$Now calculate each element.
$$ \begin{aligned} (1,1) &= 2(1)+4(-2)\\ &= 2-8\\ &= -6, \end{aligned} $$ $$ \begin{aligned} (1,2) &= 2(3)+4(5)\\ &= 6+20\\ &= 26, \end{aligned} $$ $$ \begin{aligned} (2,1) &= 3(1)+2(-2)\\ &= 3-4\\ &= -1, \end{aligned} $$ $$ \begin{aligned} (2,2) &= 3(3)+2(5)\\ &= 9+10\\ &= 19. \end{aligned} $$Hence,
$$ AB= \begin{bmatrix} -6 & 26\\ -1 & 19 \end{bmatrix}. $$Answer:
$$ \boxed{ AB= \begin{bmatrix} -6 & 26\\ -1 & 19 \end{bmatrix} } $$Question 1 (v)
Find
$$ BA. $$Solution
Multiply the rows of
$$ B $$by the columns of
$$ A. $$ $$ \begin{aligned} BA &= \begin{bmatrix} 1 & 3\\ -2 & 5 \end{bmatrix} \begin{bmatrix} 2 & 4\\ 3 & 2 \end{bmatrix}. \end{aligned} $$Now calculate each element.
$$ \begin{aligned} (1,1) &= 1(2)+3(3)\\ &= 2+9\\ &= 11, \end{aligned} $$ $$ \begin{aligned} (1,2) &= 1(4)+3(2)\\ &= 4+6\\ &= 10, \end{aligned} $$ $$ \begin{aligned} (2,1) &= (-2)(2)+5(3)\\ &= -4+15\\ &= 11, \end{aligned} $$ $$ \begin{aligned} (2,2) &= (-2)(4)+5(2)\\ &= -8+10\\ &= 2. \end{aligned} $$Hence,
$$ BA= \begin{bmatrix} 11 & 10\\ 11 & 2 \end{bmatrix}. $$Answer:
$$ \boxed{ BA= \begin{bmatrix} 11 & 10\\ 11 & 2 \end{bmatrix} } $$Question 2 (i)
Compute
$$ \begin{bmatrix} a & b\\ -b & a \end{bmatrix} + \begin{bmatrix} a & b\\ b & a \end{bmatrix}. $$Solution
Add the corresponding elements of the two matrices.
$$ \begin{aligned} & \begin{bmatrix} a & b\\ -b & a \end{bmatrix} + \begin{bmatrix} a & b\\ b & a \end{bmatrix} \\[4mm] &= \begin{bmatrix} a+a & b+b\\ -b+b & a+a \end{bmatrix} \\[4mm] &= \begin{bmatrix} 2a & 2b\\ 0 & 2a \end{bmatrix}. \end{aligned} $$Answer:
$$ \boxed{ \begin{bmatrix} 2a & 2b\\ 0 & 2a \end{bmatrix} } $$Question 2 (ii)
Compute
$$ \begin{bmatrix} a^2+b^2 & b^2+c^2\\ a^2+c^2 & a^2+b^2 \end{bmatrix} + \begin{bmatrix} 2ab & 2bc\\ -2ac & -2ab \end{bmatrix}. $$Solution
Add the corresponding elements of the two matrices.
$$ \begin{aligned} & \begin{bmatrix} a^2+b^2 & b^2+c^2\\ a^2+c^2 & a^2+b^2 \end{bmatrix} + \begin{bmatrix} 2ab & 2bc\\ -2ac & -2ab \end{bmatrix} \\[4mm] &= \begin{bmatrix} a^2+b^2+2ab & b^2+c^2+2bc\\ a^2+c^2-2ac & a^2+b^2-2ab \end{bmatrix}. \end{aligned} $$Using the identities
$$ a^2+b^2+2ab=(a+b)^2, $$ $$ b^2+c^2+2bc=(b+c)^2, $$ $$ a^2+c^2-2ac=(a-c)^2, $$and
$$ a^2+b^2-2ab=(a-b)^2, $$we obtain
$$ \boxed{ \begin{bmatrix} (a+b)^2 & (b+c)^2\\ (a-c)^2 & (a-b)^2 \end{bmatrix} }. $$Answer:
$$ \boxed{ \begin{bmatrix} (a+b)^2 & (b+c)^2\\ (a-c)^2 & (a-b)^2 \end{bmatrix} } $$Question 2 (iii)
Compute
$$ \begin{bmatrix} -1 & 4 & -6\\ 8 & 5 & 16\\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6\\ 8 & 0 & 5\\ 3 & 2 & 4 \end{bmatrix}. $$Solution
Add the corresponding elements of the two matrices.
$$ \begin{aligned} & \begin{bmatrix} -1 & 4 & -6\\ 8 & 5 & 16\\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6\\ 8 & 0 & 5\\ 3 & 2 & 4 \end{bmatrix} \\[4mm] &= \begin{bmatrix} -1+12 & 4+7 & -6+6\\ 8+8 & 5+0 & 16+5\\ 2+3 & 8+2 & 5+4 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 11 & 11 & 0\\ 16 & 5 & 21\\ 5 & 10 & 9 \end{bmatrix}. \end{aligned} $$Answer:
$$ \boxed{ \begin{bmatrix} 11 & 11 & 0\\ 16 & 5 & 21\\ 5 & 10 & 9 \end{bmatrix} } $$Question 2 (iv)
Compute
$$ \begin{bmatrix} \cos^2x & \sin^2x\\ \sin^2x & \cos^2x \end{bmatrix} + \begin{bmatrix} \sin^2x & \cos^2x\\ \cos^2x & \sin^2x \end{bmatrix}. $$Solution
Add the corresponding elements of the two matrices.
$$ \begin{aligned} & \begin{bmatrix} \cos^2x & \sin^2x\\ \sin^2x & \cos^2x \end{bmatrix} + \begin{bmatrix} \sin^2x & \cos^2x\\ \cos^2x & \sin^2x \end{bmatrix} \\[4mm] &= \begin{bmatrix} \cos^2x+\sin^2x & \sin^2x+\cos^2x\\ \sin^2x+\cos^2x & \cos^2x+\sin^2x \end{bmatrix}. \end{aligned} $$Using the identity
$$ \sin^2x+\cos^2x=1, $$we obtain
$$ \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix}. $$Answer:
$$ \boxed{ \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix} } $$Question 3 (i)
Compute the product
$$ \begin{bmatrix} a & b\\ -b & a \end{bmatrix} \begin{bmatrix} a & -b\\ b & a \end{bmatrix}. $$Solution
Multiply the rows of the first matrix by the columns of the second matrix.
$$ \begin{aligned} & \begin{bmatrix} a & b\\ -b & a \end{bmatrix} \begin{bmatrix} a & -b\\ b & a \end{bmatrix} \\[4mm] &= \begin{bmatrix} a(a)+b(b) & a(-b)+b(a)\\ (-b)a+a(b) & (-b)(-b)+a(a) \end{bmatrix}. \end{aligned} $$Simplifying each element, we get
$$ \begin{aligned} a(a)+b(b)&=a^2+b^2,\\ a(-b)+b(a)&=-ab+ab=0,\\ (-b)a+a(b)&=-ab+ab=0,\\ (-b)(-b)+a(a)&=b^2+a^2=a^2+b^2. \end{aligned} $$Hence,
$$ \boxed{ \begin{bmatrix} a & b\\ -b & a \end{bmatrix} \begin{bmatrix} a & -b\\ b & a \end{bmatrix} = \begin{bmatrix} a^2+b^2 & 0\\ 0 & a^2+b^2 \end{bmatrix} } $$Question 3 (ii)
Compute the product
$$ \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \end{bmatrix}. $$Solution
The first matrix is of order
$$ 3\times1, $$and the second matrix is of order
$$ 1\times3. $$Therefore, the product is defined and the resulting matrix is of order
$$ 3\times3. $$Multiply each element of the column matrix by each element of the row matrix.
$$ \begin{aligned} & \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 1\times2 & 1\times3 & 1\times4\\ 2\times2 & 2\times3 & 2\times4\\ 3\times2 & 3\times3 & 3\times4 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 2 & 3 & 4\\ 4 & 6 & 8\\ 6 & 9 & 12 \end{bmatrix}. \end{aligned} $$Answer:
$$ \boxed{ \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 4\\ 4 & 6 & 8\\ 6 & 9 & 12 \end{bmatrix} } $$Question 3 (iii)
Compute the product
$$ \begin{bmatrix} 1 & -2 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3\\ 2 & 3 & 1 \end{bmatrix}. $$Solution
The first matrix is of order
$$ 1\times2, $$and the second matrix is of order
$$ 2\times3. $$Hence, the product is defined and the resulting matrix is of order
$$ 1\times3. $$Multiply the row of the first matrix by each column of the second matrix.
$$ \begin{aligned} (1,1) &= 1(1)+(-2)(2)\\ &= 1-4\\ &= -3, \end{aligned} $$ $$ \begin{aligned} (1,2) &= 1(2)+(-2)(3)\\ &= 2-6\\ &= -4, \end{aligned} $$ $$ \begin{aligned} (1,3) &= 1(3)+(-2)(1)\\ &= 3-2\\ &= 1. \end{aligned} $$Therefore,
$$ \boxed{ \begin{bmatrix} 1 & -2 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3\\ 2 & 3 & 1 \end{bmatrix} = \begin{bmatrix} -3 & -4 & 1 \end{bmatrix} } $$Question 3 (iv)
Compute the product
$$ \begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5\\ 0 & 2 & 4\\ 3 & 0 & 5 \end{bmatrix}. $$Solution
Multiply each row of the first matrix by each column of the second matrix.
The elements of the first row are
$$ \begin{aligned} (1,1)&=2(1)+3(0)+4(3)=14,\\ (1,2)&=2(-3)+3(2)+4(0)=0,\\ (1,3)&=2(5)+3(4)+4(5)=42. \end{aligned} $$The elements of the second row are
$$ \begin{aligned} (2,1)&=3(1)+4(0)+5(3)=18,\\ (2,2)&=3(-3)+4(2)+5(0)=-1,\\ (2,3)&=3(5)+4(4)+5(5)=56. \end{aligned} $$The elements of the third row are
$$ \begin{aligned} (3,1)&=4(1)+5(0)+6(3)=22,\\ (3,2)&=4(-3)+5(2)+6(0)=-2,\\ (3,3)&=4(5)+5(4)+6(5)=70. \end{aligned} $$Hence,
$$ \boxed{ \begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5\\ 0 & 2 & 4\\ 3 & 0 & 5 \end{bmatrix} = \begin{bmatrix} 14 & 0 & 42\\ 18 & -1 & 56\\ 22 & -2 & 70 \end{bmatrix} } $$Question 3 (v)
Compute the product
$$ \begin{bmatrix} 2 & 1\\ 3 & 2\\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1\\ -1 & 2 & 1 \end{bmatrix}. $$Solution
The first matrix is of order
$$ 3\times2, $$and the second matrix is of order
$$ 2\times3. $$Hence, the product is defined and the resulting matrix is of order
$$ 3\times3. $$Multiply each row of the first matrix by each column of the second matrix.
The elements of the first row are
$$ \begin{aligned} (1,1)&=2(1)+1(-1)=1,\\ (1,2)&=2(0)+1(2)=2,\\ (1,3)&=2(1)+1(1)=3. \end{aligned} $$The elements of the second row are
$$ \begin{aligned} (2,1)&=3(1)+2(-1)=1,\\ (2,2)&=3(0)+2(2)=4,\\ (2,3)&=3(1)+2(1)=5. \end{aligned} $$The elements of the third row are
$$ \begin{aligned} (3,1)&=(-1)(1)+1(-1)=-2,\\ (3,2)&=(-1)(0)+1(2)=2,\\ (3,3)&=(-1)(1)+1(1)=0. \end{aligned} $$Hence,
$$ \boxed{ \begin{bmatrix} 2 & 1\\ 3 & 2\\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1\\ -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3\\ 1 & 4 & 5\\ -2 & 2 & 0 \end{bmatrix} } $$Question 3 (vi)
Compute the product
$$ \begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}. $$Solution
The first matrix is of order
$$ 2\times3, $$and the second matrix is of order
$$ 3\times2. $$Hence, the product is defined and the resulting matrix is of order
$$ 2\times2. $$Multiply each row of the first matrix by each column of the second matrix.
The elements of the first row are
$$ \begin{aligned} (1,1)&=3(2)+(-1)(1)+3(3)=6-1+9=14,\\ (1,2)&=3(-3)+(-1)(0)+3(1)=-9+0+3=-6. \end{aligned} $$The elements of the second row are
$$ \begin{aligned} (2,1)&=(-1)(2)+0(1)+2(3)=-2+0+6=4,\\ (2,2)&=(-1)(-3)+0(0)+2(1)=3+0+2=5. \end{aligned} $$Therefore,
$$ \boxed{ \begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 14 & -6\\ 4 & 5 \end{bmatrix} } $$Question 4
Given
$$ A= \begin{bmatrix} 1 & 2 & -3\\ 5 & 0 & 2\\ 1 & -1 & 1 \end{bmatrix}, \qquad B= \begin{bmatrix} 3 & -1 & 2\\ 4 & 2 & 5\\ 2 & 0 & 3 \end{bmatrix}, $$ $$ C= \begin{bmatrix} 4 & 1 & 2\\ 0 & 3 & 2\\ 1 & -2 & 3 \end{bmatrix}. $$Compute
$$ (A+B) \quad\text{and}\quad (B-C). $$Also, verify that
$$ A+(B-C)=(A+B)-C. $$Solution
Step 1: Find \(A+B\)
Add the corresponding elements of the matrices.
$$ \begin{aligned} A+B &= \begin{bmatrix} 1 & 2 & -3\\ 5 & 0 & 2\\ 1 & -1 & 1 \end{bmatrix} + \begin{bmatrix} 3 & -1 & 2\\ 4 & 2 & 5\\ 2 & 0 & 3 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 1+3 & 2+(-1) & -3+2\\ 5+4 & 0+2 & 2+5\\ 1+2 & -1+0 & 1+3 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 4 & 1 & -1\\ 9 & 2 & 7\\ 3 & -1 & 4 \end{bmatrix}. \end{aligned} $$Step 2: Find \(B-C\)
Subtract the corresponding elements of
$$ C $$from
$$ B. $$ $$ \begin{aligned} B-C &= \begin{bmatrix} 3 & -1 & 2\\ 4 & 2 & 5\\ 2 & 0 & 3 \end{bmatrix} - \begin{bmatrix} 4 & 1 & 2\\ 0 & 3 & 2\\ 1 & -2 & 3 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 3-4 & -1-1 & 2-2\\ 4-0 & 2-3 & 5-2\\ 2-1 & 0-(-2) & 3-3 \end{bmatrix} \\[4mm] &= \begin{bmatrix} -1 & -2 & 0\\ 4 & -1 & 3\\ 1 & 2 & 0 \end{bmatrix}. \end{aligned} $$Step 3: Find \(A+(B-C)\)
$$ \begin{aligned} A+(B-C) &= \begin{bmatrix} 1 & 2 & -3\\ 5 & 0 & 2\\ 1 & -1 & 1 \end{bmatrix} + \begin{bmatrix} -1 & -2 & 0\\ 4 & -1 & 3\\ 1 & 2 & 0 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 0 & 0 & -3\\ 9 & -1 & 5\\ 2 & 1 & 1 \end{bmatrix}. \end{aligned} $$Step 4: Find \((A+B)-C\)
$$ \begin{aligned} (A+B)-C &= \begin{bmatrix} 4 & 1 & -1\\ 9 & 2 & 7\\ 3 & -1 & 4 \end{bmatrix} - \begin{bmatrix} 4 & 1 & 2\\ 0 & 3 & 2\\ 1 & -2 & 3 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 0 & 0 & -3\\ 9 & -1 & 5\\ 2 & 1 & 1 \end{bmatrix}. \end{aligned} $$Verification
Since
$$ A+(B-C) = \begin{bmatrix} 0 & 0 & -3\\ 9 & -1 & 5\\ 2 & 1 & 1 \end{bmatrix}, $$and
$$ (A+B)-C = \begin{bmatrix} 0 & 0 & -3\\ 9 & -1 & 5\\ 2 & 1 & 1 \end{bmatrix}, $$we have
$$ A+(B-C)=(A+B)-C. $$Answer:
$$ \boxed{ A+B= \begin{bmatrix} 4 & 1 & -1\\ 9 & 2 & 7\\ 3 & -1 & 4 \end{bmatrix} } $$ $$ \boxed{ B-C= \begin{bmatrix} -1 & -2 & 0\\ 4 & -1 & 3\\ 1 & 2 & 0 \end{bmatrix} } $$ $$ \boxed{ A+(B-C)=(A+B)-C= \begin{bmatrix} 0 & 0 & -3\\ 9 & -1 & 5\\ 2 & 1 & 1 \end{bmatrix} } $$Question 5
Given
$$ A= \begin{bmatrix} \dfrac23 & 1 & \dfrac53\\[2mm] \dfrac13 & \dfrac23 & \dfrac43\\[2mm] \dfrac73 & 2 & \dfrac23 \end{bmatrix}, \qquad B= \begin{bmatrix} \dfrac25 & \dfrac35 & 1\\[2mm] \dfrac15 & \dfrac25 & \dfrac45\\[2mm] \dfrac75 & \dfrac65 & \dfrac25 \end{bmatrix}. $$Compute
$$ 3A-5B. $$Solution
Step 1: Find \(3A\)
Multiply each element of matrix
$$ A $$by
$$ 3. $$ $$ \begin{aligned} 3A &= 3 \begin{bmatrix} \dfrac23 & 1 & \dfrac53\\[2mm] \dfrac13 & \dfrac23 & \dfrac43\\[2mm] \dfrac73 & 2 & \dfrac23 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 & 4\\ 7 & 6 & 2 \end{bmatrix}. \end{aligned} $$Step 2: Find \(5B\)
Multiply each element of matrix
$$ B $$by
$$ 5. $$ $$ \begin{aligned} 5B &= 5 \begin{bmatrix} \dfrac25 & \dfrac35 & 1\\[2mm] \dfrac15 & \dfrac25 & \dfrac45\\[2mm] \dfrac75 & \dfrac65 & \dfrac25 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 & 4\\ 7 & 6 & 2 \end{bmatrix}. \end{aligned} $$Step 3: Find \(3A-5B\)
Subtract the corresponding elements of
$$ 5B $$from
$$ 3A. $$ $$ \begin{aligned} 3A-5B &= \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 & 4\\ 7 & 6 & 2 \end{bmatrix} - \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 & 4\\ 7 & 6 & 2 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}. \end{aligned} $$Answer:
$$ \boxed{ 3A-5B= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} } $$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
First, multiply the scalar
$$ \cos\theta $$with every element of the first matrix.
$$ \begin{aligned} \cos\theta \begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix} = \begin{bmatrix} \cos^2\theta & \sin\theta\cos\theta\\ -\sin\theta\cos\theta & \cos^2\theta \end{bmatrix}. \end{aligned} $$Now multiply the scalar
$$ \sin\theta $$with every element of the second matrix.
$$ \begin{aligned} \sin\theta \begin{bmatrix} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{bmatrix} = \begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta\\ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix}. \end{aligned} $$Now add the corresponding elements of the two matrices.
$$ \begin{aligned} & \begin{bmatrix} \cos^2\theta & \sin\theta\cos\theta\\ -\sin\theta\cos\theta & \cos^2\theta \end{bmatrix} + \begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta\\ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix} \\[4mm] &= \begin{bmatrix} \cos^2\theta+\sin^2\theta & \sin\theta\cos\theta-\sin\theta\cos\theta\\ -\sin\theta\cos\theta+\sin\theta\cos\theta & \cos^2\theta+\sin^2\theta \end{bmatrix}. \end{aligned} $$Using the identity
$$ \sin^2\theta+\cos^2\theta=1, $$we obtain
$$ \begin{aligned} &= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}. \end{aligned} $$Answer:
$$ \boxed{ \cos\theta \begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta \begin{bmatrix} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} } $$Question 7 (i)
Find the matrices
$$ X \quad\text{and}\quad Y, $$if
$$ X+Y= \begin{bmatrix} 7 & 0\\ 2 & 5 \end{bmatrix} \quad\text{and}\quad X-Y= \begin{bmatrix} 3 & 0\\ 0 & 3 \end{bmatrix}. $$Solution
We are given
$$ X+Y= \begin{bmatrix} 7 & 0\\ 2 & 5 \end{bmatrix} \tag{1} $$and
$$ X-Y= \begin{bmatrix} 3 & 0\\ 0 & 3 \end{bmatrix}. \tag{2} $$Step 1: Find \(X\)
Add equations (1) and (2).
$$ (X+Y)+(X-Y)= \begin{bmatrix} 7 & 0\\ 2 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 0\\ 0 & 3 \end{bmatrix}. $$Therefore,
$$ 2X= \begin{bmatrix} 10 & 0\\ 2 & 8 \end{bmatrix}. $$Dividing each element by
$$ 2, $$we get
$$ X= \begin{bmatrix} 5 & 0\\ 1 & 4 \end{bmatrix}. $$Step 2: Find \(Y\)
Subtract equation (2) from equation (1).
$$ (X+Y)-(X-Y)= \begin{bmatrix} 7 & 0\\ 2 & 5 \end{bmatrix} - \begin{bmatrix} 3 & 0\\ 0 & 3 \end{bmatrix}. $$Hence,
$$ 2Y= \begin{bmatrix} 4 & 0\\ 2 & 2 \end{bmatrix}. $$Dividing each element by
$$ 2, $$we obtain
$$ Y= \begin{bmatrix} 2 & 0\\ 1 & 1 \end{bmatrix}. $$Verification
Now,
$$ X+Y= \begin{bmatrix} 5 & 0\\ 1 & 4 \end{bmatrix} + \begin{bmatrix} 2 & 0\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0\\ 2 & 5 \end{bmatrix}, $$and
$$ X-Y= \begin{bmatrix} 5 & 0\\ 1 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 0\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 0\\ 0 & 3 \end{bmatrix}. $$Thus, both given conditions are satisfied.
Answer:
$$ \boxed{ X= \begin{bmatrix} 5 & 0\\ 1 & 4 \end{bmatrix} } $$ $$ \boxed{ Y= \begin{bmatrix} 2 & 0\\ 1 & 1 \end{bmatrix} } $$Question 7 (ii)
Find the matrices
$$ X \quad\text{and}\quad Y, $$if
$$ X+Y= \begin{bmatrix} 0 & 2\\ 3 & 4 \end{bmatrix} \quad\text{and}\quad X-Y= \begin{bmatrix} 0 & 2\\ 3 & 4 \end{bmatrix}. $$Solution
We are given
$$ X+Y= \begin{bmatrix} 0 & 2\\ 3 & 4 \end{bmatrix} \tag{1} $$and
$$ X-Y= \begin{bmatrix} 0 & 2\\ 3 & 4 \end{bmatrix}. \tag{2} $$Step 1: Find \(X\)
Add equations (1) and (2).
$$ (X+Y)+(X-Y) = \begin{bmatrix} 0 & 2\\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 0 & 2\\ 3 & 4 \end{bmatrix}. $$Therefore,
$$ 2X = \begin{bmatrix} 0 & 4\\ 6 & 8 \end{bmatrix}. $$Dividing each element by
$$ 2, $$we obtain
$$ X = \begin{bmatrix} 0 & 2\\ 3 & 4 \end{bmatrix}. $$Step 2: Find \(Y\)
Subtract equation (2) from equation (1).
$$ (X+Y)-(X-Y) = \begin{bmatrix} 0 & 2\\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 0 & 2\\ 3 & 4 \end{bmatrix}. $$Hence,
$$ 2Y = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}. $$Dividing each element by
$$ 2, $$we get
$$ Y = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}. $$Verification
Now,
$$ X+Y = \begin{bmatrix} 0 & 2\\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2\\ 3 & 4 \end{bmatrix}, $$and
$$ X-Y = \begin{bmatrix} 0 & 2\\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2\\ 3 & 4 \end{bmatrix}. $$Thus, both given conditions are satisfied.
Answer:
$$ \boxed{ X= \begin{bmatrix} 0 & 2\\ 3 & 4 \end{bmatrix} } $$ $$ \boxed{ Y= \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} } $$Question 8
Find the matrix
$$ X, $$if
$$ Y= \begin{bmatrix} 3 & 2\\ 1 & 4 \end{bmatrix} \quad\text{and}\quad 2X+Y= \begin{bmatrix} 1 & 0\\ -3 & 2 \end{bmatrix}. $$Solution
We are given
$$ Y= \begin{bmatrix} 3 & 2\\ 1 & 4 \end{bmatrix} $$and
$$ 2X+Y= \begin{bmatrix} 1 & 0\\ -3 & 2 \end{bmatrix}. $$Subtract
$$ Y $$from both sides.
$$ 2X= \begin{bmatrix} 1 & 0\\ -3 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 2\\ 1 & 4 \end{bmatrix}. $$Subtract the corresponding elements.
$$ \begin{aligned} 2X &= \begin{bmatrix} 1-3 & 0-2\\ -3-1 & 2-4 \end{bmatrix} \\[4mm] &= \begin{bmatrix} -2 & -2\\ -4 & -2 \end{bmatrix}. \end{aligned} $$Now divide each element by
$$ 2. $$ $$ \begin{aligned} X &= \frac12 \begin{bmatrix} -2 & -2\\ -4 & -2 \end{bmatrix} \\[4mm] &= \begin{bmatrix} -1 & -1\\ -2 & -1 \end{bmatrix}. \end{aligned} $$Verification
Now,
$$ 2X= 2 \begin{bmatrix} -1 & -1\\ -2 & -1 \end{bmatrix} = \begin{bmatrix} -2 & -2\\ -4 & -2 \end{bmatrix}. $$Adding
$$ Y, $$we get
$$ \begin{aligned} 2X+Y &= \begin{bmatrix} -2 & -2\\ -4 & -2 \end{bmatrix} + \begin{bmatrix} 3 & 2\\ 1 & 4 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 1 & 0\\ -3 & 2 \end{bmatrix}, \end{aligned} $$which is the given matrix. Hence, the value of
$$ X $$is correct.
Answer:
$$ \boxed{ X= \begin{bmatrix} -1 & -1\\ -2 & -1 \end{bmatrix} } $$Question 9
Find the values of
$$ x \quad \text{and} \quad y, $$if
$$ 2 \begin{bmatrix} 1 & 3\\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0\\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8 \end{bmatrix}. $$Solution
First, multiply every element of the first matrix by
$$ 2. $$ $$ 2 \begin{bmatrix} 1 & 3\\ 0 & x \end{bmatrix} = \begin{bmatrix} 2 & 6\\ 0 & 2x \end{bmatrix}. $$Now add the two matrices.
$$ \begin{aligned} & \begin{bmatrix} 2 & 6\\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0\\ 1 & 2 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 2+y & 6\\ 1 & 2x+2 \end{bmatrix}. \end{aligned} $$Since the two matrices are equal, their corresponding elements must be equal.
Comparing the corresponding elements, we obtain
$$ 2+y=5, $$ $$ 6=6, $$ $$ 1=1, $$ $$ 2x+2=8. $$From
$$ 2+y=5, $$we get
$$ y=3. $$From
$$ 2x+2=8, $$we obtain
$$ 2x=6, $$so
$$ x=3. $$Verification:
Substituting
$$ x=3 \quad\text{and}\quad y=3, $$the left-hand side becomes
$$ 2 \begin{bmatrix} 1 & 3\\ 0 & 3 \end{bmatrix} + \begin{bmatrix} 3 & 0\\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 6\\ 0 & 6 \end{bmatrix} + \begin{bmatrix} 3 & 0\\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8 \end{bmatrix}, $$which is equal to the right-hand side.
Answer:
$$ \boxed{x=3,\qquad y=3} $$Question 10
Solve the equation for
$$ x,\;y,\;z\;\text{and}\;t, $$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
First, multiply each matrix by its scalar.
$$ 2 \begin{bmatrix} x & z\\ y & t \end{bmatrix} = \begin{bmatrix} 2x & 2z\\ 2y & 2t \end{bmatrix}, $$and
$$ 3 \begin{bmatrix} 1 & -1\\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -3\\ 0 & 6 \end{bmatrix}. $$Also,
$$ 3 \begin{bmatrix} 3 & 5\\ 4 & 6 \end{bmatrix} = \begin{bmatrix} 9 & 15\\ 12 & 18 \end{bmatrix}. $$Hence, the given equation becomes
$$ \begin{bmatrix} 2x & 2z\\ 2y & 2t \end{bmatrix} + \begin{bmatrix} 3 & -3\\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 9 & 15\\ 12 & 18 \end{bmatrix}. $$Adding the matrices on the left-hand side, we get
$$ \begin{bmatrix} 2x+3 & 2z-3\\ 2y & 2t+6 \end{bmatrix} = \begin{bmatrix} 9 & 15\\ 12 & 18 \end{bmatrix}. $$Since equal matrices have equal corresponding elements, we obtain
$$ 2x+3=9, $$ $$ 2z-3=15, $$ $$ 2y=12, $$ $$ 2t+6=18. $$Now solve each equation.
From
$$ 2x+3=9, $$we get
$$ 2x=6, $$so
$$ x=3. $$From
$$ 2z-3=15, $$we obtain
$$ 2z=18, $$hence
$$ z=9. $$From
$$ 2y=12, $$we get
$$ y=6. $$Finally, from
$$ 2t+6=18, $$we obtain
$$ 2t=12, $$which gives
$$ t=6. $$Verification
Substituting
$$ x=3,\quad y=6,\quad z=9,\quad t=6, $$the left-hand side becomes
$$ 2 \begin{bmatrix} 3 & 9\\ 6 & 6 \end{bmatrix} + 3 \begin{bmatrix} 1 & -1\\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 18\\ 12 & 12 \end{bmatrix} + \begin{bmatrix} 3 & -3\\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 9 & 15\\ 12 & 18 \end{bmatrix}, $$which is equal to the right-hand side.
Answer:
$$ \boxed{x=3,\qquad y=6,\qquad z=9,\qquad t=6} $$Question 11
If
$$ x \begin{bmatrix} 2\\ 3 \end{bmatrix} + y \begin{bmatrix} -1\\ 1 \end{bmatrix} = \begin{bmatrix} 10\\ 5 \end{bmatrix}, $$find the values of
$$ x \quad\text{and}\quad y. $$Solution
First, multiply each column matrix by its corresponding scalar.
$$ x \begin{bmatrix} 2\\ 3 \end{bmatrix} = \begin{bmatrix} 2x\\ 3x \end{bmatrix}, $$and
$$ y \begin{bmatrix} -1\\ 1 \end{bmatrix} = \begin{bmatrix} -y\\ y \end{bmatrix}. $$Adding the two matrices, we get
$$ \begin{bmatrix} 2x-y\\ 3x+y \end{bmatrix} = \begin{bmatrix} 10\\ 5 \end{bmatrix}. $$Since equal matrices have equal corresponding elements, we obtain
$$ 2x-y=10, $$ $$ 3x+y=5. $$Step 1: Find the value of \(x\)
Add the two equations.
$$ (2x-y)+(3x+y)=10+5. $$Therefore,
$$ 5x=15, $$which gives
$$ x=3. $$Step 2: Find the value of \(y\)
Substitute
$$ x=3 $$into
$$ 2x-y=10. $$We get
$$ 6-y=10, $$so
$$ -y=4, $$and therefore
$$ y=-4. $$Verification
Substituting
$$ x=3 \quad\text{and}\quad y=-4, $$the left-hand side becomes
$$ 3 \begin{bmatrix} 2\\ 3 \end{bmatrix} + (-4) \begin{bmatrix} -1\\ 1 \end{bmatrix} = \begin{bmatrix} 6\\ 9 \end{bmatrix} + \begin{bmatrix} 4\\ -4 \end{bmatrix} = \begin{bmatrix} 10\\ 5 \end{bmatrix}, $$which is equal to the right-hand side.
Answer:
$$ \boxed{x=3,\qquad y=-4} $$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 the values of
$$ x,\;y,\;z\;\text{and}\;w. $$Solution
First, multiply the matrix on the left-hand side by
$$ 3. $$ $$ 3 \begin{bmatrix} x & y\\ z & w \end{bmatrix} = \begin{bmatrix} 3x & 3y\\ 3z & 3w \end{bmatrix}. $$Now add the two matrices on the right-hand side.
$$ \begin{aligned} & \begin{bmatrix} x & 6\\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 & x+y\\ z+w & 3 \end{bmatrix} \\[4mm] &= \begin{bmatrix} x+4 & x+y+6\\ z+w-1 & 2w+3 \end{bmatrix}. \end{aligned} $$Hence,
$$ \begin{bmatrix} 3x & 3y\\ 3z & 3w \end{bmatrix} = \begin{bmatrix} x+4 & x+y+6\\ z+w-1 & 2w+3 \end{bmatrix}. $$Comparing the corresponding elements, we obtain
$$ 3x=x+4, $$ $$ 3y=x+y+6, $$ $$ 3z=z+w-1, $$ $$ 3w=2w+3. $$Step 1: Find the value of \(x\)
$$ 3x=x+4 $$gives
$$ 2x=4, $$so
$$ x=2. $$Step 2: Find the value of \(y\)
Substitute
$$ x=2 $$into
$$ 3y=x+y+6. $$Then
$$ 3y=2+y+6, $$which gives
$$ 2y=8. $$Therefore,
$$ y=4. $$Step 3: Find the value of \(w\)
From
$$ 3w=2w+3, $$we obtain
$$ w=3. $$Step 4: Find the value of \(z\)
Substitute
$$ w=3 $$into
$$ 3z=z+w-1. $$Then
$$ 3z=z+3-1, $$or
$$ 3z=z+2. $$Hence,
$$ 2z=2, $$which gives
$$ z=1. $$Verification
Substituting
$$ x=2,\quad y=4,\quad z=1,\quad w=3, $$the left-hand side becomes
$$ 3 \begin{bmatrix} 2 & 4\\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 6 & 12\\ 3 & 9 \end{bmatrix}. $$The right-hand side is
$$ \begin{aligned} & \begin{bmatrix} 2 & 6\\ -1 & 6 \end{bmatrix} + \begin{bmatrix} 4 & 6\\ 4 & 3 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 6 & 12\\ 3 & 9 \end{bmatrix}. \end{aligned} $$Since both sides are equal, the obtained values are correct.
Answer:
$$ \boxed{x=2,\qquad y=4,\qquad z=1,\qquad w=3} $$Question 13
If
$$ F(x)= \begin{bmatrix} \cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0 & 0 & 1 \end{bmatrix}, $$show that
$$ F(x)F(y)=F(x+y). $$Proof
We have
$$ F(x)= \begin{bmatrix} \cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0 & 0 & 1 \end{bmatrix} $$and
$$ F(y)= \begin{bmatrix} \cos y & -\sin y & 0\\ \sin y & \cos y & 0\\ 0 & 0 & 1 \end{bmatrix}. $$Now multiply the two matrices.
$$ \begin{aligned} F(x)F(y) &= \begin{bmatrix} \cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0\\ \sin y & \cos y & 0\\ 0 & 0 & 1 \end{bmatrix}. \end{aligned} $$The element in the first row and first column is
$$ \cos x\cos y-\sin x\sin y = \cos(x+y). $$The element in the first row and second column is
$$ -\cos x\sin y-\sin x\cos y = -\sin(x+y). $$The element in the first row and third column is
$$ 0. $$The element in the second row and first column is
$$ \sin x\cos y+\cos x\sin y = \sin(x+y). $$The element in the second row and second column is
$$ -\sin x\sin y+\cos x\cos y = \cos(x+y). $$The element in the second row and third column is
$$ 0. $$The third row is obtained as
$$ \begin{aligned} (3,1)&=0,\\ (3,2)&=0,\\ (3,3)&=1. \end{aligned} $$Hence,
$$ F(x)F(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0\\ \sin(x+y) & \cos(x+y) & 0\\ 0 & 0 & 1 \end{bmatrix}. $$But, by the definition of
$$ F(x+y), $$we have
$$ F(x+y)= \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0\\ \sin(x+y) & \cos(x+y) & 0\\ 0 & 0 & 1 \end{bmatrix}. $$Therefore,
$$ \boxed{F(x)F(y)=F(x+y).} $$Question 14 (i)
Show that
$$ \begin{bmatrix} 5 & -1\\ 6 & 7 \end{bmatrix} \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix} \ne \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & -1\\ 6 & 7 \end{bmatrix}. $$Proof
First, find
$$ AB. $$ $$ \begin{aligned} AB &= \begin{bmatrix} 5 & -1\\ 6 & 7 \end{bmatrix} \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 5(2)+(-1)(3) & 5(1)+(-1)(4)\\ 6(2)+7(3) & 6(1)+7(4) \end{bmatrix} \\[4mm] &= \begin{bmatrix} 7 & 1\\ 33 & 34 \end{bmatrix}. \end{aligned} $$Now find
$$ BA. $$ $$ \begin{aligned} BA &= \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & -1\\ 6 & 7 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 2(5)+1(6) & 2(-1)+1(7)\\ 3(5)+4(6) & 3(-1)+4(7) \end{bmatrix} \\[4mm] &= \begin{bmatrix} 16 & 5\\ 39 & 25 \end{bmatrix}. \end{aligned} $$Since
$$ AB= \begin{bmatrix} 7 & 1\\ 33 & 34 \end{bmatrix} \ne \begin{bmatrix} 16 & 5\\ 39 & 25 \end{bmatrix} =BA, $$therefore,
$$ \boxed{ \begin{bmatrix} 5 & -1\\ 6 & 7 \end{bmatrix} \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix} \ne \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & -1\\ 6 & 7 \end{bmatrix}. } $$Question 14 (ii)
Show that
$$ \begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 0\\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \ne \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 0\\ 1 & 1 & 0 \end{bmatrix}. $$Proof
Let
$$ A= \begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 0\\ 1 & 1 & 0 \end{bmatrix}, \qquad B= \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix}. $$First, calculate
$$ AB. $$ $$ \begin{aligned} AB &= \begin{bmatrix} 5 & 8 & 14\\ 0 & -1 & 1\\ -1 & 0 & 1 \end{bmatrix}. \end{aligned} $$Now calculate
$$ BA. $$ $$ \begin{aligned} BA &= \begin{bmatrix} -1 & -1 & -3\\ 1 & 0 & 0\\ 6 & 11 & 6 \end{bmatrix}. \end{aligned} $$Since
$$ AB= \begin{bmatrix} 5 & 8 & 14\\ 0 & -1 & 1\\ -1 & 0 & 1 \end{bmatrix} \ne \begin{bmatrix} -1 & -1 & -3\\ 1 & 0 & 0\\ 6 & 11 & 6 \end{bmatrix} =BA, $$it follows that matrix multiplication is not commutative.
Hence,
$$ \boxed{ \begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 0\\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \ne \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 0\\ 1 & 1 & 0 \end{bmatrix}. } $$Question 15
Find
$$ A^2-5A+6I, $$if
$$ A= \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 & 3\\ 1 & -1 & 0 \end{bmatrix}. $$Solution
First, calculate
$$ A^2=A\times A. $$ $$ \begin{aligned} A^2 &= \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 & 3\\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 & 3\\ 1 & -1 & 0 \end{bmatrix}. \end{aligned} $$Multiplying the rows of the first matrix by the columns of the second matrix, we obtain
$$ \begin{aligned} A^2 &= \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 & 5\\ 0 & -1 & -2 \end{bmatrix}. \end{aligned} $$Now calculate
$$ 5A. $$ $$ \begin{aligned} 5A &= 5 \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 & 3\\ 1 & -1 & 0 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 10 & 0 & 5\\ 10 & 5 & 15\\ 5 & -5 & 0 \end{bmatrix}. \end{aligned} $$The identity matrix of order
$$ 3 $$is
$$ I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}. $$Therefore,
$$ 6I= \begin{bmatrix} 6 & 0 & 0\\ 0 & 6 & 0\\ 0 & 0 & 6 \end{bmatrix}. $$Now compute
$$ A^2-5A. $$ $$ \begin{aligned} A^2-5A &= \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 & 5\\ 0 & -1 & -2 \end{bmatrix} - \begin{bmatrix} 10 & 0 & 5\\ 10 & 5 & 15\\ 5 & -5 & 0 \end{bmatrix} \\[4mm] &= \begin{bmatrix} -5 & -1 & -3\\ -1 & -7 & -10\\ -5 & 4 & -2 \end{bmatrix}. \end{aligned} $$Finally, add
$$ 6I. $$ $$ \begin{aligned} A^2-5A+6I &= \begin{bmatrix} -5 & -1 & -3\\ -1 & -7 & -10\\ -5 & 4 & -2 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0\\ 0 & 6 & 0\\ 0 & 0 & 6 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 1 & -1 & -3\\ -1 & -1 & -10\\ -5 & 4 & 4 \end{bmatrix}. \end{aligned} $$Answer:
$$ \boxed{ A^2-5A+6I= \begin{bmatrix} 1 & -1 & -3\\ -1 & -1 & -10\\ -5 & 4 & 4 \end{bmatrix} } $$Question 16
If
$$ A= \begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1\\ 2 & 0 & 3 \end{bmatrix}, $$prove that
$$ A^3-6A^2+A+2I=0. $$Proof
First, calculate
$$ A^2=A\times A. $$ $$ \begin{aligned} A^2 &= \begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1\\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1\\ 2 & 0 & 3 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 5 & 0 & 8\\ 2 & 4 & 5\\ 8 & 0 & 13 \end{bmatrix}. \end{aligned} $$Now calculate
$$ A^3=A^2A. $$ $$ \begin{aligned} A^3 &= \begin{bmatrix} 5 & 0 & 8\\ 2 & 4 & 5\\ 8 & 0 & 13 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1\\ 2 & 0 & 3 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 21 & 0 & 34\\ 12 & 8 & 23\\ 34 & 0 & 55 \end{bmatrix}. \end{aligned} $$Next, compute
$$ 6A^2. $$ $$ 6A^2= 6 \begin{bmatrix} 5 & 0 & 8\\ 2 & 4 & 5\\ 8 & 0 & 13 \end{bmatrix} = \begin{bmatrix} 30 & 0 & 48\\ 12 & 24 & 30\\ 48 & 0 & 78 \end{bmatrix}. $$Also,
$$ 2I= \begin{bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{bmatrix}. $$Now calculate
$$ A^3-6A^2. $$ $$ \begin{aligned} A^3-6A^2 &= \begin{bmatrix} 21 & 0 & 34\\ 12 & 8 & 23\\ 34 & 0 & 55 \end{bmatrix} - \begin{bmatrix} 30 & 0 & 48\\ 12 & 24 & 30\\ 48 & 0 & 78 \end{bmatrix} \\[4mm] &= \begin{bmatrix} -9 & 0 & -14\\ 0 & -16 & -7\\ -14 & 0 & -23 \end{bmatrix}. \end{aligned} $$Add the matrix
$$ A. $$ $$ \begin{aligned} A^3-6A^2+A &= \begin{bmatrix} -9 & 0 & -14\\ 0 & -16 & -7\\ -14 & 0 & -23 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1\\ 2 & 0 & 3 \end{bmatrix} \\[4mm] &= \begin{bmatrix} -8 & 0 & -12\\ 0 & -14 & -6\\ -12 & 0 & -20 \end{bmatrix}. \end{aligned} $$This does not match the required identity, so let us instead compute the required expression directly:
$$ \begin{aligned} A^3-6A^2+11A &= \begin{bmatrix} 21 & 0 & 34\\ 12 & 8 & 23\\ 34 & 0 & 55 \end{bmatrix} - \begin{bmatrix} 30 & 0 & 48\\ 12 & 24 & 30\\ 48 & 0 & 78 \end{bmatrix} + \begin{bmatrix} 11 & 0 & 22\\ 0 & 22 & 11\\ 22 & 0 & 33 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 2 & 0 & 8\\ 0 & 6 & 4\\ 8 & 0 & 10 \end{bmatrix}, \end{aligned} $$which is also not equal to
$$ -6I. $$Observation:
The identity printed in the question,
$$ A^3-6A^2+A+2I=0, $$is not satisfied by the given matrix. Substituting the given matrix yields
$$ A^3-6A^2+A+2I= \begin{bmatrix} -6 & 0 & -12\\ 0 & -12 & -6\\ -12 & 0 & -18 \end{bmatrix} \neq 0. $$Hence, there appears to be a printing error in the textbook question. The given matrix does not satisfy the stated identity.
Question 17
If
$$ A= \begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix}, $$show that
$$ A^n= \begin{bmatrix} 1 & 0\\ n & 1 \end{bmatrix}, \qquad n\in\mathbb{N}. $$Proof
We shall use the principle of mathematical induction.
Step 1: Verify for \(n=1\)
For
$$ n=1, $$the left-hand side is
$$ A^1=A= \begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix}. $$The right-hand side is
$$ \begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix}. $$Hence, the result is true for
$$ n=1. $$Step 2: Induction Hypothesis
Assume that the statement is true for
$$ n=k. $$That is, assume
$$ A^k= \begin{bmatrix} 1 & 0\\ k & 1 \end{bmatrix}. $$Step 3: Prove for \(n=k+1\)
We have
$$ A^{k+1}=A^kA. $$Using the induction hypothesis,
$$ \begin{aligned} A^{k+1} &= \begin{bmatrix} 1 & 0\\ k & 1 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 1(1)+0(1) & 1(0)+0(1)\\ k(1)+1(1) & k(0)+1(1) \end{bmatrix} \\[4mm] &= \begin{bmatrix} 1 & 0\\ k+1 & 1 \end{bmatrix}. \end{aligned} $$This is exactly the required form for
$$ n=k+1. $$Conclusion
Since the statement is true for
$$ n=1 $$and true for
$$ n=k+1 $$whenever it is true for
$$ n=k, $$it follows by the principle of mathematical induction that
$$ A^n= \begin{bmatrix} 1 & 0\\ n & 1 \end{bmatrix} \qquad \text{for all }n\in\mathbb{N}. $$Hence proved.
$$ \boxed{ A^n= \begin{bmatrix} 1 & 0\\ n & 1 \end{bmatrix} } $$Question 18
If
$$ A= \begin{bmatrix} 1 & 0\\ 0 & 2 \end{bmatrix}, $$show that
$$ A^n= \begin{bmatrix} 1 & 0\\ 0 & 2^n \end{bmatrix}, \qquad n\in\mathbb{N}. $$Proof
We shall prove the result by the principle of mathematical induction.
Step 1: Verify for \(n=1\)
For
$$ n=1, $$the left-hand side is
$$ A^1=A= \begin{bmatrix} 1 & 0\\ 0 & 2 \end{bmatrix}. $$The right-hand side is
$$ \begin{bmatrix} 1 & 0\\ 0 & 2^1 \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 2 \end{bmatrix}. $$Hence, the statement is true for
$$ n=1. $$Step 2: Induction Hypothesis
Assume that the statement is true for
$$ n=k. $$That is, assume
$$ A^k= \begin{bmatrix} 1 & 0\\ 0 & 2^k \end{bmatrix}. $$Step 3: Prove for \(n=k+1\)
We have
$$ A^{k+1}=A^kA. $$Using the induction hypothesis,
$$ \begin{aligned} A^{k+1} &= \begin{bmatrix} 1 & 0\\ 0 & 2^k \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & 2 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 1\cdot1+0\cdot0 & 1\cdot0+0\cdot2\\ 0\cdot1+2^k\cdot0 & 0\cdot0+2^k\cdot2 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 1 & 0\\ 0 & 2^{k+1} \end{bmatrix}. \end{aligned} $$This is exactly the required form for
$$ n=k+1. $$Conclusion
Since the statement is true for
$$ n=1, $$and true for
$$ n=k+1 $$whenever it is true for
$$ n=k, $$it follows by the principle of mathematical induction that
$$ A^n= \begin{bmatrix} 1 & 0\\ 0 & 2^n \end{bmatrix} \qquad \text{for all }n\in\mathbb{N}. $$Hence proved.
$$ \boxed{ A^n= \begin{bmatrix} 1 & 0\\ 0 & 2^n \end{bmatrix} } $$Question 19
A trust fund has
$$ ₹30,000 $$which 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:
- ₹1800
- ₹2000
(a) Annual Interest = ₹1800
Solution
Let the amount invested in the
$$ 5\% $$bond be
$$ x $$rupees and the amount invested in the
$$ 7\% $$bond be
$$ y $$rupees.
Then,
$$ x+y=30000. $$The total annual interest is
$$ 0.05x+0.07y=1800. $$These equations can be written in matrix form as
$$ \begin{bmatrix} 1 & 1\\ 0.05 & 0.07 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 30000\\ 1800 \end{bmatrix}. $$From
$$ x+y=30000, $$we get
$$ x=30000-y. $$Substituting this value into the second equation,
$$ 0.05(30000-y)+0.07y=1800. $$Simplifying,
$$ 1500-0.05y+0.07y=1800, $$ $$ 0.02y=300, $$ $$ y=15000. $$Therefore,
$$ x=30000-15000=15000. $$Answer:
$$ \boxed{ \begin{aligned} \text{Amount invested at }5\% &= ₹15,000,\\ \text{Amount invested at }7\% &= ₹15,000. \end{aligned} } $$(b) Annual Interest = ₹2000
Solution
The required equations are
$$ x+y=30000, $$and
$$ 0.05x+0.07y=2000. $$In matrix form,
$$ \begin{bmatrix} 1 & 1\\ 0.05 & 0.07 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 30000\\ 2000 \end{bmatrix}. $$From
$$ x+y=30000, $$we obtain
$$ x=30000-y. $$Substituting into the second equation,
$$ 0.05(30000-y)+0.07y=2000. $$Therefore,
$$ 1500-0.05y+0.07y=2000, $$ $$ 0.02y=500, $$ $$ y=25000. $$Hence,
$$ x=30000-25000=5000. $$Answer:
$$ \boxed{ \begin{aligned} \text{Amount invested at }5\% &= ₹5,000,\\ \text{Amount invested at }7\% &= ₹25,000. \end{aligned} } $$Question 20
The bookshop of a particular school has
- 10 dozen Chemistry books,
- 8 dozen Physics books,
- 10 dozen Economics books.
The selling prices are
$$ ₹80,\quad ₹60,\quad \text{and}\quad ₹40 $$per book respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.
Solution
Since
$$ 1\text{ dozen}=12\text{ books}, $$the total number of books is
$$ \begin{aligned} \text{Chemistry} &=10\times12=120,\\ \text{Physics} &=8\times12=96,\\ \text{Economics} &=10\times12=120. \end{aligned} $$Represent the number of books by the row matrix
$$ A= \begin{bmatrix} 120 & 96 & 120 \end{bmatrix}, $$and the selling prices by the column matrix
$$ B= \begin{bmatrix} 80\\ 60\\ 40 \end{bmatrix}. $$Using matrix multiplication, the total amount received is
$$ AB= \begin{bmatrix} 120 & 96 & 120 \end{bmatrix} \begin{bmatrix} 80\\ 60\\ 40 \end{bmatrix}. $$Now multiply the corresponding entries.
$$ \begin{aligned} AB &= 120(80)+96(60)+120(40)\\[2mm] &= 9600+5760+4800\\[2mm] &= 20160. \end{aligned} $$Answer:
$$ \boxed{\text{The total amount received from selling all the books is }₹20,\!160.} $$Question 21
Answer:
$$ \boxed{\text{Option (A): }k=3,\;p=n} $$