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Chapter 8

Matrices — Exercise 8.2

Class - 10 ML Aggarwal Understanding ICSE Mathematics

Exercise 8.2

Question 1

Given that M=[2012] and N =[20−12]\text{M} = \begin{bmatrix*}[r] 2 & 0 \\ 1 & 2 \end{bmatrix*} \text{ and N }= \begin{bmatrix*}[r] 2 & 0 \\ -1 & 2 \end{bmatrix*}, find M + 2N.

Answer

M + 2N =[2012]+2[20−12]=[2012]+[40−24]=[2+40+01−22+4]=[60−16]\text{M + 2N }= \begin{bmatrix*}[r] 2 & 0 \\ 1 & 2 \end{bmatrix*} + 2\begin{bmatrix*}[r] 2 & 0 \\ -1 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 0 \\ 1 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 0 \\ -2 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 + 4 & 0 + 0 \\ 1 - 2 & 2 + 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 & 0 \\ -1 & 6 \end{bmatrix*} \\[1em]

Hence, the matrix M + 2N = [60−16]\begin{bmatrix*}[r] 6 & 0 \\ -1 & 6 \end{bmatrix*}.

Question 2

If A =[20−31] and B =[01−23]\text{If A }= \begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} \text{ and B }= \begin{bmatrix*}[r] 0 & 1 \\ -2 & 3 \end{bmatrix*}, find 2A - 3B.

Answer

2A - 3B =2[20−31]−3[01−23]=[40−62]−[03−69]=[4+00−3−6−(−6)2−9]=[4−30−7]\text{2A - 3B }= 2\begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} - 3\begin{bmatrix*}[r] 0 & 1 \\ -2 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 0 \\ -6 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 0 & 3 \\ -6 & 9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + 0 & 0 - 3 \\ -6 - (-6) & 2 - 9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & -3 \\ 0 & -7 \end{bmatrix*} \\[1em]

Hence, the matrix 2A - 3B = [4−30−7].\begin{bmatrix*}[r] 4 & -3 \\ 0 & -7 \end{bmatrix*}.

Question 3

Simplify : sin A[sin A−cos Acos Asin A]+cos A[cos Asin A−sin Acos A]\text{sin A}\begin{bmatrix*}[r] \text{sin A} & -\text{cos A} \\ \text{cos A} & \text{sin A} \end{bmatrix*} + \text{cos A}\begin{bmatrix*}[r] \text{cos A} & \text{sin A} \\ -\text{sin A} & \text{cos A} \end{bmatrix*}.

Answer

Given,

⇒sin A[sin A−cos Acos Asin A]+cos A[cos Asin A−sin Acos A]⇒[sin2 A−sin A.cos Asin A.cos Asin2 A]+[cos2 Acos A.sin A−cos A.sin Acos2 A]=[sin2 A+cos2 A−sin A.cos A+cos A.sin Asin A.cos A−cos A.sin Asin2 A+cos2 A]∵sin2 A+cos2 A=1=[1001]\Rightarrow \text{sin A}\begin{bmatrix*}[r] \text{sin A} & -\text{cos A} \\ \text{cos A} & \text{sin A} \end{bmatrix*} + \text{cos A}\begin{bmatrix*}[r] \text{cos A} & \text{sin A} \\ -\text{sin A} & \text{cos A} \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] \text{sin}^2\text{ A} & -\text{sin A}.\text{cos A} \\ \text{sin A}.\text{cos A} & \text{sin}^2\text{ A} \end{bmatrix*} + \begin{bmatrix*}[r] \text{cos}^2\text{ A} & \text{cos A}.\text{sin A} \\ -\text{cos A}.\text{sin A} & \text{cos}^2\text{ A} \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] \text{sin}^2\text{ A} + \text{cos}^2\text{ A} & -\text{sin A}.\text{cos A} + \text{cos A}.\text{sin A} \\ \text{sin A}.\text{cos A} - \text{cos A}.\text{sin A} & \text{sin}^2\text{ A} + \text{cos}^2\text{ A} \end{bmatrix*} \\[1em] \because \text{sin}^2\text{ A} + \text{cos}^2\text{ A} = 1 \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em]

Hence, on simplifying, the resultant matrix = [1001].\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}.

Question 4

If A =[12−23], B =[−2−112] and C=[032−1]\text{If A } = \begin{bmatrix*}[r] 1 & 2 \\ -2 & 3 \end{bmatrix*}, \text{ B } = \begin{bmatrix*}[r] -2 & -1 \\ 1 & 2 \end{bmatrix*} \text{ and C} = \begin{bmatrix*}[r] 0 & 3 \\ 2 & -1 \end{bmatrix*}, find A + 2B - 3C.

Answer

A + 2B - 3C =[12−23]+2[−2−112]−3[032−1]=[12−23]+[−4−224]−[096−3]=[1+(−4)−02+(−2)−9−2+2−63+4−(−3)]=[−3−9−610]\text{A + 2B - 3C }= \begin{bmatrix*}[r] 1 & 2 \\ -2 & 3 \end{bmatrix*} + 2\begin{bmatrix*}[r] -2 & -1 \\ 1 & 2 \end{bmatrix*} - 3\begin{bmatrix*}[r] 0 & 3 \\ 2 & -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 2 \\ -2 & 3 \end{bmatrix*} + \begin{bmatrix*}[r] -4 & -2 \\ 2 & 4 \end{bmatrix*} - \begin{bmatrix*}[r] 0 & 9 \\ 6 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + (-4) - 0 & 2 + (-2) - 9 \\ -2 + 2 - 6 & 3 + 4 - (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 & -9 \\ -6 & 10 \end{bmatrix*}

Hence, the matrix A + 2B - 3C = [−3−9−610].\begin{bmatrix*}[r] -3 & -9 \\ -6 & 10 \end{bmatrix*}.

Question 5

If A = [0−112] and B=[12−11]\begin{bmatrix*}[r] 0 & -1 \\ 1 & 2 \end{bmatrix*} \text{ and B} = \begin{bmatrix*}[r] 1 & 2 \\ -1 & 1 \end{bmatrix*}, find the matrix X if

(i) 3A + X = B

(ii) X - 3B = 2A.

Answer

(i) 3A + X = B

⇒ X = B - 3A

⇒X=[12−11]−3[0−112]=[12−11]−[0−336]=[1−02−(−3)−1−31−6]=[15−4−5]\Rightarrow X = \begin{bmatrix*}[r] 1 & 2 \\ -1 & 1 \end{bmatrix*} - 3\begin{bmatrix*}[r] 0 & -1 \\ 1 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 2 \\ -1 & 1 \end{bmatrix*} - \begin{bmatrix*}[r] 0 & -3 \\ 3 & 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 - 0 & 2 - (-3) \\ -1 - 3 & 1 - 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 5 \\ -4 & -5 \end{bmatrix*}

Hence, matrix X = [15−4−5].\begin{bmatrix*}[r] 1 & 5 \\ -4 & -5 \end{bmatrix*}.

(ii) X - 3B = 2A

⇒ X = 2A + 3B

X=2[0−112]+3[12−11]=[0−224]+[36−33]=[0+3−2+62+(−3)4+3]=[34−17]X = 2\begin{bmatrix*}[r] 0 & -1 \\ 1 & 2 \end{bmatrix*} + 3\begin{bmatrix*}[r] 1 & 2 \\ -1 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & -2 \\ 2 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] 3 & 6 \\ -3 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 + 3 & -2 + 6 \\ 2 + (-3) & 4 + 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 3 & 4 \\ -1 & 7 \end{bmatrix*}

Hence, matrix X = [34−17].\begin{bmatrix*}[r] 3 & 4 \\ -1 & 7 \end{bmatrix*}.

Question 6

Solve the matrix equation [2150]−3X=[−7426]\begin{bmatrix} 2 & 1 \\ 5 & 0 \end{bmatrix} - 3X = \begin{bmatrix*}[r] -7 & 4 \\ 2 & 6 \end{bmatrix*}

Answer

Given,

⇒[2150]−3X=[−7426]⇒3X=[2150]−[−7426]⇒3X=[2−(−7)1−45−20−6]⇒3X=[9−33−6]⇒X=13[9−33−6]⇒X=[3−11−2]\Rightarrow \begin{bmatrix} 2 & 1 \\ 5 & 0 \end{bmatrix} - 3X = \begin{bmatrix*}[r] -7 & 4 \\ 2 & 6 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix} 2 & 1 \\ 5 & 0 \end{bmatrix} - \begin{bmatrix*}[r] -7 & 4 \\ 2 & 6 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix*}[r] 2 - (-7) & 1 - 4 \\ 5 - 2 & 0 - 6 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix*}[r] 9 & -3 \\ 3 & -6 \end{bmatrix*} \\[1em] \Rightarrow X = \dfrac{1}{3}\begin{bmatrix*}[r] 9 & -3 \\ 3 & -6 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] 3 & -1 \\ 1 & -2 \end{bmatrix*}

Hence, matrix X = [3−11−2].\begin{bmatrix*}[r] 3 & -1 \\ 1 & -2 \end{bmatrix*} .

Question 7

If [14−23]+2M=3[320−3],\begin{bmatrix*}[r] 1 & 4 \\ -2 & 3 \end{bmatrix*} + 2\text{M} = 3\begin{bmatrix*}[r] 3 & 2 \\ 0 & -3 \end{bmatrix*}, \\[1em] find the matrix M.

Answer

Given,

⇒[14−23]+2M=3[320−3]⇒[14−23]+2M=[960−9]⇒2M=[960−9]−[14−23]⇒2M=[9−16−40−(−2)−9−3]⇒M=12[822−12]⇒M=[411−6]\Rightarrow \begin{bmatrix*}[r] 1 & 4 \\ -2 & 3 \end{bmatrix*} + 2\text{M} = 3\begin{bmatrix*}[r] 3 & 2 \\ 0 & -3 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 & 4 \\ -2 & 3 \end{bmatrix*} + 2\text{M} = \begin{bmatrix*}[r] 9 & 6 \\ 0 & -9 \end{bmatrix*} \\[1em] \Rightarrow 2\text{M} = \begin{bmatrix*}[r] 9 & 6 \\ 0 & -9 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 4 \\ -2 & 3 \end{bmatrix*} \\[1em] \Rightarrow 2\text{M} = \begin{bmatrix*}[r] 9 - 1 & 6 - 4 \\ 0 - (-2) & -9 - 3 \end{bmatrix*} \\[1em] \Rightarrow \text{M} = \dfrac{1}{2}\begin{bmatrix*}[r] 8 & 2 \\ 2 & -12 \end{bmatrix*} \\[1em] \Rightarrow \text{M} = \begin{bmatrix*}[r] 4 & 1 \\ 1 & -6 \end{bmatrix*} \\[1em]

Hence, matrix M = [411−6].\begin{bmatrix*}[r] 4 & 1 \\ 1 & -6 \end{bmatrix*}.

Question 8

Given A=[2−620],B=[−3240] and C=[4002].A = \begin{bmatrix*}[r] 2 & -6 \\ 2 & 0 \end{bmatrix*}, B = \begin{bmatrix*}[r] -3 & 2 \\ 4 & 0 \end{bmatrix*} \text{ and } C = \begin{bmatrix*}[r] 4 & 0 \\ 0 & 2 \end{bmatrix*}.

Find the matrix X such that A + 2X = 2B + C.

Answer

Putting values of A, B and C in A + 2X = 2B + C,

⇒[2−620]+2X=2[−3240]+[4002]⇒2X=[−6480]+[4002]−[2−620]⇒2X=[−6+4−24+0−(−6)8+0−20+2−0]⇒X=12[−41062]⇒X=[−2531]\Rightarrow \begin{bmatrix*}[r] 2 & -6 \\ 2 & 0 \end{bmatrix*} + 2X = 2\begin{bmatrix*}[r] -3 & 2 \\ 4 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 0 \\ 0 & 2 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] -6 & 4 \\ 8 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 0 \\ 0 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & -6 \\ 2 & 0 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] -6 + 4 - 2 & 4 + 0 - (-6) \\ 8 + 0 - 2 & 0 + 2 - 0 \end{bmatrix*} \\[1em] \Rightarrow X = \dfrac{1}{2}\begin{bmatrix*}[r] -4 & 10 \\ 6 & 2 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -2 & 5 \\ 3 & 1 \end{bmatrix*}

Hence, matrix X = [−2531]\begin{bmatrix*}[r] -2 & 5 \\ 3 & 1 \end{bmatrix*}.

Question 9

Find X and Y if

X + Y=[7025] and X - Y=[3003].\text{X + Y} = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} \text{ and X - Y} = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}.

Answer

Given,

X + Y = [7025]\begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}        (...Eq1)

X - Y = [3003]\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}         (...Eq2)

Adding both the equations,

⇒X+Y+X−Y=[7025]+[3003]⇒2X=[7+30+02+05+3]⇒X=12[10028]⇒X=[5014]\Rightarrow X + Y + X - Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \\[1em] \Rightarrow 2X = \begin{bmatrix} 7 + 3 & 0 + 0 \\ 2 + 0 & 5 + 3 \end{bmatrix} \\[1em] \Rightarrow X = \dfrac{1}{2}\begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \\[1em]

From Eq 1 we get,

⇒Y=[7025]−X=[7025]−[5014]=[7−50−02−15−4]=[2011]∴X=[5014],Y=[2011].\Rightarrow Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} - X \\[1em] = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} - \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 - 5 & 0 - 0 \\ 2 - 1 & 5 - 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \\[1.5em] \therefore X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} , Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}.

Hence, X=[5014] and Y=[2011].X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \text{ and } Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}.

Question 10

If 2[x79y−5]+[6−745]=[1072215],2\begin{bmatrix*}[r] x & 7 \\ 9 & y - 5 \end{bmatrix*} + \begin{bmatrix*}[r] 6 & -7 \\ 4 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 7 \\ 22 & 15 \end{bmatrix*}, find the values of x and y.

Answer

Given,

⇒2[x79y−5]+[6−745]=[1072215]⇒[2x14182y−10]+[6−745]=[1072215]⇒[2x+614+(−7)18+42y−10+5]=[1072215]⇒2x+6=10 and 2y−5=15⇒2x=4 and 2y=20\Rightarrow 2\begin{bmatrix*}[r] x & 7 \\ 9 & y - 5 \end{bmatrix*} + \begin{bmatrix*}[r] 6 & -7 \\ 4 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 7 \\ 22 & 15 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x & 14 \\ 18 & 2y - 10 \end{bmatrix*} + \begin{bmatrix*}[r] 6 & -7 \\ 4 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 7 \\ 22 & 15 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + 6 & 14 + (-7) \\ 18 + 4 & 2y - 10 + 5 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 7 \\ 22 & 15 \end{bmatrix*} \\[1em] \Rightarrow 2x + 6 = 10 \text{ and } 2y - 5 = 15 \\[1em] \Rightarrow 2x = 4 \text{ and } 2y = 20 \\[1em]

∴ x = 2 and y = 10.

Hence, the value of x = 2 and y = 10.

Question 11

If 2[345x]+[1y01]=[z0105],2\begin{bmatrix*}[r] 3 & 4 \\ 5 & x \end{bmatrix*} + \begin{bmatrix*}[r] 1 & y \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] z & 0 \\ 10 & 5 \end{bmatrix*}, find the values of x, y and z.

Answer

Given,

⇒2[345x]+[1y01]=[z0105]⇒[68102x]+[1y01]=[z0105]⇒[6+18+y10+02x+1]=[z0105]⇒[78+y102x+1]=[z0105]⇒7=z,8+y=0 and 2x+1=5⇒z=7,y=−8 and 2x=4⇒z=7,y=−8 and x=2.\Rightarrow 2\begin{bmatrix*}[r] 3 & 4 \\ 5 & x \end{bmatrix*} + \begin{bmatrix*}[r] 1 & y \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] z & 0 \\ 10 & 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 & 8 \\ 10 & 2x \end{bmatrix*} + \begin{bmatrix*}[r] 1 & y \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] z & 0 \\ 10 & 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 + 1 & 8 + y \\ 10 + 0 & 2x + 1 \end{bmatrix*} = \begin{bmatrix*}[r] z & 0 \\ 10 & 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 7 & 8 + y \\ 10 & 2x + 1 \end{bmatrix*} = \begin{bmatrix*}[r] z & 0 \\ 10 & 5 \end{bmatrix*} \\[1em] \Rightarrow 7 = z, 8 + y = 0 \text{ and } 2x + 1 = 5 \\[0.5em] \Rightarrow z = 7, y = -8 \text{ and } 2x = 4 \\[0.5em] \Rightarrow z = 7, y = -8 \text{ and } x = 2.

∴ x = 2, y = -8 and z = 7.

Hence, the values are x = 2, y = -8 and z = 7.

Question 12

If [52−1y+1]−2[12x−13−2]=[3−8−72],\begin{bmatrix*}[r] 5 & 2 \\ -1 & y + 1 \end{bmatrix*} - 2\begin{bmatrix*}[r] 1 & 2x - 1 \\ 3 & -2 \end{bmatrix*} = \begin{bmatrix*}[r] 3 & -8 \\ -7 & 2 \end{bmatrix*}, find the values of x and y.

Answer

Given,

⇒[52−1y+1]−2[12x−13−2]=[3−8−72]⇒[52−1y+1]−[24x−26−4]=[3−8−72]⇒[5−22−4x+2−1−6y+1−(−4)]=[3−8−72]⇒[34−4x−7y+5]=[3−8−72]⇒4−4x=−8 and y+5=2⇒4x=8+4 and y=2−5⇒4x=12 and y=−3\Rightarrow \begin{bmatrix*}[r] 5 & 2 \\ -1 & y + 1 \end{bmatrix*} - 2\begin{bmatrix*}[r] 1 & 2x - 1 \\ 3 & -2 \end{bmatrix*} = \begin{bmatrix*}[r] 3 & -8 \\ -7 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 5 & 2 \\ -1 & y + 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 4x - 2 \\ 6 & -4 \end{bmatrix*} = \begin{bmatrix*}[r] 3 & -8 \\ -7 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 5 - 2 & 2 - 4x + 2 \\ -1 - 6 & y + 1 - (-4) \end{bmatrix*} = \begin{bmatrix*}[r] 3 & -8 \\ -7 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 & 4 - 4x \\ -7 & y + 5 \end{bmatrix*} = \begin{bmatrix*}[r] 3 & -8 \\ -7 & 2 \end{bmatrix*} \\[1em] \Rightarrow 4 - 4x = -8 \text{ and } y + 5 = 2 \\[0.5em] \Rightarrow 4x = 8 + 4 \text{ and } y = 2 - 5 \\[0.5em] \Rightarrow 4x = 12 \text{ and } y = -3 \\[0.5em]

∴ x = 3 and y = -3.

Hence, the value of x = 3 and y = -3.

Question 13

If [a342]+[2b1−2]−[11−2c]=[5073],\begin{bmatrix*}[r] a & 3 \\ 4 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] 2 & b \\ 1 & -2 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 1 \\ -2 & c \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 0 \\ 7 & 3 \end{bmatrix*}, find the values of a, b and c.

Answer

Given,

⇒[a342]+[2b1−2]−[11−2c]=[5073]⇒[a+2−13+b−14+1−(−2)2+(−2)−c]=[5073]⇒[a+1b+27−c]=[5073]⇒a+1=5,b+2=0 and −c=3\Rightarrow \begin{bmatrix*}[r] a & 3 \\ 4 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] 2 & b \\ 1 & -2 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 1 \\ -2 & c \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 0 \\ 7 & 3 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a + 2 - 1 & 3 + b - 1 \\ 4 + 1 -(-2) & 2 + (-2) - c \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 0 \\ 7 & 3 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a + 1 & b + 2 \\ 7 & -c \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 0 \\ 7 & 3 \end{bmatrix*} \\[1em] \Rightarrow a + 1 = 5, b + 2 = 0 \text{ and } -c = 3 \\[1em]

∴ a = 4, b = -2 and c = -3.

Hence, the values are a = 4, b = -2 and c = -3.

Question 14

If A=[2a−35],B=[−237b],C=[c9−1−11]A = \begin{bmatrix*}[r] 2 & a \\ -3 & 5 \end{bmatrix*}, B = \begin{bmatrix*}[r] -2 & 3 \\ 7 & b \end{bmatrix*}, C = \begin{bmatrix*}[r] c & 9 \\ -1 & -11 \end{bmatrix*} and 5A + 2B = C, find the values of a, b and c.

Answer

Given, 5A + 2B = C

⇒5[2a−35]+2[−237b]=[c9−1−11]⇒[105a−1525]+[−46142b]=[c9−1−11]⇒[10+(−4)5a+6−15+1425+2b]=[c9−1−11]⇒[65a+6−125+2b]=[c9−1−11]⇒6=c,5a+6=9 and 25+2b=−11⇒c=6,5a=3 and 2b=−11−25⇒c=6,a=35 and 2b=−36⇒c=6,a=35 and b=−18∴a=35,b=−18 and c=6.\Rightarrow 5\begin{bmatrix*}[r] 2 & a \\ -3 & 5 \end{bmatrix*} + 2\begin{bmatrix*}[r] -2 & 3 \\ 7 & b \end{bmatrix*} = \begin{bmatrix*}[r] c & 9 \\ -1 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 10 & 5a \\ -15 & 25 \end{bmatrix*} + \begin{bmatrix*}[r] -4 & 6 \\ 14 & 2b \end{bmatrix*} = \begin{bmatrix*}[r] c & 9 \\ -1 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 10 + (-4) & 5a + 6 \\ -15 + 14 & 25 + 2b \end{bmatrix*} = \begin{bmatrix*}[r] c & 9 \\ -1 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 & 5a + 6 \\ -1 & 25 + 2b \end{bmatrix*} = \begin{bmatrix*}[r] c & 9 \\ -1 & -11 \end{bmatrix*} \\[1em] \Rightarrow 6 = c, 5a + 6 = 9 \text{ and } 25 + 2b = -11 \\[0.5em] \Rightarrow c = 6, 5a = 3 \text{ and } 2b = -11 - 25 \\[0.5em] \Rightarrow c = 6, a = \dfrac{3}{5} \text{ and } 2b = -36 \\[0.5em] \Rightarrow c = 6, a = \dfrac{3}{5} \text{ and } b = -18 \\[0.5em] \therefore a = \dfrac{3}{5}, b = -18 \text{ and } c = 6.

Hence, the values are a = 35,\bold{\dfrac{3}{5}}, b = -18 and c = 6.

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