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Chapter 8

Matrices — Chapter Test

Class - 10 ML Aggarwal Understanding ICSE Mathematics

Chapter Test

Question 1

Find the values of a and b if,

[a+3b2+20−6]=[2a+13b0b2−5b].\begin{bmatrix*}[r] a + 3 & b^2 + 2 \\ 0 & -6 \end{bmatrix*} = \begin{bmatrix*}[r] 2a + 1 & 3b \\ 0 & b^2 - 5b \end{bmatrix*}.

Answer

By definition of equality of matrices we get,

⇒ a + 3 = 2a + 1 or a = 2

⇒ b2 + 2 = 3b     (...Eq 1)

⇒ b2 - 5b = -6     (...Eq 2)

Solving Eq 1 for b,

⇒ b2 + 2 = 3b
⇒ b2 - 3b + 2 = 0
⇒ b2 - 2b - b + 2 = 0
⇒ b(b - 2) - 1(b - 2) = 0
⇒ (b - 1)(b - 2) = 0
⇒ b = 1 or b = 2.

Checking whether the value of b = 1 satisfies Eq 2

⇒ b2 - 5b = -6

L.H.S. = b2 - 5b = (1)2 - 5(1) = -4.

L.H.S. ≠\neq R.H.S., so b = 1 is not the solution.

Checking whether the value of b = 2 satisfies Eq 2

⇒ b2 - 5b = -6

L.H.S. = b2 - 5b
= (2)2 - 5(2)
= 4 - 10
= -6 = R.H.S..

∴ a = 2 and b = 2.

Hence, the values are a = 2 and b = 2.

Question 2

Find a, b, c and d if 3[abcd]=[4a+bc+d3]+[a6−12d]3\begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} = \begin{bmatrix*}[r] 4 & a + b \\ c + d & 3 \end{bmatrix*} + \begin{bmatrix*}[r] a & 6 \\ -1 & 2d \end{bmatrix*}

Answer

Given,

3[abcd]=[4a+bc+d3]+[a6−12d]⇒[3a3b3c3d]=[4+aa+b+6c+d−13+2d]3\begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} = \begin{bmatrix*}[r] 4 & a + b \\ c + d & 3 \end{bmatrix*} + \begin{bmatrix*}[r] a & 6 \\ -1 & 2d \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 3a & 3b \\ 3c & 3d \end{bmatrix*} = \begin{bmatrix*}[r] 4 + a & a + b + 6 \\ c + d - 1 & 3 + 2d \end{bmatrix*} \\[0.5em]

By definition of equality of matrices

⇒ 3a = 4 + a     (...Eq 1)

⇒ 3b = a + b + 6     (...Eq 2)

⇒ 3c = c + d - 1     (...Eq 3)

⇒ 3d = 3 + 2d     (...Eq 4)

Solving Eq 1 we get,

⇒ 3a = 4 + a
⇒ 3a - a = 4
⇒ 2a = 4
⇒ a = 2.

Solving Eq 2 by putting value of a from Eq 1 we get,

⇒ 3b = a + b + 6
⇒ 3b = 2 + b + 6
⇒ 3b = b + 8
⇒ 3b - b = 8
⇒ 2b = 8
⇒ b = 4.

Solving Eq 4 we get,

⇒ 3d = 3 + 2d
⇒ 3d - 2d = 3
⇒ d = 3.

Solving Eq 3 by putting value of d from Eq 4 we get,

⇒ 3c = c + d - 1
⇒ 3c = c + 3 - 1
⇒ 3c - c = 2
⇒ 2c = 2
⇒ c = 1.

∴ a = 2, b = 4, c = 1 and d = 3.

Hence, the value of a = 2, b = 4, c = 1 and d = 3.

Question 3

Determine the matrices A and B when

A + 2B = [126−3] and 2A - B=[2−12−1].\begin{bmatrix*}[r] 1 & 2 \\ 6 & -3 \end{bmatrix*} \text{ and 2A - B} = \begin{bmatrix*}[r] 2 & -1 \\ 2 & -1 \end{bmatrix*}.

Answer

Given,

A + 2B=[126−3]....[Eq 1] 2A - B=[2−12−1]....[Eq 2] \text{A + 2B} = \begin{bmatrix*}[r] 1 & 2 \\ 6 & -3 \end{bmatrix*} \qquad \text{....[Eq 1] } \\[1em] \text{2A - B} = \begin{bmatrix*}[r] 2 & -1 \\ 2 & -1 \end{bmatrix*} \qquad \text{....[Eq 2] } \\[1em]

Multiplying Eq 1 by 2,

⇒2A+4B=[2412−6]\Rightarrow 2A + 4B = \begin{bmatrix*}[r] 2 & 4 \\ 12 & -6 \end{bmatrix*} \\[1em]

Subtracting Eq 2 from above equation we get,

⇒2A+4B−(2A−B)=[2412−6]−[2−12−1]⇒2A−2A+4B−(−B)=[2−24−(−1)12−2−6−(−1)]⇒5B=[0510−5]⇒B=15[0510−5]⇒B=[012−1].\Rightarrow 2A + 4B - (2A - B) = \begin{bmatrix*}[r] 2 & 4 \\ 12 & -6 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & -1 \\ 2 & -1 \end{bmatrix*} \\[1em] \Rightarrow 2A - 2A + 4B - (-B) = \begin{bmatrix*}[r] 2 - 2 & 4 - (-1) \\ 12 - 2 & -6 - (-1) \end{bmatrix*} \\[1em] \Rightarrow 5B = \begin{bmatrix*}[r] 0 & 5 \\ 10 & -5 \end{bmatrix*} \\[1em] \Rightarrow B = \dfrac{1}{5}\begin{bmatrix*}[r] 0 & 5 \\ 10 & -5 \end{bmatrix*} \\[1em] \Rightarrow B = \begin{bmatrix*}[r] 0 & 1 \\ 2 & -1 \end{bmatrix*}. \\[1em]

Putting value of matrix B in Eq 1,

⇒A+2[012−1]=[126−3]⇒A+[024−2]=[126−3]⇒A=[126−3]−[024−2]⇒A=[1−02−26−4−3−(−2)]⇒A=[102−1].∴A=[102−1] and B=[012−1].\Rightarrow A + 2 \begin{bmatrix*}[r] 0 & 1 \\ 2 & -1 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \\ 6 & -3 \end{bmatrix*} \\[1em] \Rightarrow A + \begin{bmatrix*}[r] 0 & 2 \\ 4 & -2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \\ 6 & -3 \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] 1 & 2 \\ 6 & -3 \end{bmatrix*} - \begin{bmatrix*}[r] 0 & 2 \\ 4 & -2 \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] 1 - 0 & 2 - 2 \\ 6 - 4 & -3 - (-2) \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] 1 & 0 \\ 2 & -1 \end{bmatrix*}. \\[1em] \therefore A = \begin{bmatrix*}[r] 1 & 0 \\ 2 & -1 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] 0 & 1 \\ 2 & -1 \end{bmatrix*}.

Hence, the value of A=[102−1]and B=[012−1].\text{A} = \begin{bmatrix*}[r] 1 & 0 \\ 2 & -1 \end{bmatrix*} \text{and B} = \begin{bmatrix*}[r] 0 & 1 \\ 2 & -1 \end{bmatrix*}.

Question 4(i)

Find the matrix B if A = [4123]\begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} and A2 = A + 2B.

Answer

Given, A2 = A + 2B.

⇒[4123][4123]=[4123]+2B⇒[4×4+1×24×1+1×32×4+3×22×1+3×3]=[4123]+2B⇒[16+24+38+62+9]=[4123]+2B⇒[1871411]=[4123]+2B⇒2B=[1871411]−[4123]⇒2B=[18−47−114−211−3]⇒2B=[146128]∴B=[7364].\Rightarrow \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} + 2B \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 4 \times 4 + 1 \times 2 & 4 \times 1 + 1 \times 3 \\ 2 \times 4 + 3 \times 2 & 2 \times 1 + 3 \times 3 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} + 2B \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 16 + 2 & 4 + 3 \\ 8 + 6 & 2 + 9 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} + 2B \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 18 & 7 \\ 14 & 11 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} + 2B \\[0.5em] \Rightarrow 2B = \begin{bmatrix*}[r] 18 & 7 \\ 14 & 11 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} \\[0.5em] \Rightarrow 2B = \begin{bmatrix*}[r] 18 - 4 & 7 - 1 \\ 14 - 2 & 11 - 3 \end{bmatrix*} \\[0.5em] \Rightarrow 2B = \begin{bmatrix*}[r] 14 & 6 \\ 12 & 8 \end{bmatrix*} \\[0.5em] \therefore B = \begin{bmatrix*}[r] 7 & 3 \\ 6 & 4 \end{bmatrix*}.

Hence, the matrix B = [7364].\begin{bmatrix*}[r] 7 & 3 \\ 6 & 4 \end{bmatrix*}.

Question 4(ii)

If A=[12−34],B=[01−25] and C =[−20−11],A = \begin{bmatrix*}[r] 1 & 2 \\ -3 & 4 \end{bmatrix*}, B = \begin{bmatrix*}[r] 0 & 1 \\ -2 & 5 \end{bmatrix*} \text{ and C } = \begin{bmatrix*}[r] -2 & 0 \\ -1 & 1 \end{bmatrix*}, find A(4B - 3C).

Answer

We need to find the value of A(4B - 3C).

A(4B - 3C)=[12−34](4[01−25]−3[−20−11])=[12−34]([04−820]−[−60−33])=[12−34][0−(−6)4−0−8−(−3)20−3]=[12−34][64−517]=[1×6+2×(−5)1×4+2×17−3×6+4×(−5)−3×4+4×17]=[6−104+34−18−20−12+68]=[−438−3856].\text{A(4B - 3C)} = \begin{bmatrix*}[r] 1 & 2 \\ -3 & 4 \end{bmatrix*} \Big(4\begin{bmatrix*}[r] 0 & 1 \\ -2 & 5 \end{bmatrix*} - 3\begin{bmatrix*}[r] -2 & 0 \\ -1 & 1 \end{bmatrix*}\Big) \\[0.5em] = \begin{bmatrix*}[r] 1 & 2 \\ -3 & 4 \end{bmatrix*} \Big(\begin{bmatrix*}[r] 0 & 4 \\ -8 & 20 \end{bmatrix*} - \begin{bmatrix*}[r] -6 & 0 \\ -3 & 3 \end{bmatrix*}\Big) \\[0.5em] = \begin{bmatrix*}[r] 1 & 2 \\ -3 & 4 \end{bmatrix*} \begin{bmatrix*}[r] 0 - (-6) & 4 - 0 \\ -8 - (-3) & 20 - 3 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 1 & 2 \\ -3 & 4 \end{bmatrix*} \begin{bmatrix*}[r] 6 & 4 \\ -5 & 17 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 1 \times 6 + 2 \times (-5) & 1 \times 4 + 2 \times 17 \\ -3 \times 6 + 4 \times (-5) & -3 \times 4 + 4 \times 17 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 6 - 10 & 4 + 34 \\ -18 - 20 & -12 + 68 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] -4 & 38 \\ -38 & 56 \end{bmatrix*}.

Hence, the value of A(4B - 3C) = [−438−3856].\begin{bmatrix*}[r] -4 & 38 \\ -38 & 56 \end{bmatrix*}.

Question 5

If A=[1324],B=[1224],C=[4115] and I=[1001]A = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}, B = \begin{bmatrix*}[r] 1 & 2 \\ 2 & 4 \end{bmatrix*}, C = \begin{bmatrix*}[r] 4 & 1 \\ 1 & 5 \end{bmatrix*} \text{ and } I = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}. Find A(B + C) - 14I.

Answer

B + C = [1224]+[4115]=[5339]\begin{bmatrix*}[r] 1 & 2 \\ 2 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 1 \\ 1 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 3 \\ 3 & 9 \end{bmatrix*}

Substituting values we get :

A(B+C)−14I=[1324][5339]−14[1001]=[1×5+3×31×3+3×92×5+4×32×3+4×9]−[140014]=[5+93+2710+126+36]−[140014]=[14302242]−[140014]=[14−1430−022−042−14]=[0302228].A(B + C) - 14I = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 5 & 3 \\ 3 & 9 \end{bmatrix*} - 14\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 5 + 3 \times 3 & 1 \times 3 + 3 \times 9 \\ 2 \times 5 + 4 \times 3 & 2 \times 3 + 4 \times 9 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 + 9 & 3 + 27 \\ 10 + 12 & 6 + 36 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 & 30 \\ 22 & 42 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 - 14 & 30 - 0 \\ 22 - 0 & 42 - 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 30 \\ 22 & 28 \end{bmatrix*}.

Hence, A(B + C) - 14I = [0302228].\begin{bmatrix*}[r] 0 & 30 \\ 22 & 28 \end{bmatrix*}.

Question 6

If A = [3205] and B =[1012],\begin{bmatrix*}[r] 3 & 2 \\ 0 & 5 \end{bmatrix*} \text{ and B } = \begin{bmatrix*}[r] 1 & 0 \\ 1 & 2 \end{bmatrix*}, find each of the following and state if they are equal :

(i) (A + B)(A - B)

(ii) A2 - B2

Answer

(i) We need to find the value of (A + B)(A - B)

(A+B)(A−B)=([3205]+[1012])([3205]−[1012])=[3+12+00+15+2][3−12−00−15−2]=[4217][22−13]=[4×2+2×(−1)4×2+2×31×2+7×(−1)1×2+7×3]=[8−28+62−72+21]=[614−523].(A + B)(A - B) = \Big(\begin{bmatrix*}[r] 3 & 2 \\ 0 & 5 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 0 \\ 1 & 2 \end{bmatrix*}\Big)\Big(\begin{bmatrix*}[r] 3 & 2 \\ 0 & 5 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 0 \\ 1 & 2 \end{bmatrix*}\Big) \\[0.5em] = \begin{bmatrix*}[r] 3 + 1 & 2 + 0 \\ 0 + 1 & 5 + 2 \end{bmatrix*} \begin{bmatrix*}[r] 3 - 1 & 2 - 0 \\ 0 - 1 & 5 - 2 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 4 & 2 \\ 1 & 7 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 2 \\ -1 & 3 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 4 \times 2 + 2 \times (-1) & 4 \times 2 + 2 \times 3 \\ 1 \times 2 + 7 \times (-1) & 1 \times 2 + 7 \times 3 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 8 - 2 & 8 + 6 \\ 2 - 7 & 2 + 21 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 6 & 14 \\ -5 & 23 \end{bmatrix*}.

Hence, the value of (A + B)(A - B) = [614−523]\begin{bmatrix*}[r] 6 & 14 \\ -5 & 23 \end{bmatrix*}.

(ii) We need to find the value of A2 - B2

A2−B2=[3205][3205]−[1012][1012]=[3×3+2×03×2+2×50×3+5×00×2+5×5]−[1×1+0×11×0+0×21×1+2×11×0+2×2]=[916025]−[1034]=[9−116−00−325−4]=[816−321].A^2 - B^2 = \begin{bmatrix*}[r] 3 & 2 \\ 0 & 5 \end{bmatrix*} \begin{bmatrix*}[r] 3 & 2 \\ 0 & 5 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 0 \\ 1 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 1 & 2 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 3 \times 3 + 2 \times 0 & 3 \times 2 + 2 \times 5 \\ 0 \times 3 + 5 \times 0 & 0 \times 2 + 5 \times 5 \end{bmatrix*} - \begin{bmatrix*}[r] 1 \times 1 + 0 \times 1 & 1 \times 0 + 0 \times 2 \\ 1 \times 1 + 2 \times 1 & 1 \times 0 + 2 \times 2 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 9 & 16 \\ 0 & 25 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 0 \\ 3 & 4 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 9 - 1 & 16 - 0 \\ 0 - 3 & 25 - 4 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 8 & 16 \\ -3 & 21 \end{bmatrix*} .

Hence, the value of A2−B2=[816−321] and (A + B)(A - B) ≠A2−B2.A^2 - B^2 = \begin{bmatrix*}[r] 8 & 16 \\ -3 & 21 \end{bmatrix*} \text{ and (A + B)(A - B) } \neq A^2 - B^2.

Question 7

If A = [3−5−42],\begin{bmatrix*}[r] 3 & -5 \\ -4 & 2 \end{bmatrix*}, find A2 - 5A - 14I, where I is unit matrix of order 2 × 2.

Answer

We need to find the value of A2 - 5A - 14I.

A2−5A−14I=[3−5−42][3−5−42]−5[3−5−42]−14[1001]=[3×3+(−5)×(−4)3×(−5)+(−5)×2(−4)×3+2×(−4)(−4)×(−5)+2×2]−[15−25−2010]−[140014]=[9+20−15−10−12−820+4]−[15−25−2010]−[140014]=[29−25−2024]−[15−25−2010]−[140014]=[29−15−14−25−(−25)−0−20−(−20)−024−10−14]=[29−29−25+25−20+2024−24]=[0000]A^2 - 5A -14I = \begin{bmatrix*}[r] 3 & -5 \\ -4 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 3 & -5 \\ -4 & 2 \end{bmatrix*} - 5\begin{bmatrix*}[r] 3 & -5 \\ -4 & 2 \end{bmatrix*} - 14\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 3 \times 3 + (-5) \times (-4) & 3 \times (-5) + (-5) \times 2 \\ (-4) \times 3 + 2 \times (-4) & (-4) \times (-5) + 2 \times 2 \end{bmatrix*} - \begin{bmatrix*}[r] 15 & -25 \\ -20 & 10 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 9 + 20 & -15 - 10 \\ -12 - 8 & 20 + 4 \end{bmatrix*} - \begin{bmatrix*}[r] 15 & -25 \\ -20 & 10 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 29 & -25 \\ -20 & 24 \end{bmatrix*} - \begin{bmatrix*}[r] 15 & -25 \\ -20 & 10 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 29 - 15 - 14 & -25 - (-25) - 0 \\ -20 - (-20) - 0 & 24 - 10 - 14 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 29 - 29 & -25 + 25 \\ -20 + 20 & 24 - 24 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[0.5em]

Hence, the value of A2−5A−14I=[0000].A^2 - 5A - 14I = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}.

Question 8

If A=[33pq] and A2=O,A = \begin{bmatrix*}[r] 3 & 3 \\ p & q \end{bmatrix*} \text{ and } A^2 = O, find p and q.

Answer

Given, A2 = O.

⇒[33pq][33pq]=[0000]⇒[3×3+3×p3×3+3×qp×3+q×pp×3+q×q]=[0000]⇒[9+3p9+3q3p+qp3p+q2]=[0000]\Rightarrow \begin{bmatrix*}[r] 3 & 3 \\ p & q \end{bmatrix*} \begin{bmatrix*}[r] 3 & 3 \\ p & q \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 3 \times 3 + 3 \times p & 3 \times 3 + 3 \times q \\ p \times 3 + q \times p & p \times 3 + q \times q \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 9 + 3p & 9 + 3q \\ 3p + qp & 3p + q^2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[0.5em]

By definition of equality of matrices we get,

⇒ 9 + 3p = 0 or 3p = -9 or p = -3

⇒ 9 + 3q = 0 or 3q = -9 or q = -3

⇒ 3p + qp = 0     (Eq 1)

⇒ 3p + q2 = 0     (Eq 2)

Checking whether p = -3 and q = -3 satisfy Eq 1,

⇒ 3p + qp = 0

L.H.S. = 3(-3) + (-3)(-3) = -9 + 9 = 0 = R.H.S.

Checking whether p = -3 and q = -3 satisfy Eq 2,

⇒ 3p + q2 = 0

L.H.S. = 3(-3) + (-3)2 = -9 + 9 = 0 = R.H.S.

Since, p = -3 and q = -3 satisfies Eq 1 and Eq 2,

∴ p = -3 and q = -3.

Hence, the values are p = -3 and q = -3.

Question 9

If [−1001][abcd]=[100−1],\begin{bmatrix*}[r] -1 & 0 \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*}, find a, b, c and d.

Answer

Given,

[−1001][abcd]=[100−1]⇒[−1×a+0×c−1×b+0×d0×a+1×c0×b+1×d]=[100−1]⇒[−a−bcd]=[100−1]\begin{bmatrix*}[r] -1 & 0 \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] -1 \times a + 0 \times c & -1 \times b + 0 \times d \\ 0 \times a + 1 \times c & 0 \times b + 1 \times d \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] -a & -b \\ c & d \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \\[0.5em]

By definition of equality of matrices we get,

⇒ -a = 1 or a = -1

⇒ -b = 0 or b = 0

⇒ c = 0

⇒ d = -1.

Hence, the value of a = -1, b = 0, c = 0 and d = -1.

Question 10

Find a and b if [a−bb−4b+4a−2][2002]=[−2−2140].\begin{bmatrix*}[r] a - b & b - 4 \\ b + 4 & a - 2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 0 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] -2 & -2 \\ 14 & 0 \end{bmatrix*}.

Answer

Given,

[a−bb−4b+4a−2][2002]=[−2−2140]⇒[(a−b)×2+(b−4)×0(a−b)×0+(b−4)×2(b+4)×2+(a−2)×0(b+4)×0+(a−2)×2]=[−2−2140]⇒[2a−2b2b−82b+82a−4]=[−2−2140]\begin{bmatrix*}[r] a - b & b - 4 \\ b + 4 & a - 2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 0 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] -2 & -2 \\ 14 & 0 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] (a - b) \times 2 + (b - 4) \times 0 & (a - b) \times 0 + (b - 4) \times 2 \\ (b + 4) \times 2 + (a - 2) \times 0 & (b + 4) \times 0 + (a - 2) \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] -2 & -2 \\ 14 & 0 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2a - 2b & 2b - 8 \\ 2b + 8 & 2a - 4 \end{bmatrix*} = \begin{bmatrix*}[r] -2 & -2 \\ 14 & 0 \end{bmatrix*} \\[0.5em]

By definition of equality of matrices we get,

⇒ 2a - 4 = 0 or 2a = 4 or a = 2

⇒ 2b - 8 = -2 or 2b = -2 + 8 = 6 or b = 3

⇒ 2a - 2b = -2     (Eq 1)

Checking whether a = 2 and b = 3 satisfies Eq 1,

⇒ 2a - 2b = -2

L.H.S. = 2a - 2b = 2(2) - 2(3) = 4 - 6 = -2 = R.H.S.

∴ a = 2 and b = 3.

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

Question 11

If A = [sec 60°cos 90°-3 tan 45°sin 90°] and B =[0cot 45°−23 sin 90°],\begin{bmatrix*}[r] \text{sec 60°} & \text{cos 90°} \\ \text{-3 tan 45°} & \text{sin 90°} \end{bmatrix*} \text{ and B } = \begin{bmatrix*}[r] 0 & \text{cot 45°} \\ -2 & \text{3 sin 90°} \end{bmatrix*}, find

(i) 2A - 3B

(ii) A2

(iii) BA

Answer

(i) Given,

A=[sec 60°cos 90°-3 tan 45°sin 90°]=[20−31]B=[0cot 45°−23 sin 90°]=[01−23]2A - 3B=2[20−31]−3[01−23]=[40−62]−[03−69]=[4−00−3−6−(−6)2−9]=[4−30−7].\text{A} = \begin{bmatrix*}[r] \text{sec 60°} & \text{cos 90°} \\ \text{-3 tan 45°} & \text{sin 90°} \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} \\[1em] \text{B} = \begin{bmatrix*}[r] 0 & \text{cot 45°} \\ -2 & \text{3 sin 90°} \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 1 \\ -2 & 3 \end{bmatrix*} \\[1em] \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*}.

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

(ii) Given,

A2=[20−31][20−31]=[2×2+0×(−3)2×0+0×1(−3)×2+1×(−3)(−3)×0+1×1]=[4+00+0−6−30+1]=[40−91].\text{A}^2 = \begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 2 \times 2 + 0 \times (-3) & 2 \times 0 + 0 \times 1 \\ (-3) \times 2 + 1 \times (-3) & (-3) \times 0 + 1 \times 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 4 + 0 & 0 + 0 \\ -6 - 3 & 0 + 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 4 & 0 \\ -9 & 1 \end{bmatrix*}.

Hence, the value of A2=[40−91].\text{A}^2 = \begin{bmatrix*}[r] 4 & 0 \\ -9 & 1 \end{bmatrix*}.

(iii)

BA =[01−23][20−31]=[0×2+1×(−3)0×0+1×1−2×2+3×(−3)−2×0+3×1]=[−31−133].\text{BA } = \begin{bmatrix*}[r] 0 & 1 \\ -2 & 3 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 0 \times 2 + 1 \times (-3) & 0 \times 0 + 1 \times 1 \\ -2 \times 2 + 3 \times (-3) & -2 \times 0 + 3 \times 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] -3 & 1 \\ -13 & 3 \end{bmatrix*}.

Hence, the value of matrix BA = [−31−133].\begin{bmatrix*}[r] -3 & 1 \\ -13 & 3 \end{bmatrix*}.

Question 12

Given matrix, X = [1183] and I=[1001],\begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*} \text{ and } I = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}, prove that X2 = 4X + 5I.

Answer

Given,

X2 = 4X + 5I

Solving for L.H.S.,

X2=[1183][1183]=[1×1+1×81×1+1×38×1+3×88×1+3×3]=[1+81+38+248+9]=[943217].X^2 = \begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 1 \times 8 & 1 \times 1 + 1\times 3 \\ 8 \times 1 + 3 \times 8 & 8 \times 1 + 3 \times 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 8 & 1 + 3 \\ 8 + 24 & 8 + 9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 4 \\ 32 & 17 \end{bmatrix*}.

Solving for R.H.S.,

4X+5I=4[1183]+5[1001]=[443212]+[5005]=[4+54+032+012+5]=[943217].4X + 5I = 4\begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*} + 5\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 4 \\ 32 & 12 \end{bmatrix*} + \begin{bmatrix*}[r] 5 & 0 \\ 0 & 5 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + 5 & 4 + 0 \\ 32 + 0 & 12 + 5 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 4 \\ 32 & 17 \end{bmatrix*}.

Since, L.H.S. = R.H.S.

Hence, proved that X2 = 4X + 5I.

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