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CBSE 12th Mathematics - Matrices Four and Six Marks Questions

Published on: 30/09/2019

Matrices

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Questions:

Generic 5

9 x 5 = 45

If \(A=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \) Prove that , A =\(\left[ \begin{matrix} { 3 }^{ x-1 } & { 3 }^{ x-1 } & { 3 }^{ x-1 } \\ { 3 }^{ x-1 } & { 3 }^{ x-1 } & { 3 }^{ x-1 } \\ { 3 }^{ x-1 } & { 3 }^{ x-1 } & { 3 }^{ x-1 } \end{matrix} \right] \) for every positive integer n.

If, A = \(\left[ \begin{matrix} a & 0 \\ 1 & 1 \end{matrix} \right] \)and B = \(\left[ \begin{matrix} 1 & 0 \\ 5 & 1 \end{matrix} \right] \) find all those values of a for which A = B

Let f(x) = x² - 5x+6 find f(A) If, A =\(\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] \)

Using elementary column operations, find the inverse of the following matrix :
\(\left[ \begin{matrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix} \right] \)

If \(A=\left( \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right) \), then show that A³ - 4A² - 3A + 11I = 0

If \(A=\left( \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right) \), prove that A³ - 6A² + 7A + 2I = 0

If \(A=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\) and \(l=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), prove that \((al+bA)^{ 3 }={ a }^{ 3 }l+{ 3a }^{ 2 }bA.\)

If \(A=\left[ \begin{matrix} \cos { \alpha } & \sin { \alpha } \\ -\sin { \alpha } & \cos { \alpha } \end{matrix} \right] ,\) then show that \({ A }^{ 2 }=\begin{bmatrix} \cos { 2\alpha } & \sin { 2\alpha } \\ -\sin { 2\alpha } & \cos { 2\alpha } \end{bmatrix}\)

Using elementary transformations, find the inverse of the matrix

\(\left[ \begin{matrix} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{matrix} \right] \).

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Answers Key:

Generic 5

9 x 5 = 45

\({ A }^{ -1 }=\frac { 1 }{ 8 } \left[ \begin{matrix} 5 & -1 \\ -7 & 3 \end{matrix} \right] \)

No values of \(\alpha\) can be found for which A² = B is true

\(f(A)\left[ \begin{matrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{matrix} \right] \)

Let \(A=\left[ \begin{matrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix} \right] \)
A = AI
\(\therefore \left[ \begin{matrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix} \right] =A\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\({ C }_{ 1 }\leftrightarrow { C }_{ 2 }\)
\(\left[ \begin{matrix} 1 & -1 & 2 \\ 2 & 1 & 3 \\ 1 & 3 & 1 \end{matrix} \right] =A\left[ \begin{matrix} 0 & 1 & 0 \\1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\({ C }_{ 2 }\rightarrow { C }_{ 2 }+{ C }_{ 1 }\)
\({ C }_{ 3 }\rightarrow { C }_{ 3 }-{ 2C }_{ 1 }\)
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 2 & 3 & -1 \\ 1 & 4 & -1 \end{matrix} \right] =A\left[ \begin{matrix} 0 & 1 & 0 \\ 1 & 1 & -2 \\ 0 & 0 & 1 \end{matrix} \right] \)
\({ C }_{ 1 }\rightarrow { C }_{ 1 }+{ 2C }_{ 3 }\)
\({ C }_{ 2}\rightarrow { C }_{ 2 }+{ 2C }_{ 3 }\)
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ -1 & 2 & -1 \end{matrix} \right] =A\left[ \begin{matrix} 0 & 1 & 0 \\ -3 & -3 & -2 \\ 2 & 2 & 1 \end{matrix} \right] \)
\({ C }_{ 3}\rightarrow { C }_{ 3 }+{ C }_{ 2 }\)
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 2 & 1 \end{matrix} \right] =A\left[ \begin{matrix} 0 & 1 & 1 \\ -3 & -3 & -5 \\ 2 & 2 & 3 \end{matrix} \right] \)
\({ C }_{ 1 }\rightarrow { C }_{ 1 }+{ C }_{ 3 }\)
\({ C }_{ 2 }\rightarrow { C }_{ 2 }-{ 2C }_{ 3 }\)
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] =A\left[ \begin{matrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{matrix} \right] \)
\(\Rightarrow { A }^{ -1 }=\left[ \begin{matrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{matrix} \right] \)

Getting \({ A }^{ 2 }=\left( \begin{matrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{matrix} \right) \)
and \({ A }^{ 3 }=\left( \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right) \)
\(\therefore\) A³ - 4A² - 3A + 11I
\(=\left( \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right) -\left( \begin{matrix} 36 & 28 & 20 \\ 4 & 16 & 4 \\ 32 & 36 & 36 \end{matrix} \right) -\left( \begin{matrix} 3 & 9 & 6 \\ 6 & 0 & -3 \\ 3 & 6 & 9 \end{matrix} \right) +\left( \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) =0\)
Alternative Method :
\(A=\left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] \)
\({ A }^{ 2 }=\left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] \left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{matrix} \right] \)
\({ A }^{ 3 }=\left[ \begin{matrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{matrix} \right] \left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right] \)
\(\therefore\) A³ - 4A² - 4A + 11A
\(=\left[ \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right] -4\left[ \begin{matrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{matrix} \right] -3\left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] +\left[ \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right] -\left[ \begin{matrix} 36 & 28 & 20 \\ 4 & 16 & 4 \\ 32 & 36 & 36 \end{matrix} \right] -\left[ \begin{matrix} 3 & 9 & 6 \\ 6 & 0 & -3 \\ 3 & 6 & 9 \end{matrix} \right] +\left[ \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
= 0

A² = \(\left( \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right) \); A³ = \(\left( \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right) \)
L.H.S = A³ - 6A² + 7A + 2I
\(=\left( \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right) -\left( \begin{matrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{matrix} \right) +\left( \begin{matrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{matrix} \right) +\left( \begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) =0\)
= RHS

\(\mathrm{LHS}=(a I+b A)^{3}=\left\{\left[\begin{array}{ll} a & 0 \\ 0 & a \end{array}\right]+\left[\begin{array}{ll} 0 & b \\ 0 & 0 \end{array}\right]\right\}^{3} \)
\(=\left[\left.\begin{array}{ll} a & b \\ 0 & a \end{array}\right|^{3}=\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\right. \)
\(=\left[\begin{array}{cc} a^{2} & 2 a b \\ 0 & a^{2} \end{array}\right]\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]=\left[\begin{array}{cc} a^{3}+0 & a^{2} b+2 a^{2} b \\ 0+0 & 0+a^{3} \end{array}\right]=\left[\begin{array}{cc} a^{3} & 3 a^{2} b \\ 0 & a^{3} \end{array}\right] \)
\(\mathrm{RHS}=a^{3} I+3 a^{2} b A=\left[\begin{array}{cc} a^{3} & 0 \\ 0 & a^{3} \end{array}\right]+\left[\begin{array}{cc} 0 & 3 a^{2} b \\ 0 & 0 \end{array}\right]=\left[\begin{array}{cc} a^{3} & 3 a^{2} b \\ 0 & a^{3} \end{array}\right] \)
\(\mathrm{LHS}=\mathrm{RHS} \)
\(\text {Hence, }(a I+b l)^{3}=a^{3} I+3 a^{2} b A \)

\(\begin{bmatrix} \cos { 2\alpha } & \sin { 2\alpha } \\ -\sin { 2\alpha } & \cos { 2\alpha } \end{bmatrix}\)

Use A=lA. Proceed. [Refer 2]

\(\left[ \begin{matrix} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{matrix} \right] \)