# Matrices Problems

## Matrices Problems

i) Inverse Matrix: If $$A\,=\,\left( \begin{matrix} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \\\end{matrix} \right)$$, then A⁻¹ is equal to

Solution: $$A\,=\,\left( \begin{matrix} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \\\end{matrix} \right)$$,

|A| = 1 and adj (A) = A

∴ $${{A}^{-1}}\,=\,\frac{adj\,\left( A \right)}{\left| A \right|}\,=\,\frac{A}{1}\,=\,A$$.

ii) Unity Matrix: Element of a matrix A of order 10 x 10 are defined as aij = ωi + j (Where, ω is cube root of unity), then tr (A) of the matrix is

Solution: tr (A) = (a₁₁ + a₂₂ + … + a₁₀ a₁₀)

= (ω² + ω⁴ + ω⁶ + … + ω²⁰)

= ω² (1 + ω² + ω⁴ + … + ω¹⁸)

= ω² ((1 + ω² + ω) + … + (1 + ω + ω²) + 1)

= ω² (0 + … + 0 + 1) = ω²

iii) Multiplication Matrix: If $$A\,=\,\left( \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\\end{matrix} \right)$$, then Aⁿ is equal to

Solution:  $$A\,=\,\left( \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\\end{matrix} \right)$$,

A² = A. A

= $$\left( \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\\end{matrix} \right)\left( \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\\end{matrix} \right)$$,

= $$\,\left( \begin{matrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \\\end{matrix} \right)$$,

Similarly, $$\,{{A}^{n}}\,=\,\left( \begin{matrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \\\end{matrix} \right)$$.

iv) Adjoint Matrix: If $$A\,=\,\left( \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\\end{matrix} \right)$$ and A (adj A) = λI, then the value of λ is

Solution: $$A\,=\,\left( \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\\end{matrix} \right)$$ and $$adj\,\left( A \right)\,=\,\left( \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\\end{matrix} \right)$$,

∴ $$A.\,\left( adj\,A \right)\,=\,\left( \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\\end{matrix} \right)\left( \begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\\end{matrix} \right)$$,

= $$\,\left( \begin{matrix} {{\cos }^{2}}\theta \,+\,{{\sin }^{2}}\theta & -\,\sin 2\theta \cos 2\theta \,+\,\sin 2\theta \cos 2\theta \, \\ -\,\sin 2\theta \cos 2\theta \,+\,\sin 2\theta \cos 2\theta & {{\sin }^{2}}\theta \,+\,{{\cos }^{2}}\theta \\\end{matrix} \right)$$,

= $$\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\\end{matrix} \right)$$,

λ = I.

v) Inverse Matrix: If $$A\,=\,\left( \begin{matrix} 3 & 2 \\ 0 & 1 \\\end{matrix} \right)$$, then (A⁻¹)³ is equal to

Solution: Here, |A| = 3, $$adj\,A\,=\,\left( \begin{matrix} 1 & -2 \\ 0 & 3 \\\end{matrix} \right)$$,

∴ $${{A}^{-1}}\,=\,\frac{1}{3}\,\left( \begin{matrix} 1 & -2 \\ 0 & 3 \\\end{matrix} \right)$$,

$${{\left( {{A}^{-1}} \right)}^{3}}\,=\,\frac{1}{27}\,{{\left( \begin{matrix} 1 & -2 \\ 0 & 3 \\\end{matrix} \right)}^{3}}$$,

$${{\left( {{A}^{-1}} \right)}^{3}}\,=\,\frac{1}{27}\,\left( \begin{matrix} 1 & -26 \\ 0 & 27 \\\end{matrix} \right)$$.

vi) Transpose Matrix: Find the values of x, y and z. If the matrix $$A\,=\,\left( \begin{matrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \\\end{matrix} \right)$$ satisfies the equation A’A = 1 are

Solution: Given A’A = I

$${{\left( \begin{matrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \\\end{matrix} \right)}^{‘}}\left( \begin{matrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \\\end{matrix} \right)\,=\,I$$,

$${{\left( \begin{matrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \\\end{matrix} \right)}^{‘}}\left( \begin{matrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \\\end{matrix} \right)\,=\,\left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\\end{matrix} \right)$$,

$$\left( \begin{matrix} 0\,+\,{{x}^{2}}\,+\,{{x}^{2}} & 0\,+\,xy\,-\,xy & 0\,-\,xz\,+\,xz \\ 0\,+\,yx\,-\,yx & 4{{y}^{2}}\,+\,{{y}^{2}}\,+\,{{y}^{2}} & 2yz\,-\,yz\,-\,yz \\ 0\,-\,zx\,+\,zx & 2yz\,-\,yz\,-\,yz & {{z}^{2}}\,+\,{{z}^{2}}\,+\,{{z}^{2}} \\\end{matrix} \right)\,=\,\left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\\end{matrix} \right)$$,

$$\left( \begin{matrix} 2{{x}^{2}} & 0 & 0 \\ 0 & 6{{y}^{2}} & 0 \\ 0 & 0 & 3{{z}^{2}} \\\end{matrix} \right)\,=\,\left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\\end{matrix} \right)$$,

On comparing the corresponding elements, we have

2x² = 1, 6y² = 1, 3z² = 1

x² = ½, y² = ⅙, z² = ⅓.

$$x\,=\,\pm \frac{1}{\sqrt{2}}$$, $$y\,=\,\pm \frac{1}{\sqrt{6}}$$, $$z\,=\,\pm \frac{1}{\sqrt{3}}$$.