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

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

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

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

(adj A) = I

(adj A) = λ I

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.

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