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