If (i) A=\(\begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha \end{bmatrix}\),then verify that A'A=I
(ii) A= \(\begin{bmatrix} \sin\alpha & \cos\alpha\\ -\cos \alpha & \sin\alpha \end{bmatrix}\),then verify that A'A=I
(ii) A= \(\begin{bmatrix} \sin\alpha & \cos\alpha\\ -\cos \alpha & \sin\alpha \end{bmatrix}\),then verify that A'A=I
Solution and Explanation
(i) \(A=\) \(\begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha \end{bmatrix}\)
\(\therefore A'=\) \(\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}\)
A'A= \(\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}\) \(\begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha \end{bmatrix}\)
= \(\begin{bmatrix} (\cos\alpha) (cos\alpha) + (- \sin\alpha)( -\sin\alpha) & (\cos\alpha)(\sin\alpha)+(-\sin\alpha)(\cos\alpha)\\ (\sin\alpha)(\cos\alpha)+(\cos\alpha)(-\sin\alpha) & (\sin\alpha)(\sin\alpha)+(\cos\alpha)(\cos\alpha) \end{bmatrix}\)
= \(\begin{bmatrix} \cos^2α+\sin^2α & \sinα\cosα-\sinα\cosα\\ \ sinα\cosα-\sinα\cosα & \sin^2α+\cos^2α \end{bmatrix}\)
= \(\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}= I\)
Hence we verified that: A'A=I
(ii) \(\begin{bmatrix} \sin\alpha & \cos\alpha\\ -\cos\alpha & \sin\alpha \end{bmatrix}\)
so A'= \(\begin{bmatrix} \sin\alpha & -\cos\alpha\\ \cos\alpha & \sin\alpha \end{bmatrix}\)
A'A= \(\begin{bmatrix} \sin\alpha & -\cos\alpha\\ \cos\alpha & \sin\alpha \end{bmatrix}\)\(\begin{bmatrix} \sin\alpha & \cos\alpha\\ -\cos\alpha & \sin\alpha \end{bmatrix}\)
= \(\begin{bmatrix} (\sin\alpha)(\sin\alpha)+(-\cos\alpha)(-\cos\alpha) & (\sin\alpha)(\cos\alpha)+(-\cos\alpha)(\sin\alpha)\\ (\cos\alpha)(\sin\alpha)+(\sin\alpha)(-\cos\alpha) & (\cos\alpha)(\cos\alpha)+(\sin\alpha)(\sin\alpha) \end{bmatrix}\)
= \(\begin{bmatrix} \sin^2α\cos^2α & \sinα\cosα-\sin\alpha\cos\alpha & \\ \sinα\cosα-\sinα\cosα & \cos^2α+\sin^2α \end{bmatrix}\)
= \(\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}= I\)
Hence we verified that: \(A'A=I\)
Top Questions on Matrices
- Let \( A = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} \). The determinant of \( A^3 \) is:
- Let \[ A = \begin{pmatrix} 1 & 4 \\ -2 & 1 \end{pmatrix} \quad \text{and} \quad C = \begin{pmatrix} 3 & 4 & 2 \\ 12 & 16 & 8 \\ -6 & -8 & -4 \end{pmatrix}. \] Then, find the matrix $B$ if $AB = C$.
- The matrix $A = \begin{pmatrix} \sqrt{3} & 0 & 0 \\ 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{5} \end{pmatrix}$ is a/an :
- If \[ \begin{bmatrix} 3 & -1 \\ 0 & 1 \\ 2 & -3 \end{bmatrix} A = \begin{bmatrix} 2 & -5 \\ -17 \end{bmatrix} \] then find matrix \( A \).
- If \[ \begin{bmatrix} 2x-1 & 3x \\ 0 & y^2-1 \end{bmatrix} = \begin{bmatrix} x+3 & 12 \\ 0 & 35 \end{bmatrix} \] then the value of \( (x - y) \) is :
Questions Asked in CBSE CLASS XII exam
- What type of cell is mercury cell? Why is it more advantageous than dry cell?
- CBSE CLASS XII - 2025
- CBSE Compartment XII - 2025
- Electrochemistry
A compound (A) with molecular formula $C_4H_9I$ which is a primary alkyl halide, reacts with alcoholic KOH to give compound (B). Compound (B) reacts with HI to give (C) which is an isomer of (A). When (A) reacts with Na metal in the presence of dry ether, it gives a compound (D), C8H18, which is different from the compound formed when n-butyl iodide reacts with sodium. Write the structures of A, (B), (C) and (D) when (A) reacts with alcoholic KOH.
- CBSE CLASS XII - 2025
- Haloalkanes And Haloarenes
- State Faraday’s second law of electrolysis. How much electricity is required in terms of Faraday for the reduction of 1 mole of $Cr_2\text{O}_7^{2-} \text{ to Cr}^{3+}$?
- CBSE CLASS XII - 2025
- Electrochemistry
- Calculate the \(\Delta G\) of the following cell:
\[ \text{Mg}(s) | \text{Mg}^{2+} (aq) || \text{Cu}^{2+} (aq) | \text{Cu}(s) \] \[ \left[ E^\circ_{\text{Mg}^{2+}/\text{Mg}} = -2.37 \, \text{V}, E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \, \text{V}, F = 96500 \, \text{C mol}^{-1} \right] \]- CBSE CLASS XII - 2025
- Electrochemistry
- Calculate the emf of the following cell at 25°C : \[ \text{Zn(s)} | \text{Zn}^{2+} (0.1 \, \text{M}) \parallel \text{H}^+ (0.01 \, \text{M}) | \text{Hg}(l) | \text{Hg}_2^{2+} (1 \, \text{bar}), \, \text{Pt(s)} \] \[ \left[ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, \text{V}, \, E^\circ_{\text{H}_2/\text{H}^+} = 0.00 \, \text{V}, \, \log 10 = 1 \right] \]
- CBSE CLASS XII - 2025
- Electrochemistry
Concepts Used:
Transpose of a Matrix
The matrix acquired by interchanging the rows and columns of the parent matrix is called the Transpose matrix. The transpose matrix is also defined as - “A Matrix which is formed by transposing all the rows of a given matrix into columns and vice-versa.”
The transpose matrix of A is represented by A’. It can be better understood by the given example:
Now, in Matrix A, the number of rows was 4 and the number of columns was 3 but, on taking the transpose of A we acquired A’ having 3 rows and 4 columns. Consequently, the vertical Matrix gets converted into Horizontal Matrix.
Hence, we can say if the matrix before transposing was a vertical matrix, it will be transposed to a horizontal matrix and vice-versa.
Read More: Transpose of a Matrix