Match List-I with List-II Choose the correct answer from the options given below:
List-1 List-II (A) If $\begin{bmatrix}\lambda-1 & 0 \\ 0 & \lambda-1 \end{bmatrix} $, then $\lambda$ is (I) 0 (B) If A=$ \begin{bmatrix}1 & 2 \\2 & 4 \end{bmatrix} $, then $\Delta$ is (II) 1 (C) If A = $ \begin{bmatrix}1 & 0 \\0 & \frac{1}{2} \end{bmatrix} $, then $|A^{-1}|$ is (III) -2 (D) If $ \begin{bmatrix}a+1 & 1 \\1 & 2 \end{bmatrix} = \begin{bmatrix}-1 & 1 \\1 & 2 \end{bmatrix} $, then a is (IV) 2
Matrix Lambda, Determinant, Inverse, Equality Matching
This solution provides a step-by-step explanation for matching List-I items with corresponding List-II values, focusing on matrix properties like the value of lambda, determinants, inverse determinants, and matrix equality.
Lambda Value Determination
For item (A), we are given a matrix structure involving $\lambda$: $\begin{bmatrix}\lambda-1 & 0 \\ 0 & \lambda-1 \end{bmatrix}$. The correct match suggests $\lambda = 2$. Let's verify this interpretation.
When $\lambda = 2$, the matrix becomes:
$ \begin{bmatrix} 2-1 & 0 \\ 0 & 2-1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $
This resulting matrix is the identity matrix ($I$). Therefore, the value $\lambda = 2$ (List-II item (II)) is associated with this matrix structure.
Determinant Calculation for Matrix
Item (B) presents matrix $A = \begin{bmatrix}1 & 2 \\ 2 & 4 \end{bmatrix}$ and asks for the value of $\Delta$. The symbol $\Delta$ typically represents the determinant of the matrix.
The determinant of a 2x2 matrix $\begin{bmatrix}p & q \\ r & s \end{bmatrix}$ is calculated as $ps - qr$.
For matrix A:
$ \Delta = |A| = (1 \times 4) - (2 \times 2) = 4 - 4 = 0 $
The determinant is 0. This corresponds to List-II item (I).
Inverse Determinant of Matrix
For item (C), we have the matrix $A = \begin{bmatrix}1 & 0 \\ 0 & \frac{1}{2} \end{bmatrix}$ and need to find $|A^{-1}|$, the determinant of its inverse.
First, calculate the determinant of A:
$ |A| = \left(1 \times \frac{1}{2}\right) - (0 \times 0) = \frac{1}{2} $
The property of determinants states that $|A^{-1}| = \frac{1}{|A|}$.
$ |A^{-1}| = \frac{1}{\frac{1}{2}} = 2 $
Thus, $|A^{-1}| = 2$, which matches List-II item (IV).
Matrix Equality and Variable 'a'
Item (D) involves the equality of two matrices:
$ \begin{bmatrix}a+1 & 1 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix}-1 & 1 \\ 1 & 2 \end{bmatrix} $
Matrix equality requires that all corresponding elements are identical.
Comparing the element at the first row, first column (position (1,1)) yields:
$ a+1 = -1 $
Solving for 'a':
$ a = -1 - 1 \\ a = -2 $
Therefore, $a = -2$. This matches List-II item (III).
List Matches Summary
Based on the calculations and reasoning above, the correct pairings are:
- (A) matches with (II)
- (B) matches with (I)
- (C) matches with (IV)
- (D) matches with (III)
This set of matches corresponds to the first option provided.
Important Questions from Algebra (Notes)
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In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If she/he attempts all 60 questions and secures 130 marks, the number of questions she/he attempts wrongly, are?
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If \[ \begin{vmatrix} \dfrac{1}{a(b+c)} & \dfrac{1}{b(c+a)} & \dfrac{1}{c(a+b)} \\ \dfrac{bc}{a(b+c)} & \dfrac{ca}{b(c+a)} & \dfrac{ab}{c(a+b)} \\ 1 & 1 & 1 \end{vmatrix} = k, \text{ then the value of } k \text{ is:} \]
