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Matrices

Multiple Choice Questions

151.

The multiplicative inverse of A = $[\begin{matrix} \cos (θ) & - \sin (θ) \\ \sin (θ) & \cos (θ) \end{matrix}]$ is

$[\begin{matrix} - \cos (θ) & \sin (θ) \\ - \sin (θ) & - \cos (θ) \end{matrix}]$
$[\begin{matrix} \cos (θ) & \sin (θ) \\ - \sin (θ) & \cos (θ) \end{matrix}]$
$[\begin{matrix} - \cos (θ) & - \sin (θ) \\ \sin (θ) & - \cos (θ) \end{matrix}]$
$[\begin{matrix} \cos (θ) & \sin (θ) \\ \sin (θ) & - \cos (θ) \end{matrix}]$

152.

If A = $[\begin{matrix} 1 & 1 & 0 \\ 2 & 1 & 5 \\ 1 & 2 & 1 \end{matrix}]$ , then a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃ is equal to

153.

If A = $[\begin{matrix} 2 & 2 \\ - 3 & 2 \end{matrix}], B = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}]$ , then (B^{- 1}A^{- 1})^{- 1} is equal to

$[\begin{matrix} 2 & - 2 \\ 2 & 3 \end{matrix}]$
$[\begin{matrix} 2 & 2 \\ - 2 & 3 \end{matrix}]$
$[\begin{matrix} 2 & - 3 \\ 2 & 2 \end{matrix}]$
$[\begin{matrix} 1 & - 1 \\ - 2 & 3 \end{matrix}]$

154.

If matrix A = $[\begin{matrix} 1 & 2 \\ 4 & 3 \end{matrix}]$ such that AX = $ϕ$ ,then X is equal to

$\frac{1}{5} [\begin{matrix} 1 & 3 \\ 2 & - 1 \end{matrix}]$
$\frac{1}{5} [\begin{matrix} 4 & 2 \\ 4 & - 1 \end{matrix}]$
$\frac{1}{5} [\begin{matrix} - 3 & 2 \\ 4 & - 1 \end{matrix}]$
$\frac{1}{5} [\begin{matrix} - 1 & 2 \\ - 1 & 4 \end{matrix}]$

155.

The inverse of the matrix $[\begin{matrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & - 1 \end{matrix}]$ is

$- \frac{1}{3} [\begin{matrix} - 3 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 2 & - 3 \end{matrix}]$
$- \frac{1}{3} [\begin{matrix} - 3 & 0 & 0 \\ 3 & - 1 & 0 \\ - 9 & - 2 & 3 \end{matrix}]$
$- \frac{1}{3} [\begin{matrix} 3 & 0 & 0 \\ 3 & - 1 & 0 \\ - 9 & - 2 & 3 \end{matrix}]$
$- \frac{1}{3} [\begin{matrix} - 3 & 0 & 0 \\ - 3 & - 1 & 0 \\ - 9 & - 2 & 3 \end{matrix}]$

156.

For a invertible matrix A if A(adjA) = $[\begin{matrix} 10 & 0 \\ 0 & 10 \end{matrix}]$ then $|A|$ =

100
- 100
10
- 10

157.

If the inverse of the matrix $[\begin{matrix} α & 14 & - 1 \\ 2 & 3 & 1 \\ 6 & 2 & 3 \end{matrix}]$ does not exist, then the value of $α$ is

158.
The system $(\begin{matrix} 1 & - 1 & 2 \\ 3 & 5 & - 3 \\ 2 & 6 & a \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 3 \\ b \\ 2 \end{matrix})$ has no solutions, if

a = - 5, $b \neq 5$

a = - 5, b = 5

$a \neq - 5, b = 5$

$a \neq - 5, b \neq 5$

a = - 5, $b \neq 5$

$[\begin{matrix} 1 & - 1 & 2 \\ 3 & 5 & - 3 \\ 2 & 6 & a \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 3 \\ b \\ 2 \end{matrix}] The augment matrix of given system is [A | B] = [\begin{matrix} 1 & - 1 & 2 \\ 3 & 5 & - 3 \\ 2 & 6 & a \end{matrix} \begin{matrix} 3 \\ b \\ 2 \end{matrix}] R_{2} \to R_{2} - 3 R_{1}, gives [A | B] = [\begin{matrix} 1 & - 1 & 2 \\ 0 & 8 & - 9 \\ 2 & 6 & a \end{matrix} \begin{matrix} 3 \\ b - 9 \\ 2 \end{matrix}] R_{3} \to R_{3} - 2 R_{1}, gives [A | B] = [\begin{matrix} 1 & - 1 & 2 \\ 0 & 8 & - 9 \\ 0 & 8 & a - 4 \end{matrix} \begin{matrix} 3 \\ b - 9 \\ - 4 \end{matrix}] R_{3} \to R_{3} - R_{2}, gives [A | B] = [\begin{matrix} 1 & - 1 & 2 \\ 0 & 8 & - 9 \\ 0 & 0 & a + 5 \end{matrix} \begin{matrix} 3 \\ b - 9 \\ 5 - b \end{matrix}] We know that, For no solution Rank A < Rank [A | B] ∴ a + 5 = 0 and 5 - b \neq 0 a = - 5, b \neq 5$

159.

If $[\begin{matrix} 3 & - 1 \\ 0 & 6 \end{matrix}] [\begin{matrix} 3 x \\ 1 \end{matrix}] + [\begin{matrix} - 2 x \\ 3 \end{matrix}] = [\begin{matrix} 8 \\ 9 \end{matrix}]$ , then the value of x is