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Matrices

Long Answer Type

201.

Using elementary row operations, find the inverse of the following matrix:

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

Short Answer Type

202.

For a 2 x 2 matrix, A = [ a_ij ] whose elements are given by a_ij = $\frac{i}{j}$ , write the value of _a12 .

203.

For what value of x, the matrix $[\begin{matrix} 5 - x & x + 1 \\ 2 & 4 \end{matrix}]$ is singular?

204.

Write A^-1 for A = $[\begin{matrix} 2 & 5 \\ 1 & 3 \end{matrix}]$

Long Answer Type

205.

Using elementary transformations, find the inverse of the matrx

$(\begin{matrix} 1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0 \end{matrix})$

Short Answer Type

206.

If $(\begin{matrix} 2 & 3 \\ 5 & 7 \end{matrix}) (\begin{matrix} 1 & - 3 \\ - 2 & 4 \end{matrix}) = (\begin{matrix} - 4 & 6 \\ - 9 & x \end{matrix})$ , write the value of x.

207.

Simplify: $\cos θ [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] + \sin θ [\begin{matrix} \sin θ & - \cos θ \\ \cos θ & \sin θ \end{matrix}]$

Long Answer Type

208.
Using elementary operations, find the inverse of the following matrix:
$(\begin{matrix} - 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix})$

Consider the given matrix.

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

We know that, A = I_n A

Perform sequence of elementary row operations on A on the left hand side and the term I_n on the right hand side till we obtain the results,

I_n = BA

Thus, B = A^{- 1}

$Here, I_{3} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] Thus, we have, [\begin{matrix} - 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix}] = [\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}] A R_{1} \leftrightarrow R_{2} [\begin{matrix} 1 & 2 & 3 \\ - 1 & 1 & 2 \\ 3 & 1 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A R_{2} \to R_{2} + R_{1} R_{3} \to R_{3} - 3 R_{1} [\begin{matrix} 1 & 2 & 3 \\ 0 & 3 & 5 \\ 3 & - 5 & - 8 \end{matrix}] = [\begin{matrix} 0 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & - 3 & 1 \end{matrix}] A R_{1} \to R_{1} + R_{2} [\begin{matrix} 1 & 5 & 8 \\ 0 & 3 & 5 \\ 0 & - 5 & - 8 \end{matrix}] = [\begin{matrix} 1 & 2 & 0 \\ 1 & 1 & 0 \\ 0 & - 3 & 1 \end{matrix}] A R_{1} \to R_{1} + R_{3}$

$[\begin{matrix} 1 & 0 & 0 \\ 0 & 3 & 5 \\ 0 & - 5 & - 8 \end{matrix}] = [\begin{matrix} 1 & - 1 & 1 \\ 1 & 1 & 0 \\ 0 & - 3 & 1 \end{matrix}] A R_{2} \to \frac{R_{2}}{3} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & \frac{5}{3} \\ 0 & - 5 & - 8 \end{matrix}] = [\begin{matrix} 1 & - 1 & 1 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ 0 & - 3 & 1 \end{matrix}] A R_{3} \to R_{3} + 5 R_{2} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & \frac{1}{3} \end{matrix}] = [\begin{matrix} 1 & - 1 & 1 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ \frac{5}{3} & \frac{- 4}{3} & 1 \end{matrix}] A [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & - 1 & 1 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ 5 & - 4 & 3 \end{matrix}] A$

$R_{2} \to R_{2} - \frac{5}{3} R_{3} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & - 1 & 1 \\ - 8 & 7 & - 5 \\ 5 & - 4 & 3 \end{matrix}] A$

Thus, the inverse of the matrix A is given by

$[\begin{matrix} 1 & - 1 & 1 \\ - 8 & 7 & - 5 \\ 5 & - 4 & 3 \end{matrix}]$

Multiple Choice Questions

209.

If P = $open square brackets table row 1 straight alpha 3 row 1 3 3 row 2 4 4 end table close square brackets$ is the adjoint of a 3 x3 matrix A and |A| = 4, then α is equal to

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210.

Let A = $open square brackets table row 1 0 0 row 2 1 0 row 3 2 1 end table close square brackets$ . If u₁ and u₂ are column matrices such that Au1 = $open square brackets table row 1 row 0 row 0 end table close square brackets$ and Au2 = $open square brackets table row 0 row 1 row 0 end table close square brackets$ , then u₁ +u₂ is equal to

$open square brackets table row cell negative 1 end cell row 1 row 0 end table close square brackets$
$open square brackets table row cell negative 1 end cell row cell space space space 1 end cell row cell negative 1 end cell end table close square brackets$
$open square brackets table row cell negative 1 end cell row cell negative 1 end cell row 0 end table close square brackets$
$open square brackets table row cell negative 1 end cell row cell negative 1 end cell row 0 end table close square brackets$