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Matrices

Multiple Choice Questions

131.

The sum of the real roots of the equation $|\begin{matrix} x & - 6 & - 1 \\ 2 & - 3 x & x - 3 \\ - 3 & 2 x & x + 2 \end{matrix}| = 0$ , is equal to

132.

If $A = [\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}]$ , then A² - 5A is equal to

2I
- 2I
3I
null matrix

133.

If A = $[\begin{matrix} 2 & 1 \\ - 1 & 2 \end{matrix}], B = [\begin{matrix} 1 & - 2 \\ 2 & 1 \end{matrix}], C = [\begin{matrix} 1 & - 3 \\ 2 & 1 \end{matrix}]$ , then

A + B = B + A and A + (B + C) = (A + B) + C
A + B = B + A and AC = BC
A + B = B + A and AB = BC
AC = BC and A = BC

134.

$A = [\begin{matrix} - 2 & 4 \\ - 1 & 2 \end{matrix}]$ , then A² is equal to

null matrix
unit matrix
$[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$
$[\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}]$

135.

If A = [x y z], B = $[\begin{matrix} a & h & g \\ h & b & f \\ g & f & c \end{matrix}] and C = [\begin{matrix} x \\ y \\ z \end{matrix}]$ . Then, ABC = O, if

[ax² + by² + cz² + 2gxy + 2fyz + 2czx] = 0
[ax² + cy² + bz² + xy + yz + zx] = 0
[ax² + by² + cz² + 2hxy + 2by + 2cz] = 0
[ax² + by² + cz² + 2zx + 2hxy + 2fyz] = 0

136.

A = $[\begin{matrix} 0 & 3 & 3 \\ - 3 & 0 & - 4 \\ - 3 & 4 & 0 \end{matrix}] and B = [\begin{matrix} x \\ y \\ z \end{matrix}]$ , then B'(AB) is

null matrix
singular matrix
unit matrix
symmetric matrix

137.

A square matrix is an orthogonal matrix, if

AA' = 0
A + A' = I
AA' = I
None of these

138.

$A = [\begin{matrix} 1 & 2 \\ 3 & 2 \\ - 1 & 0 \end{matrix}], B = [\begin{matrix} 1 & 3 & 2 \\ 4 & - 1 & 3 \end{matrix}]$ , then order of AB is

2 x 2
3 x 3
1 x 3
3 x 2

139.

If A + I = $[\begin{matrix} 3 & - 2 \\ 4 & 1 \end{matrix}]$ , then (A + I)(A - I) is equal to

$[\begin{matrix} - 5 & - 4 \\ 8 & - 9 \end{matrix}]$
$[\begin{matrix} - 5 & 4 \\ - 8 & 9 \end{matrix}]$
$[\begin{matrix} 5 & 4 \\ 8 & 9 \end{matrix}]$
$[\begin{matrix} - 5 & - 4 \\ - 8 & - 9 \end{matrix}]$

140.
If A = $[\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}]$ and B = $[\begin{matrix} 1 & a \\ 4 & b \end{matrix}]$ and (A + B)² = A² + B². Then, a and b are respectively

1, - 1

2, - 3

- 1, 1

3, - 2

1, - 1

$Given, A = [\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}] and B = [\begin{matrix} 1 & a \\ 4 & b \end{matrix}] A + B = [\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}] + [\begin{matrix} 1 & a \\ 4 & b \end{matrix}] = [\begin{matrix} 2 & - 1 + a \\ 6 & - 1 + b \end{matrix}] \Rightarrow {(A + B)}^{2} = [\begin{matrix} 2 & - 1 + a \\ 6 & - 1 + b \end{matrix}] [\begin{matrix} 2 & - 1 + a \\ 6 & - 1 + b \end{matrix}] = [\begin{matrix} 4 - 6 + 6 a & - 2 + 2 a + 1 - b - a + ab \\ 12 - 6 + 6 b & - 6 + 6 a + 1 + b^{2} - 2 b \end{matrix}] = [\begin{matrix} - 2 + 6 a & - 1 + a - b + ab \\ 6 + 6 b & - 5 + 6 a - 2 b + b^{2} \end{matrix}]$

$and A^{2} = [\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}] [\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}] = [\begin{matrix} 1 - 2 & - 1 + 1 \\ 2 - 2 & - 2 + 1 \end{matrix}] = [\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix}] Also, B^{2} = [\begin{matrix} 1 & a \\ 4 & b \end{matrix}] [\begin{matrix} 1 & a \\ 4 & b \end{matrix}] = [\begin{matrix} 1 + 4 a & a + ab \\ 4 + 4 b & 4 a + b^{2} \end{matrix}] Given, {(A + B)}^{2} = A^{2} + B^{2} ∴ [\begin{matrix} - 2 + 6 a & - 1 + a - b + ab \\ 6 + 6 b & - 5 + 6 a - 2 b + b^{2} \end{matrix}] = [\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix}] + [\begin{matrix} 1 + 4 a & a + ab \\ 4 + 4 b & 4 a + b^{2} \end{matrix}] \Rightarrow [\begin{matrix} - 2 + 6 a & - 1 + a - b + ab \\ 6 + 6 b & - 5 + 6 a - 2 b + b^{2} \end{matrix}] = [\begin{matrix} 4 a & a + ab \\ 4 + 4 b & - 1 + 4 a + b^{2} \end{matrix}]$

On comparing both sides, we get

- 2 + 6a = 4a and 6 + 6b = 4 + 4b

$\Rightarrow$ a = 1 and b = - 1

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Multiple Choice Questions

140. If A = 1- 12- 1 and B = 1a4b and (A + B)2 = A2 + B2. Then, a and b are respectively 1, - 1 2, - 3 - 1, 1 3, - 2

140.
If A = $[\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}]$ and B = $[\begin{matrix} 1 & a \\ 4 & b \end{matrix}]$ and (A + B)² = A² + B². Then, a and b are respectively

1, - 1

2, - 3

- 1, 1

3, - 2