If a_ij = \({1\over2}(3i-2j)\) and A = [a_ij]_2x2 is

(a)

\(\begin{bmatrix} {1\over 2}& 2 \\ -{1\over2} & 1 \end{bmatrix}\)

(b)

\(\begin{bmatrix} {1\over 2}& -{1\over2} \\ 2& 1 \end{bmatrix}\)

(c)

\(\begin{bmatrix} 2& 2\\ {1\over 2}& -{1\over2} \end{bmatrix}\)

(d)

\(\begin{bmatrix} -{1\over 2}& {1\over2} \\ 1& 2 \end{bmatrix}\)

What must be the matrix X, if 2x +\(\begin{bmatrix} 1& 2 \\ 3 & 4 \end{bmatrix}=\begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix}?\)

(a)

\(\begin{bmatrix} 1& 3 \\ 2 &-1 \end{bmatrix}\)

(b)

\(\begin{bmatrix} 1& -3 \\ 2 &-1 \end{bmatrix}\)

(c)

\(\begin{bmatrix} 2& 6 \\ 4 &-2 \end{bmatrix}\)

(d)

\(\begin{bmatrix} 2& -6 \\ 4 &-2 \end{bmatrix}\)

If A is a square matrix, then which of the following is not symmetric?

(a)

A + A^T

(b)

AA^T

(c)

A^T A

(d)

A − A^T

If A = \(\begin{bmatrix}a & x \\ y& a \end{bmatrix}\) and if xy = 1, then det (A A^T ) is equal to

(a)

(a −1)²

(b)

(a² +1)²

(c)

a² −1

(d)

(a² −1)²

The value of x, for which the matrix A = \(\begin{bmatrix} e^{x-2}& e^{7+x} \\ e^{2+x} & e^{2x+3} \end{bmatrix}\) is singular

(a)

9

(b)

8

(c)

7

(d)

6

If the points (x,−2), (5, 2), (8, 8) are collinear, then x is equal to

(a)

-3

(b)

\({1\over 3}\)

(c)

1

(d)

3

If \(\begin{vmatrix}2a & x_1 &y_1 \\ 2b & x_2 & y_2 \\ 2c & x_3 &y_3 \end{vmatrix}={abc\over 2}\neq 0,\) then the area of the triangle whose vertices are \(\begin{pmatrix} {x_1\over a}, {y_1\over a} \end{pmatrix}\), \(\begin{pmatrix} {x_2\over b}, {y_2\over b} \end{pmatrix}\), \(\begin{pmatrix} {x_3\over c}, {y_3\over c} \end{pmatrix}\) is

(a)

\({1\over 4}\)

(b)

\({1\over 4} abc\)

(c)

\({1\over 8}\)

(d)

\({1\over 8}abc\)

If the square of the matrix \(\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\) is the unit matrix of order 2, then \(\alpha ,\beta \) and \(\gamma\) should satisfy the relation.

(a)

1 + \(\alpha ^2+\beta \gamma=0\)

(b)

1 - \(\alpha ^2-\beta \gamma=0\)

(c)

1 - \(\alpha ^2+\beta \gamma=0\)

(d)

1 + \(\alpha ^2-\beta \gamma=0\)

If \(\triangle\) = \(\begin{vmatrix} a&b &c \\ x & y & z \\ p &q &r \end{vmatrix}\), then \(\begin{vmatrix} ka&kb &kc \\ kx & ky & kz \\k p &kq &kr \end{vmatrix}\) is

(a)

\(\triangle\)

(b)

k\(\triangle\)

(c)

3k\(\triangle\)

(d)

k³\(\triangle\)

A root of the equation \(\begin{vmatrix} 3-x&-6 &3 \\ -6 & 3-x & 3 \\ 3 &3 &-6-x \end{vmatrix}=0 \ is\)

(a)

6

(b)

3

(c)

0

(d)

-6