Linear Algebra · Engineering Mathematics · GATE IN

The figure shows a shape $$ABC$$ and its minor image $${A_1}$$$${B_1}$$$${C_1}$$ across the horizontal axis ($$x$$-axis). The coordinate transformation matrix that maps $$ABC$$ to $${A_1}$$$${B_1}$$$${C_1}$$ is

The eigen values of the matrix $$A = \left[ {\matrix{ 1 & { - 1} & 5 \cr 0 & 5 & 6 \cr 0 & { - 6} & 5 \cr } } \right]$$ are

If $$V$$ is a non-zero vector of dimension $$3 \times 1,$$ then the matrix $$\,A = V{V^T}$$ has a rank $$=$$ ________

Let $$A$$ be an $$n \times n$$ matrix with rank $$r\left( {0 < r < n} \right).$$ Then $$AX=0$$ has $$p$$ independent solutions, where $$p$$ is

A scalar valued function is defined as $$f\left( x \right){x^T}Ax + {b^T}x + c,$$ where $$A$$ is a symmetric positive definite matrix with dimension $$n \times n;$$ $$b$$ and $$x$$ are vectors of dimension $$n \times 1$$. The minimum value of $$f(x)$$ will occur when $$x$$ equals.

For the matrix $$A$$ satisfying the equation given below, the eigen values are $$$\left[ A \right]\left[ {\matrix{ 1 & 2 & 3 \cr 7 & 8 & 9 \cr 4 & 5 & 6 \cr } } \right] = \left[ {\matrix{ 1 & 2 & 3 \cr 4 & 5 & 6 \cr 7 & 8 & 9 \cr } } \right]$$$

The dimension of the null space of the matrix $$\left[ {\matrix{ 0 & 1 & 1 \cr 1 & { - 1} & 0 \cr { - 1} & 0 & { - 1} \cr } } \right]$$ is

Given that $$A = \left[ {\matrix{ { - 5} & { - 3} \cr 2 & 0 \cr } } \right]$$ and $${\rm I} = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right],$$ the value of $${A^3}$$ is

The matrix $$M = \left[ {\matrix{ { - 2} & 2 & { - 3} \cr 2 & 1 & 6 \cr { - 1} & { - 2} & 0 \cr } } \right]$$ has eigen values $$-3, -3, 5.$$ An eigen vector corresponding to the eigen value $$5$$ is $${\left[ {\matrix{ 1 & 2 & { - 1} \cr } } \right]^T}.$$ One of the eigen vector of the matrix $${M^3}$$ is

A real $$n \times n$$ matrix $$A$$ $$ = \left[ {{a_{ij}}} \right]$$ is defined as
follows $$\left\{ {\matrix{ {{a_{ij}} = i,} & {\forall i = j} \cr { = 0,} & {otherwise} \cr } .} \right.$$

The sum of all $$n$$ eigen values of $$A$$ is

$$X$$ and $$Y$$ are non-zero square matrices of size $$n \times n$$. If $$XY = {O_{n \times n}}$$ then

The eigen values of a $$2 \times 2$$ matrix $$X$$ are $$-2$$ and $$-3$$. The eigen values of matrix $${\left( {X + 1} \right)^{ - 1}}\left( {X + 5{\rm I}} \right)$$ are

Let $$A = \left[ {{a_{ij}}} \right],\,\,1 \le i,j \le n$$ with $$n \ge 3$$ and
$${{a_{ij}} = i.j.}$$ Then the rank of $$A$$ is

Identity which one of the following is an eigen vectors of the matrix $$A = \left[ {\matrix{ 1 & 0 \cr { - 1} & { - 2} \cr } } \right]$$

Let $$A$$ be $$3 \times 3$$ matrix with rank $$2.$$ Then $$AX=O$$ has

The necessary condition to diagonalize a matrix is that

The rank of matrix $$A = \left[ {\matrix{ 1 & 2 & 3 \cr 3 & 4 & 5 \cr 4 & 6 & 8 \cr } } \right]$$ is

Consider the matrix $$A = \left( {\matrix{ 2 & 1 & 1 \cr 2 & 3 & 4 \cr { - 1} & { - 1} & { - 2} \cr } } \right)$$ whose eigen values are $$1, -1$$ and $$3$$. Then trace of $$\left( {{A^3} - 3{A^2}} \right)$$ is ________.

One pair of eigenvectors corresponding to the two eigen values of the matrix $$\left[ {\matrix{ 0 & { - 1} \cr 1 & {0 - } \cr } } \right]$$

Let $$A$$ be $$n \times n$$ real matrix such that $${A^2} = {\rm I}$$ and $$Y$$ be an $$n$$-diamensional vector. Then the linear system of equations $$Ax=y$$ has

A system of linear simultaneous equations is given as $$AX=b$$
where $$A = \left[ {\matrix{ 1 & 0 & 1 & 0 \cr 0 & 1 & 0 & 1 \cr 1 & 1 & 0 & 0 \cr 0 & 0 & 0 & 1 \cr } } \right]\,\,\& \,\,b = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr 1 \cr } } \right]$$

Which of the following statement is true?

A system of linear simultaneous equations is given as $$AX=b$$
where $$A = \left[ {\matrix{ 1 & 0 & 1 & 0 \cr 0 & 1 & 0 & 1 \cr 1 & 1 & 0 & 0 \cr 0 & 0 & 0 & 1 \cr } } \right]\,\,\& \,\,b = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr 1 \cr } } \right]$$

Then the rank of matrix $$A$$ is

For a given $$2x2$$ matrix $$A,$$ it is observved that $$A\left[ {\matrix{ 1 \cr { - 1} \cr } } \right] = - 1\left[ {\matrix{ 1 \cr { - 1} \cr } } \right]$$ and
$$A\left[ {\matrix{ 1 \cr { - 2} \cr } } \right] = - 2\left[ {\matrix{ 1 \cr { - 2} \cr } } \right]$$ then the matrix $$A$$ is