Matrix Trace - Class XII

A matrix is not invertible if its determinant is zero. Det(A) = (4 * 2) - (1 * (3 - k)) = 8 - 3 + k = 5 + k. Setting 5 + k = 0 gives k = -5.

We know that in a skew-symmetric matrix, diagonal elements are always 0.
So, its trace is 0.

Given, $A=\left[ \begin{matrix} 1 & -5 & 7 \ 0 & 7 & 9 \ 11 & 8 & 9 \end{matrix} \right] $
$tr(A)=sum\ of\ diagonal\ matrix$
$tr(A)=1+7+9=17$

By definition of trace of a scalar matrix of order n, 
$tr(A)=a _{11}+ a _{22}+a _{33}+.....+a _{nn}$

$=k+k+....k $ (upto n times)

$tr(A)=nk$

By definition of trace of a matrix of order n, 
$tr(A)=a _{11}+ a _{22}+a _{33}+.....+a _{nn}$
$\displaystyle =: \sum _{ i=j }  { a } _{ ij } =: \sum _{ i } a _{ ij }$
Hence, option 'B' is correct.

We know that for a skew-symmetric matrix the sum of diagonal elements is zero.
$a _{ij}=0    \forall i=j$
So,$ tr (A)=0$

$A=\begin{bmatrix} 1 & -5 & 7 \ 0 & 7 & 9 \ 11 & 8 & 9 \end{bmatrix}$
$trace(A)=$sum of diagonal elements$=1+7+9=17$

Ans: A

Given $A=[a _{i,j}]$ and  $a _{i,j}=i(i+j)$ 

The trace of a product of matrices, tr(AB), is not generally equal to the product of their traces, tr(A) * tr(B). It depends on the specific entries of the matrices.

$A+2B=\begin{bmatrix} 1 & 2 & 0 \ 6 & -3 & 3 \ -5 & 3 & 1 \end{bmatrix}$         .....(i)

$2A-B=\begin{bmatrix} 2 & -1 & 5 \ 2 & -1 & 6 \ 0 & 1 & 2 \end{bmatrix}$         .....(ii)

$\Rightarrow 4A-2B=\begin{bmatrix} 4 & -2 & 10 \ 4 & -2 & 12 \ 0 & 2 & 4 \end{bmatrix}$      ....(iii)

Adding (i) and (iii), we get
$5A=\begin{bmatrix} 5 & 0 & 10 \ 10 & -5 & 15 \ -5 & 5 & 5 \end{bmatrix}$

$\Rightarrow A=\begin{bmatrix} 1 & 0 & 2 \ 2 & -1 & 3 \ -1 & 1 & 1 \end{bmatrix}$
So, $tr(A)=1$

Now, by eq(ii),
$B=\begin{bmatrix} 2 & 0 & 4 \ 4 & -2 & 6 \ -2 & 2 & 2 \end{bmatrix}-\begin{bmatrix} 2 & -1 & 5 \ 2 & -1 & 6 \ 0 & 1 & 2 \end{bmatrix}$

$\Rightarrow B=\begin{bmatrix} 0 & 1 & -1 \ 2 & -1 & 0 \ -2 & 1 & 0 \end{bmatrix}$
So, $tr(B)=-1$
Now, $tr(A)-tr(B)=1+1=2$

Given $A = \begin{bmatrix}2 & 3 & 4\ 5 & -3 & 8\ 9 & 2 & 16\end{bmatrix}$

$tr(A)=2-3+16=15$

Given $tr(A)=4$
$\Rightarrow a _{11}+a _{22}=4$

$a _{ ij }={ { 0,1,2,3,4}  }$

So, diagonal entries of A can be 0 and 4 , 4 and 0, 1 and 3, 3 and 1, 2 and 2,
Hence, 5 matrices are possible

The expression is a geometric series of traces. Given the matrices, the sum converges to a specific value based on the properties of the trace and matrix multiplication.

The matrix product BC results in the identity matrix, meaning powers of BC remain the identity matrix. Summing the resulting traces forms a geometric series whose sum evaluates precisely to 6.

Equating the traces of P and Q gives an equation involving a, b, and c. Using the properties of triangle sides and the Law of Cosines, one can solve for cos A.

Let M1 * A * M2 = M3. Then A = M1^-1 * M3 * M2^-1. Calculate the inverse of the matrices and perform the multiplication to find A, then sum the diagonal elements.

This is identical to the previous series summation problems involving matrix traces.

This has a typo in the series - the 4th term should be A(BC)³/8, but it repeats A(BC)²/8. Assuming the intended pattern (powers of BC increase): BC = I, tr(A) = 3. Series = 3[1 + 1/2 + 1/4 + 1/8 + ...] = 3 x 2 = 6. Answer A is correct despite the typographical error.

Given,  $A = \begin{bmatrix}1 & -5 & 7\ 0 & 7 & 9\ 11 & 8 & 9\end{bmatrix}$ .

Given $A = [a _{ij}]$ is a scalar matrix of order $n\times n$ such that $a _{ii} = k$ for all $i$.

As $A$ is a skew symmetric matrix
                   $A' = -A$
$\Rightarrow  a _{ii}=0 : \forall : i \Rightarrow trace : (A) =0$
Also             $|A|=|A'|=|-A|=(-1)^3|A|$
$\Rightarrow  2|A|=0\Rightarrow |A|=0$
Hence, option B.

If $A = 0,tr(A)=0$.

$tr(A)=1+\omega^n+\omega^{2n}$
if $n=3k$ i.e Mutiple of $3$.
$\Rightarrow tr(A)=1+\omega^{3k}+\omega^{6k}=1+1+1=3$
if $n\neq 3k$ i.e not a multiple of $3$.
then $n=3k+1$ or $n=3k+2$
$\Rightarrow tr(A)=1+\omega^{3k+1}+\omega^{2(3k+1)}=1+\omega+\omega^2=0$
Hence, options B and C.

${ T } _{ r }(A)={ \alpha  }^{ 2 }+{ \beta  }^{ 2 }+{ \gamma  }^{ 2 }$

Given, $B=diag(1    2    5)$
$\Rightarrow B^{2}=diag(1     4    25)$

Now, $AB^{2}=\begin{bmatrix}1 &0  &0 \ 0 &24  &0 \ 0 &0  &50 \end{bmatrix}$

$\Rightarrow tr(AB^{2})=1+24+50=75$

$A=\left| \begin{matrix} 2 & 1 \ 4 & 1 \end{matrix} \right| ,\quad B=\left| \begin{matrix} 3 & 4 \ 2 & 3 \end{matrix} \right| ,\quad C=\left| \begin{matrix} 3 & -4 \ -2 & 3 \end{matrix} \right| \ AB\quad =\quad \left| \begin{matrix} 2 & 1 \ 4 & 1 \end{matrix} \right| \left| \begin{matrix} 3 & 4 \ 2 & 3 \end{matrix} \right| \ \qquad =\quad \left| \begin{matrix} 6+2 & 8+3 \ 12+2 & 16+3 \end{matrix} \right| \ \quad \quad \quad =\quad \left| \begin{matrix} 8 & 11 \ 14 & 19 \end{matrix} \right| \ ABC\quad =\quad \left| \begin{matrix} 8 & 11 \ 14 & 19 \end{matrix} \right| \left| \begin{matrix} 3 & -4 \ -2 & 3 \end{matrix} \right| \ \quad \quad \quad \quad \quad =\quad \left| \begin{matrix} 24-22 & -32+33 \ 42-38 & -56+57 \end{matrix} \right| \ \qquad \quad \quad =\quad \left| \begin{matrix} 2 & 1 \ 4 & 1 \end{matrix} \right| \ { (BC) }^{ 2 }\quad =\quad \left| \begin{matrix} 3 & 4 \ 2 & 3 \end{matrix} \right| \left| \begin{matrix} 3 & -4 \ -2 & 3 \end{matrix} \right| \ \qquad \quad \quad =\quad { \left| \begin{matrix} 1 & 0 \ 0 & 1 \end{matrix} \right|  }^{ 2 }$

Here, $A=\begin{bmatrix} { \omega  }^{ 2 } & { \omega  }^{ 3 } & ... & { \omega  }^{ 11 } \ { \omega  }^{ 3 } & { \omega  }^{ 4 } & ... & { \omega  }^{ 12 } \ ... & ... & ... & ... \ { \omega  }^{ 11 } & { \omega  }^{ 12 } &  & { \omega  }^{ 20 } \end{bmatrix}$

$tr(A)={ \omega  }^{ 2 }+{ \omega  }^{ 4 }+{ \omega  }^{ 6 }+.....+{ \omega  }^{ 20 }$

$=\displaystyle \frac { { \omega  }^{ 2 }(1-{ \omega  }^{ 20 }) }{ 1-{ \omega  }^{ 2 } } $

 $A=\left[\begin{matrix}2&0&7\0&1&0\1&-2&1\end{matrix}\right]$ and $B=\left[\begin{matrix}-x&14x&7x\0&1&0\x&-4x&-2x\end{matrix}\right]$
$AB=\left[\begin{matrix}2&0&7\0&1&0\1&-2&1\end{matrix}\right]\left[\begin{matrix}-x&14x&7x\0&1&0\x&-4x&-2x\end{matrix}\right]=\begin{bmatrix} 5x & 14x & 0 \ 0 & 1 & 0 \ 0 & 10x-2 & 5x \end{bmatrix}$
but,$AB=(AB)^{-1}\Rightarrow (AB)^2=I$
$\Rightarrow (AB)^2=\begin{bmatrix} 5x & 14x & 0 \ 0 & 1 & 0 \ 0 & 10x-2 & 5x \end{bmatrix}\begin{bmatrix} 5x & 14x & 0 \ 0 & 1 & 0 \ 0 & 10x-2 & 5x \end{bmatrix}=\begin{bmatrix} 25x^{ 2 } & 70x^{ 2 }+14x & 0 \ 0 & 1 & 0 \ 0 & (5x+1)(10x-2) & 25x^{ 2 } \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow x=\displaystyle\frac{-1}{5}$
$Tr.\left(AB+(AB)^2+(AB)^3+...+(AB)^{100}\right)=Tr.\left(AB+I+(AB)+...+I\right)$
$tr(AB)=10x+1=-1$ and $tr(I)=3$
$\therefore Tr.\left(AB+I+(AB)+...+I\right)=(-1+3-1+3......+3)=50(3-1)=100$
Hence, option C.

Given, $A=\left[\begin{matrix}1 & \displaystyle\frac{3}{2}\1 & 2\end{matrix}\right], B = \left[\begin{matrix}4 & -3\-2 & 2\end{matrix}\right] $

So, $AB=\left[ \begin{matrix} 1 & 0 \ 0 & 1 \end{matrix} \right] $
$\Rightarrow AB=I$

Now, $(AB)^{r} C _r =I \left[\begin{matrix}r.3^r & 2^r\0 & (r-1)3^r\end{matrix}\right]$

$\Rightarrow (AB)^{r} C _r=\left[\begin{matrix}r.3^r & 2^r\0 & (r-1)3^r\end{matrix}\right]$

$\sum _{ r=1 }^{ 50 }{ tr.\left( (AB)^{ r }C _{ r } \right)  } =\sum _{ r=1 }^{ 50 }{ tr\left( C _{ r } \right)  } =\sum _{ r=1 }^{ 50 }{ (2r } -1)3^{ r }$

It is an arithmetico-geometric series
$S _{50}=1.3+3.3^{2}+5.3^{3}+......+99.3^{50}$
$3S _{50}=        3^{2}+3.3^{3}+......+97.3^{50}+99.3^{51}$

$\Rightarrow -2S=3+2.3^{2}+2.{3}^{3}+.....+2.3^{50}-99.3^{51}$

$\Rightarrow -2S=3-99.3^{ 51 }+2(3^{ 2 }+{ 3 }^{ 3 }+.....+3^{ 50 })$

$\Rightarrow -2S=3-99.3^{ 51 }+3^{ 51 }-3^{ 2 }$

$\Rightarrow S=3(1+49.3^{ 50 })$

Given $A=\left[\begin{matrix}3x^2\1\6x\end{matrix}\right], B=[a,b,c]$ 

Here, $AB=\left[ \begin{matrix} 3x^{ 2 } \ 1 \ 6x \end{matrix} \right] [a,b,c]$

$\Rightarrow AB=\left[ \begin{matrix} 3ax^{ 2 } & 3bx^{ 2 } & 3cx^{ 2 } \ a & b & c \ 6ax & 6bx & 6cx \end{matrix} \right] $

$tr(AB)=3ax^{2}+b+6cx$

Now, given $C=\left[\begin{matrix}(x+2)^2&5x^2&2x\5x^2&2x&(x+2)^2\2x&(x+2)^2&5x^2\end{matrix}\right]$

$tr(C)=(x+2)^2+2x+5x^{2}$
$\Rightarrow tr(C)=6x^{2}+6x+4$

Since, we have $tr(AB)=tr(C)$
$\Rightarrow 3ax^{2}+b+6cx=6x^{2}+6x+4$

On comparing, we get $a=2,b=4,c=1$
So, $a+b+c=7$

$f(x,y) = x^2 + y^2 - 2xy, \space (x,y \in R)$ and
$\quad A = \begin{bmatrix}f(x _1,y _1) & f(x _1,y _2) & f(x _1,y _3) \ f(x _2,y _1) & f(x _2,y _2) & f(x _2,y _3) \ f(x _3,y _1) & f(x _3,y _2) & f(x _3,y _3) \end{bmatrix}$ such that trace $(A) = 0$
$\Rightarrow A=\begin{bmatrix} (x _{ 1 }-y _{ 1 })^{ 2 } & \quad (x _{ 1 }-y _{ 2 })^{ 2 } & \quad (x _{ 1 }-y _{ 3 })^{ 2 } \ (x _{ 2 }-y _{ 1 })^{ 2 } & \quad (x _{ 2 }-y _{ 2 })^{ 2 } & \quad (x _{ 2 }-y _{ 3 })^{ 2 } \ (x _{ 3 }-y _{ 1 })^{ 2 } & \quad (x _{ 3 }-y _{ 2 })^{ 2 } & \quad (x _{ 3 }-y _{ 3 })^{ 2 } \end{bmatrix}$
Trace$(A)=0$
$\Rightarrow (x _1-y _1)^2+(x _2-y _2)^2+(x _3-y _3)^2=0$
$\Rightarrow x _1=y _1; x _2=y _2; x _3=y _3$
$A=\begin{bmatrix} 0 & \quad (y _{ 1 }-y _{ 2 })^{ 2 } & \quad (y _{ 1 }-y _{ 3 })^{ 2 } \ (y _{ 2 }-y _{ 1 })^{ 2 } & 0 & \quad (y _{ 2 }-y _{ 3 })^{ 2 } \ (y _{ 3 }-y _{ 1 })^{ 2 } & \quad (y _{ 3 }-y _{ 2 })^{ 2 } & 0 \end{bmatrix}$
$|A|=2(y _1-y _2)^2(y _2-y _3)^2(y _3-y _1)^2\geq 0$
$\therefore |A|\geq 0$
Hence, option A.