CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Adjoint of a Matrix

If Tm,Tn,Tk...

Question

If $T_{m}, T_{n}, T_{k}$ are $m^{t h}, n^{t h}, & k^{t h}$ terms of an AP then prove that $∣ ∣ ∣ ∣ \begin{matrix} T_{m} & m & 1 T_{n} & n & 1 T_{k} & k & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$

Open in App

Solution

Let the first term of $A P$ be $a$ and common difference be $d$

$⟹ T_{m} = a + (m - 1) d, T_{n} = a + (n - 1) d, T_{k} = a + (k - 1) d$

$∣ ∣ ∣ ∣ \begin{matrix} T_{m} & m & 1 T_{n} & n & 1 T_{k} & k & 1 \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a + (m - 1) d & m & 1 a + (n - 1) d & n & 1 a + (k - 1) d & k & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}$

$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} (m - n) d & (m - n) & 0 (n - k) d & (n - k) & 0 a + (k - 1) d & k & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = (m - n) (n - k) ∣ ∣ ∣ ∣ \begin{matrix} d & 1 & 0 d & 1 & 0 a + (k - 1) d & k & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$

Suggest Corrections

thumbs-up

0

Similar questions

T_{m,} T_{m + n}

T_{m - n}

are respectively the

m t h, (m + n) t h

(m - n) t h

terms of an AP, then prove that

T_{m + n} + T_{m - n} = 2 T_{m}

m

m^{t h}

term of an AP is equal to n times the

n^{t h}

term to prove that

(m + n)^{t h}

t_{n}

n^{t h}

S_{n}

n

t_{m} = \frac{1}{n}

t_{n} = \frac{1}{m}

S_{m n}

t_{m + n} + t_{m - n} = 2 t_{m}

t_{m}

n^{t h}

Q. If m times the mth term of an AP is equal to n times the nth term prove that (m+n)th term of AP is 0

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Adjoint and Inverse of a Matrix

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Adjoint of a Matrix

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon