Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 7 Matrices and Determinants Ex 7.2 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2

Question 1.

without expanding the determinant,

Answer:

= s (a^{2} + b^{2} + c^{2}) × 0

since two columns are equal.

= 0

Question 2.

Answer:

Question 3.

Prove that

Answer:

= 2abc [0 – b(0 – ac) + c(ab – 0)]

= 2 abc [ abc + abc ]

= 2 abc × 2abc

Δ = 4 a^{2}b^{2}c^{2}

Question 4.

Prove that

Answer:

= a [b(1 + c) + c (1)] – 0 – c [0 – b]

= a[b + bc + c] + bc

= ab + abc + ac + bc

= abc + ab + bc + ac

= abc

Question 5.

Answer:

Question 6.

show that \(\left| \begin{matrix} x\quad +\quad 2a & y\quad +\quad 2b & z\quad +\quad 2c \\ x & y & z \\ a & b & c \end{matrix} \right| \) = 0

Answer:

Let Δ = \(\left| \begin{matrix} x\quad +\quad 2a & y\quad +\quad 2b & z\quad +\quad 2c \\ x & y & z \\ a & b & c \end{matrix} \right| \)

Question 7.

Write the general form of a 3 × 3 skew- symmetric matrix and prove that its determinant is 0.

Answer:

A square matrix A = [ a_{ij} ]3 × 3 is a skew – symmetric matrix if a_{ij} = – a_{ij} for all i,j and the elements on the main diagonal of a skew – symmetric matrix are zero.

= 0 – a_{12} (0 + a_{13} a_{23}) + a_{13 }(a_{12} a_{23} – 0)

= – a_{12} a_{13} a_{23} + a_{13} a_{12} a_{23}

= 0

Hence the determinant of a skew – symmetric matrix is 0.

Question 8.

Prove that a, b, c are in G. P or α is a root of ax^{2} + 2bx + c = 0.

Answer:

Question 9.

Answer:

Question 10.

If a, b, c are p^{th}, q^{th} and r^{th} terms of an A.P, find the value of \(\left| \begin{matrix} a & b & c \\ p & q & r \\ 1 & 1 & 1 \end{matrix} \right| \)

Answer:

Given a, b, c are p^{th}, q^{th} and r^{th} terms of an A.P.

t_{p} = a = A + (p – 1)D,

t_{q} = b = A + (q – 1)D,

t_{r} = c = A + (r – 1) D

where A – first term , D – Common difference of the AP.

Question 11.

Answer:

Question 12.

If a , b , c , are all positive, and are p^{th}, q^{th} and r^{th} terms of a G.P., show that \(\left| \begin{matrix} log\quad a & p & 1 \\ log\quad b & q & 1 \\ log\quad c & r & 1 \end{matrix} \right| \) – 0

Answer:

Given a, b, c are the p^{th}, q^{th} and r^{th} terms of a G.P.

∴ a = AR^{p-1}, b = AR^{q-1}, c = AR^{r-1}

where A is the first term , R – common ratio.

Question 13.

Find the value of

Answer:

Question 14.

Answer:

Question 15.

Without expanding, evaluate the following determinants:

(i) \(\left| \begin{matrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x \end{matrix} \right| \)

(ii) \(\left| \begin{matrix} x\quad +\quad y & y\quad +\quad z & z\quad +\quad x \\ z & x & y \\ 1 & 1 & 1 \end{matrix} \right| \)

Answer:

(i) \(\left| \begin{matrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x \end{matrix} \right| \)

(ii) \(\left| \begin{matrix} x\quad +\quad y & y\quad +\quad z & z\quad +\quad x \\ z & x & y \\ 1 & 1 & 1 \end{matrix} \right| [

Question 16.

If A is a Square, matrix, and |A| = 2, find the value of |A A^{T}|.

Answer:

|A| = 2 (Given) |A^{T}| = 2

Now |AA^{T}| = |A| |A^{T}| = 2 × 2 = 4.

Question 17.

If A and B are square matrices of order 3 such that |A| = -1 and |B| = 3, find the value of |3 AB|.

Answer:

Given |A| = -1 : |B| = 3

Given A and B are square matrices of order 3.

∴ |kAB| = k^{3} |AB|

Here k = 3 ∴ |3AB| = 3^{3} |AB|

= 27 |AB|

= 27 (-1) (3)

= -81

Question 18.

If λ = – 2, determine the value of

Answer:

Expanding along the first row

Δ = 0 + 4 [4 × 0 – (- 1 ) ( 13)] + [4 × -13 – 0 × – 1]

= 4 [0 + 13] + 1 [- 52 + 0]

= 52 – 52 = 0

Question 19.

Determine the roots of the equation

[latex]\left| \begin{matrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & { 5x }^{ 2 } \end{matrix} \right| \) = 0

Answer:

\(\left| \begin{matrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & { 5x }^{ 2 } \end{matrix} \right| \) = 0 ………… (1)

Put x = -1 then (1) ⇒

∴ x = – 1 satisfies equation (1)

Hence x = – 1 is a root of equation (1)

Put x = 2 then ……….. (1)

[Property 4: If two rows (columns) of a determinant are identical then its determinant value is zero.]

∴ x = 2 satisfies equation (1)

Hence x = 2 is a root of equation (1)

Hence the required roots are x = -1 , 2

Question 20.

Verify that det (AB) = (det A) (det B) for

Answer:

= 4 [-10 (0 – 9 × 19) – 5 (0 + 17 × 19) + 1 (32 × 9 + 17 × 26)]

= 4 [1710 – 5 × 323 + 288 + 442]

= 4 [1710 – 1615 + 730]

= 4 [2440 – 1615]

= 4 × 825

det (AB) = 3300 …….. (1)

= 4(0 – 21) – 3 (- 5 – 14) – 2 (3 – 0)

= -84 – 3 × – 19 – 6

= -84 + 57 – 6

= -90 + 57

det A = -33 ………… (2)

= 1 (20 – 0) – 3 (- 10 – 0) + 3 (-14 – 36)

= 20 + 30 + 3 × – 50

= 50 – 150

det A = – 100 ……….. (3)

From equations (2) and (3)

(det A) (det B) = – 33 × – 100

(detA) (det B) = 3300 ………… (4)

From equations (1) and (4), we have

det (AB) = (det A) (det B)

Question 21.

Using cofactors of elements of second row, evaluate |A|, where A = \(\left[ \begin{matrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{matrix} \right] \)

Answer:

|A| = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23}

= 2 × 7 + 0 × 7 + 1 × – 7

= 14 – 7

|A| = 7