मराठी

A Matrix a of Order 3 × 3 is Such that |A| = 4. Find the Value of |2 A|. - Mathematics

Advertisements
Advertisements

प्रश्न

A matrix A of order 3 × 3 is such that |A| = 4. Find the value of |2 A|.

उत्तर

\[\left| KA \right| = K^n \left| A \right| \] 
Here, n is the order of A . 
\[\text{ Given: }\left| A \right| = 4\] 
\[ \Rightarrow \left| 2A \right| = 2^3 \times 4 = 32\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Determinants - Exercise 6.6 [पृष्ठ ९१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 6 Determinants
Exercise 6.6 | Q 42 | पृष्ठ ९१

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

If A  = `[(1,1,-2),(2,1,-3),(5,4,-9)]`, Find |A|


Using the property of determinants and without expanding, prove that:

`|(x, a, x+a),(y,b,y+b),(z,c, z+ c)| = 0`


On expanding by first row, the value of the determinant of 3 × 3 square matrix
  \[A = \left[ a_{ij} \right]\text{ is }a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13}\] , where [Cij] is the cofactor of aij in A. Write the expression for its value on expanding by second column.

 

Which of the following is not correct in a given determinant of A, where A = [aij]3×3.


If A is a matrix of order 3 and |A| = 8, then |adj A| = __________ .


Solve the following system of linear equations using matrix method: 
3x + y + z = 1
2x + 2z = 0
5x + y + 2z = 2


Without expanding, show that Δ = `|("cosec"^2theta, cot^2theta, 1),(cot^2theta, "cosec"^2theta, -1),(42, 40, 2)|` = 0


Show that Δ = `|(x, "p", "q"),("p", x, "q"),("q", "q", x)| = (x - "p")(x^2 + "p"x - 2"q"^2)` 


If Δ = `|(0, "b" - "a", "c" - "a"),("a" - "b", 0, "c" - "b"),("a" - "c", "b" - "c", 0)|`, then show that ∆ is equal to zero.


If x, y ∈ R, then the determinant ∆ = `|(cosx, -sinx, 1),(sinx, cosx, 1),(cos(x + y), -sin(x + y), 0)|` lies in the interval.


The determinant ∆ = `|(sqrt(23) + sqrt(3), sqrt(5), sqrt(5)),(sqrt(15) + sqrt(46), 5, sqrt(10)),(3 + sqrt(115), sqrt(15), 5)|` is equal to ______.


The value of the determinant ∆ = `|(sin^2 23^circ, sin^2 67^circ, cos180^circ),(-sin^2 67^circ, -sin^2 23^circ, cos^2 180^circ),(cos180^circ, sin^2 23^circ, sin^2 67^circ)|` = ______.


If a + b + c ≠ 0 and `|("a", "b","c"),("b", "c", "a"),("c", "a", "b")|` 0, then prove that a = b = c.


Prove tha `|("bc" - "a"^2, "ca" - "b"^2, "ab" - "c"^2),("ca" - "b"^2, "ab" - "c"^2, "bc" - "a"^2),("ab" - "c"^2, "bc" - "a"^2, "ca" - "b"^2)|` is divisible by a + b + c and find the quotient.


If x + y + z = 0, prove that `|(x"a", y"b", z"c"),(y"c", z"a", x"b"),(z"b", x"c", y"a")| = xyz|("a", "b", "c"),("c", "a", "b"),("b", "c", "a")|`


Let f(t) = `|(cos"t","t", 1),(2sin"t", "t", 2"t"),(sin"t", "t", "t")|`, then `lim_("t" - 0) ("f"("t"))/"t"^2` is equal to ______.


If f(x) = `|(0, x - "a", x - "b"),(x + "b", 0, x - "c"),(x + "b", x + "c", 0)|`, then ______.


If x, y, z are all different from zero and `|(1 + x, 1, 1),(1, 1 + y, 1),(1, 1, 1 + z)|` = 0, then value of x–1 + y–1 + z–1 is ______.


There are two values of a which makes determinant, ∆ = `|(1, -2, 5),(2, "a", -1),(0, 4, 2"a")|` = 86, then sum of these number is ______.


If A is a matrix of order 3 × 3, then |3A| = ______.


`|(0, xyz, x - z),(y - x, 0, y  z),(z - x, z - y, 0)|` = ______.


If A and B are matrices of order 3 and |A| = 5, |B| = 3, then |3AB| = 27 × 5 × 3 = 405.


`"A" = abs ((1/"a", "a"^2, "bc"),(1/"b", "b"^2, "ac"),(1/"c", "c"^2, "ab"))` is equal to ____________.


If A, B, and C be the three square matrices such that A = B + C, then Det A is equal to


`abs ((1 + "a", "b", "c"),("a", 1 + "b", "c"),("a", "b", 1 + "c")) =` ____________


The value of the determinant `abs ((1,0,0),(2, "cos x", "sin x"),(3, "sin x", "cos x"))` is ____________.


If `"abc" ne 0  "and" abs ((1 + "a", 1, 1),(1, 1 + "b", 1),(1,1,1 + "c")) = 0, "then"  1/"a" + 1/"b" + 1/"c" =` ____________.


Let A be a square matrix of order 2 x 2, then `abs("KA")` is equal to ____________.


Find the 5th term of expansion of `(x^2 + 1/x)^10`?


Value of `|(2, 4),(-1, 2)|` is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×