Given that A- - 11 1- 0 11 - B - - 0 -2- -3 4 - I- - 1 0- 0 1 - Find A-3B-4I-

The correct option is C [ 15 amp; 5; 9 amp;27 ] Given, A=[ 11 amp; 1; 0 amp; 11 ],B=[ 0 amp; 2; 3 amp; 4 ],I=[ ...

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

Byju's Answer

Standard XII

Mathematics

Scalar Multiplication of a Matrix

Given that ...

Question

Given that $A = [\begin{matrix} 11 & 1 0 & 11 \end{matrix}]$ , $B = [\begin{matrix} 0 & - 2 - 3 & 4 \end{matrix}]$ , $I = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]$ . Find $A + 3 B + 4 I .$

[\begin{matrix} 0 & 2 - 3 & 0 \end{matrix}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[\begin{matrix} 15 & 2 - 13 & - 16 \end{matrix}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[\begin{matrix} 15 & - 5 - 9 & 27 \end{matrix}]

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

[\begin{matrix} 11 & 2 - 4 & - 5 \end{matrix}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $[\begin{matrix} 15 & - 5 - 9 & 27 \end{matrix}]$
Given, $A = [\begin{matrix} 11 & 1 0 & 11 \end{matrix}], B = [\begin{matrix} 0 & - 2 - 3 & 4 \end{matrix}], I = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]$
Therefore,
$3 B = 3 [\begin{matrix} 0 & - 2 - 3 & 4 \end{matrix}] = [\begin{matrix} 0 & - 6 - 9 & 12 \end{matrix}]$ and $4 I = 4 [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] = [\begin{matrix} 4 & 0 0 & 4 \end{matrix}]$
Now $A + 3 B + 4 I =$ $[\begin{matrix} 11 & 1 0 & 11 \end{matrix}] + [\begin{matrix} 0 & - 6 - 9 & 12 \end{matrix}] + [\begin{matrix} 4 & 0 0 & 4 \end{matrix}]$
$\Rightarrow$ $= [\begin{matrix} 11 + 0 + 4 & 1 - 6 + 0 0 - 9 + 0 & 11 + 12 + 4 \end{matrix}]$
$\Rightarrow$ $= [\begin{matrix} 15 & - 5 - 9 & 27 \end{matrix}]$
$∴$ Option C is correct.

Suggest Corrections

Similar questions

Q. Find variance for following data:

Marks	$0 - 4$	$5 - 9$	$10 - 14$	$15 - 19$	$20 - 24$	$25 - 29$	$30 - 34$	$35 - 39$
Frquency	$2$	$5$	$7$	$13$	$21$	$16$	$8$	$3$

Prove that:
(i) $\frac{1}{3 + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{3}} + \frac{1}{\sqrt{3} + 1} = 1$
(ii) $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} = 2$

$\frac{2}{3} + \frac{- 4}{5} + \frac{7}{15} + \frac{- 11}{20} = ?$

(a) $- \frac{1}{5}$

(b) $- \frac{4}{15}$

(d) $- \frac{7}{30}$

Q. Show that

{sin}^{- 1} \frac{12}{13} + {cos}^{- 1} \frac{4}{5} + {cot}^{- 1} \frac{63}{16}

= \frac{π}{2}

Compute each of the following:

(a) $15 + (- 5) + 15 + (- 7)$

(b) $34 + (- 20) + 16 + (- 10)$

(d) $- 3 + (- 2) + (- 5) + 7$

Join BYJU'S Learning Program

Given that A=[111011], B=[0−2−34], I=[1001]. Find A+3B+4I.

The correct option is C [15−5−927]Given, A=[111011],B=[0−2−34],I=[1001]Therefore,3B=3[0−2−34]=[0−6−912] and 4I=4[1001]=[4004]Now A+3B+4I= [111011]+[0−6−912]+[4004]⇒ =[11+0+41−6+00−9+011+12+4]⇒ =[15−5−927]∴ Option C is correct.

Given that $A = [\begin{matrix} 11 & 1 0 & 11 \end{matrix}]$ , $B = [\begin{matrix} 0 & - 2 - 3 & 4 \end{matrix}]$ , $I = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]$ . Find $A + 3 B + 4 I .$